Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.8 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8

Question 1.
Find the area of the region bounded by 3x – 2y + 6 = 0, x = -3, x = 1 and x axis.
Solution:
3x – 2y + 6 = 0
2y = 3x + 6
y = \(\frac { 1 }{ 2 }\)(3x + 6)

Question 2.
Find the area of the region bounded by 2x – y + 1 = 0, y = -1, y = 3 and y axis.
Solution:
Given straight line is 2x – y + 1 = 0
y = 2x + 1, x = \(\frac { y-1 }{ 2 }\)

= 1 + 1
= 2 sq. units
Area required = 2 sq. units

Question 3.
Find the area of the region bounded by the curve 2 + x – x² + y = 0, x axis, x = -3 and x = 3
Solution:
Given curve is
2 + x – x² + y = 0
y = x² – x – 2
y = (x – 2)(x + 1)

Question 4.
Find the area of the region bounded by the line y = 2x + 5 and the parabola y = x² – 2x.
Solution:
First, we find the point of intersection of
y = 2x + 5 and y = x² – 2x
x² – 2x = 2x + 5
x² – 4x – 5 = 0
(x – 5) (x + 1) = 0
x = 5, x = – 1
when x = 5, y = 15
x = -1. y = 3
(5, 15) (-1, 3) are intersecting points.

54 – 18 = 36
Area required = 36 sq. units

Question 5.
Find the area of the region bounded between the curves y = sin x and y = cos x and the lines x = 0 and x = π
Solution:
First find the intersecting point of two curves
sin x = cos x
tan x = 1
x = \(\frac { π }{ 4 }\)

Area required = 2√2 sq. units

Question 6.
Find the area of the region bounded by y = tan x, y = cot x and the lines x = 0, x = \(\frac { π }{ 2}\), y = 0.
Solution:
First find the intersecting point of y = tan x and y = cot x
tan x = cot x
\(\frac { tan x }{ cot x }\) = 1
tan²x = 1
tan x = 1

= log √2 + log √2
= 2 log √2
= log(√2)²
= log 2 sq. units

Question 7.
Find the area of the region bounded by the parabola y² = x and the line y = x – 2.
Solution:
First find the intersecting point of y² = x and y = x – 2
y = y² – 2
y² – y – 2 = 0
y = 2, y = -1
Intersecting points are (4, 2), (1, -1)

Required Area = \(\frac { 9 }{ 2 }\) sq. units

Question 8.
Father of a family wishes to divide his square field bounded by x = 0, x = 4, y = 4 and y = 0 along the curve y² = 4x and x² = 4y into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided among them.
Solution:
Given curve y² = 4x and x² = 4y
Draw these two curves
Also draw the square bounded by the lines
x = 0, x = 4, y = 4 and y = 0
To prove Area A₁ = Area A₂ = Area A₃
Now the point of intersection of the curves y² = 4x and x² = 4y is given by
(\(\frac { y^2 }{ 4 }\))² = 4y
y⁴ = 64y ⇒ y (y³ – 64) = 0
y = 0, y = 4
when y = 0 ⇒ x = 0
y = 4 ⇒ x = 4
Point of intersection are O (0, 0) and B (4, 4)
Now, the area of the region bounded by the curves y² = 4x and x² = 4y is
A₂ = \(\int_{0}^{4}\)(\(\sqrt { 4x }\) – \(\frac { x^2 }{ 4 }\)) dx

Now the area of the region bounded by the curves x² = 4y, x = 4 and x axis is

Similarly the area of the region bounded by the curve y² = 4x, y axis and y = 4 is

Hence we see that
A₁ = A₂ = A₃ = \(\frac { 16 }{ 3 }\) sq. units

Question 9.
The curve y = (x – 2)² + 1 has a minimum point at P. A point Q on the curve is such that the slope of PQ is 2. Find the area bounded by the curve and the chord PQ.
Solution:
Given curve is y = (x – 2)² + 1
(i.e) (y – 1) = (x – 2)²
Vertex of the parabola is (2, 1)
Minimum point P is (2, 1)
Slope of PQ is 2.

Equation of PQ is y – 1 = 2 (x – 2)
y – 1 = 2x – 4
y = 2x – 3
Intersecting point of y = 2x – 3 and y = (x – 2)² + 1
2x – 3 = (x – 2)² + 1
2x – 4 = (x – 2)²
2(x – 2) = (x – 2)²
x – 2 = 2
x = 4
when x = 4, y = 5

Required Area = \(\frac { 4 }{ 3 }\) sq. units

Question 10.
Find the area of the region common to the circle x² + y² = 16 and the parabola y² = 6x
Solution:
First find the intersecting point of the curves
x² + y² = 16 and y² = 6x
x² + 6x = 16
x² + 6x – 16 = 0
(x + 8) (x – 2) = 0
x = -8, x = 2
x = -8 is impossible
x = 2, y = 2√3
Radius of the circle x² + y² = 16 is 4

Area OABC = 2 (Area of OAB)
= 2 (Area of the curve y² = 6x in [0, 2] + Area of the curve x² + y² = 16 in [2, 4])

Required Area = \(\frac { 4 }{ 3 }\) [4π + √3] sq. units

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 4 Transition and Inner Transition Elements Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements

12th Chemistry Guide Transition and Inner Transition Elements Text Book Questions and Answers

Part – I – Text Book Evaluation

I. Choose the correct answer

1. Sc (Z = 21) is a transition element but Zinc (Z = 30) is not because
a) both Sc³⁺ and Zn²⁺ ions are colourless and form white compounds
b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled
c) last electron as assumed to be added to 4s level in case of zinc
d) both Sc and Zn do not exhibit variable oxidation states
Answer:
b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled

2. Which of the following d block element has half-filled penultimate d sub shell as well as half-filled valence sub shell?
a) Cr
b) Pd
c) Pt
d) none of these
Answer:
a) Cr

3. Among the transition metals of 3d series, the one that has highest negative \(\left(\mathrm{M}^{2+} / \mathrm{M}\right)\) standard electrode potential is
a) Ti
b) Cu
c) Mn
d) Zn
Answer:
a) Ti

4. Which one of the following ions has the same number of unpaired electrons as present in V³⁺?
a) Ti³⁺
b) Fe³⁺
c) Ni²⁺
d) Cr³⁺
Answer:
c) Ni²⁺

5. The magnetic moment of Mn²⁺ ion is
a) 5.92BM
b) 2.80BM
c) 8.95BM
d) 3.90BM
Answer:
a) 5.92BM

6. The catalytic behaviour of transition metals and their compounds is ascribed mainly due to
a) their magnetic behaviour
b) their unfilled d orbitals
c) their ability to adopt variable oxidation states
d) their chemical reactivity
Answer:
c) their ability to adopt variable oxidation states

7. The correct order of increasing oxidizing power in the series

Answer:
a)VO₂⁺ < Cr₂O₇^2- < MnO₄^–

8. In acid medium, potassium permanganate oxidizes oxalic acid to
a) oxalate
b) Carbon dioxide
c) acetate
d) acetic acid
Answer:
b) Carbon dioxide

9. Which of the following statements is not true?
a) on passing H₂S, through acidified K₂Cr₂O₇ solution, a milky colour is observed
b) Na₂Cr₂O₇ is preferred over K₂Cr₂O₇ in volumetric analysis
c) K₂Cr₂O₇ solution in acidic medium is orange in colour
d) K₂Cr₂O₇ solution becomes yellow on increasing the pH beyond 7
Answer:
b) Na₂Cr₂O₇ is preferred over K₂Cr₂O₇ in volumetric analysis

10. Permanganate ion changes to ________ in acidic medium
a) MnO₄^2-
b) Mn²⁺
c) Mn³⁺
d) MnO₂
Answer:
b) Mn²⁺

11. How many moles of I₂ are liberated when 1 mole of potassium dichromate react with potassium iodide?
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3

12. The number of moles of acidified KMnO₄ required to oxidize 1 mole of ferrous oxalate (FeC₂O₄) is
a) 5
b) 3
c) 0.6
d) 1.5
Answer:
c) 0.6

13. Which one of the following statements related to lanthanons is incorrect?
a) Europium shows +2 oxidation state
b) The basicity decreases as the ionic radius decreases from Pr to Lu.
c) All the lanthanons are much more reactive than aluminium.
d) Ce⁴⁺ solutions are widely used as oxidising agents in volumetric analysis.
Answer:
c) All the lanthanons are much more reactive than aluminium.

14. Which of the following lanthanoid ions is diamagnetic?
a) Eu²⁺
b) Yb²⁺
c) Ce²⁺
d) Sm²⁺
Answer:
b) Yb²⁺

15. Which of the following oxidation states is most common among the lanthanoids?
a) 4
b) 2
c) 5
d) 3
Answer:
d) 3

16. Assertion : Ce⁴⁺ is used as an oxidizing agent in volumetric analysis
Reason : Ce⁴⁺ has the tendency of attaining +3 oxidation state.
a) Both assertion and reason are true and reason is the correct explanation of assertion.
b) Both assertion and reason are true but reason is not the correct explanation of assertion.
c) Assertion is true but reason is false
d) Both assertion and reason are false.
Answer:
a) Both assertion and reason are true but reason is the correct explanation of assertion.

17. The most common oxidation state of actinoids is (PTA -2)
a) +2
b) +3
c) +4
d) +6
Answer:
b) +3

18. The actinoid elements which show the highest oxidation state of +7 are
a) Np, Pu, Am
b) Li, Fm, Th
c) U, Th, Md
d) Es, No, Lr
Ans : a) Np, Pu, Am

19. Which one of the following is not correct? (PTA -5)
a) La (OH)₃, is less basic than Lu (OH)₃
b) In lanthanoid series ionic radius of Ln³⁺ ions decreases
c) La is actually an element of transition metal series rather than lanthanide series
d) Atomic radii of Zr and Hf are same because of lanthanide contraction,
Answer:
a) La(OH)₃, is less basic than Lu(OH)₃

II. Answer the following questions

Question 1.
What are transition metals? Give four examples (PTA – 3)
Answer:

1. Transition metal is an element whose atom has an incomplete d sub shell or which can give rise to cations with an incomplete d sub shell.
2. They occupy the central position of the periodic table, between s-block and p-block elements.
3. Their properties are transitional between highly reactive metals of s-block and elements of p-block which are mostly non-metals.
4. Example – Iron, Copper, Tungsten, Titanium.

Question 2.
Explain the oxidation states of 4d series elements.
Answer:

• The oxidation states of 4d series elements vary from +3 for Y to +8 for Ru.
• The highest oxidation state of 4d elements are found in their compounds with higher electronegative elements like O, F and Cl.
• Ex: In RuO₄, the oxidation state of Ru is +8.

Question 3.
What are inner transition elements?
Answer:

1. The inner transition elements have two series of elements.

• Lanthanoids
• Actinoids

2. Lanthanoid series consists of 14 elements from Cerium (₅₈Ce) to Lutetium (₇₁Lu) following Lanthanum (₅₇La). These elements are characterised by the preferential filling of 4f orbitals. .

3. Actinoids consists of 14 elements from Thorium (₉₀Th) to Lawrencium (₁₀₃Lr) following Actinium (₈₉Ac). These elements are characterized by the preferential filling of 5f orbital.

The elements which in their elemental or ionic form have partly filled f orbitals are called f block elements.
As the f orbitals lie inner to the penultimate shell, therefore these elements having partially filled f orbitals, are also called inner transition elements.

Question 4.
Justify the position of lanthanides and actinides in the periodic table. (PTA – 1)
Answer:
Lanthanides:

• The actual position of lanthanides in the periodic table is at group number 3 and period number 6.
• In the sixth period, the electrons are preferentially filled in inner 4f-sub shell.
• The fourteen elements following lanthanum (Ce to Lu) show similar chemical properties.
• Hence they are grouped together and placed at the bottom of the periodic table.
• This position is justified as follows:
  (1) General electronic configuration:
  [ Xe] 4f^1-145d^0,16S²
  (2) Common Oxidation state: +3
  (3) They have similar physical and chemical properties.

Actinides:

• The actual position of actinides in the periodic table is at group number 3 and period number 7.
• In the seventh period, the electrons are perferentially filled in inner 5f-Sub shell.
• The fourteen elements following actinides (Th to Lr) show similar chemical properties.
• Hence they are grouped together and placed at the bottom of the periodic table.
• This position is justified as follows:
  (1) General electronic configuration:
  [Rn] 5f^2-146d^0-27S²
  (2) Common oxidation state: +3
  (3) They have similar physical and chemical properties.

Question 5.
What are actinides? Give three examples.
Answer:
The fourteen elements following actinium is from thorium to lawrencium are called actinoids.
Examples: Thorium, uranium, Neptunium.

Question 6.
Describe the preparation of potassium dichromate. (PTA – 2)
Answer:

1. The fourteen elements following actinium ,i.e., from thorium (Th) to lawrentium (Lr) are called actinoids.
2. All the actinoids are radioactive and most of them have short half lives.
3. The heavier members being extremely unstable and not of natural occurrence. They are produced synthetically by the artificial transformation of naturally occuring elements by nuclear reactions.
4. Example – Thorium, Uranium, Plutonium, Californium.

Question 7.
What is lanthanide contraction and what are the effects of lanthanide contraction? (PTA – 3)
Answer:
As we move across 4f series, the atomic and ionic radii of lanthanoids show gradual decrease with increase in atomic number. This decrease in ionic size is called lanthanide contraction.

Consequences:
Basicity differences:
As we from Ce³⁺ to Lu³⁺, basic character of LU³⁺ ions decrease. Due to the decrease in the size of LU³⁺ ions, the ionic character of Ln-OH bond decreases (covalent character increases)

Similarities among lanthanoids:
In the complete f-series very small change in radii of Lanthanoids are observed and their chemical properties a requite similar.

Second and third transition series resemble each other due to lanthanide contraction.

Question 8.
Complete the following.
Answer:

Question 9.
What are interstitial compounds?
Answer:
An interstitial is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They are non-stoichiometric compounds.
Ex: Tic, ZrH_1.92

Question 10.
Calculate the number of unpaired electrons in Ti³⁺, Mn^2+, and calculate the spin-only magnetic moment.
Answer:
Electronic configuration of Ti³⁺ = [Ar]3d¹
It has only one unpaired electron, ie. n = 1
Spin only magnetic moment
\(=\sqrt{\mathrm{n}(\mathrm{n}+2)}=\sqrt{1(1+2)}=\sqrt{3}\) = 1.732BM

Electronic configuration of Mn²⁺ = [Ar] 3d⁵
It has five unpaired electrons ie. n = 5
Spin only magnetic moment
\(=\sqrt{n(n+2)}=\sqrt{5(5+2)}=\sqrt{35}\) = 5.92BM

Question 11.
Write the electronic configuration of Ce⁴⁺ and Co²⁺
Answer:
Electronic cofiguration of Ce⁴⁺ = [Xe] 4f⁰5d⁰6s⁰
Electronic configuration of Co²⁺ = [Ar] 3d⁷.

Question 12.
Explain briefly +2 states become more and more stable in the first half of the first-row transition elements with increasing atomic number.
Answer:
In the elements of the first half of the first row, with the removal of valance 4s electrons (+2 oxidation state) the 3d orbital get gradually occupied. Since the number of empty d orbital decreases, the stability of the cations (M²⁺) increases from Sc²⁺ to Mn²⁺.

Question 13.
Which is more stable? Fe³⁺ or Fe²⁺ – explain.
Answer:

• Electronic configuration of Fe³⁺ = [Ar] 3d⁵. It consists of 5 unpaired electrons. Half filled and stable.
• Electronic configuration of Fe²⁺ = [Ar]3d⁶. It consists of 4 unpaired electrons, partially filled d subshell.
• Hence Fe³⁺ is more stable than Fe²⁺.

Question 14.
Explain the variation in E°_{M³⁺/M²⁺} 3d series.
Answer:
Transition metals in their high oxidation states tend to be oxidising. The standard reduction potential for the M³⁺/M²⁺ half cell gives the relative stability between M³⁺ and M²⁺.

The negative values for titanium, vanadium and chromium indicate that the higher oxidation state is preferred.

The high reduction potential of M³⁺/M²⁺ indicates Mn²⁺ is more stable than Mn³⁺

Question 15.
Compare lanthanides and actinides.
Answer:

Question 16.
Explain why Cr²⁺ is slrongly reducing while Mn³⁺ is strongly oxidizing. (PTA – 5)
Answer:
E° value for Cr³⁺/Cr²⁺ is negative (-0.41 V), where as E° Value for Mn³⁺/Mn²⁺ is positive (+1.57V).

Hence Cr²⁺ ions can easily undergo oxidation to give Cr³⁺ ions and therefore, acts as strong reducing agent. On the otherhand, Mn³⁺ can easily undergo reduction to give Mn²⁺ and hence acts as oxidising agent.

Question 17.
Compare the ionization enthalpies of first series of the transition elements.
Answer:

• Ionisation energy of transition element is intermediate between those of s and p block elements.
• As we move from left to right in a transition metal series, the ionisation enthalpy increase.
• This is due to the increase in nuclear charge corresponding to the filling of d electrons.
• The following figure shows the trends in ionisation enthaply of first series of transiton elements.

Question 18.
Actinoid contraction is greater from element to element than the lanthanoid contraction why?
Answer:
This is due to poor shielding by 5f electrons in actinoids as compared to that by 4f electrons in lanthanoids.

Question 19.
Out of Lu(OH)₃ and La(OH)₃ which is more basic and why?
Answer:

Question 20.
Why europium (II) is more stable than cerium (II)
Answer:

Question 21.
Why do zirconium and Hafnium exhibit similar properties?
Answer:
This is because Zr and Hf have similar atomic sizes which is due to lanthanoid contraction.

Question 22.
Which is stronger reducing agent Cr²⁺ or Fe²⁺?
Answer:
Cr²⁺ is a stronger reducing agent than Fe.
Reason :
E°(Cr³⁺/Cr²⁺) is negative (-0.41 V) where as E°(Fe³⁺/Fe²⁺) is positive (+0.77V). Thus Cr³⁺ is easily oxidised to Cr³⁺ but Fe²⁺ cannot be easily oxidised to Fe³⁺. Hence Cr²⁺ is a stronger reducing agent than Fe²⁺.

Question 23.
The E°_M⁺/M value for copper is positive. Suggest a possible reason for this.
Answer:
E°_M⁺/M for any metal depends upon the sum of the enthalpy changes taking place in the following steps.
M_(s) + ∆H_(a) → M_(g) (∆H_(a) = enthalpy of atomisation)
M_(g) + ∆H_(i) → M²⁺_(g) (∆H_(i) = ionisation enthalpy)
M²⁺_(g) + ∆_aq → M²⁺_(aq) (∆H_(Hyd) = hydration enthalpy)

Coppes possesses a high enthalpy of atomisation and low enthalpy of hydration. Hence E° Cu²⁺/ Cu is positive.

Question 24.
Describe the variable oxidation on state of 3d series elements.
Answer:

• First transition metal Sc exhibits only +3 oxidation state
• All other transition elements exhibit variable oxidation states by losing electrons from (n-l)d orbital and ns orbital
• Because the energy difference between these two orbitals is very small.

(ex)

• At the beginning of the series, +3 oxidation is stable.
• Number of oxidation states increases with the number of electrons available.
• Number of oxidation states decreases as the number of paired electrons increases.
• Hence the first and last elements show less number of oxidation states and the middle elements show more number of oxidation states.
• (Ex): First element Sc has only one oxidation state +3. Middle elements Mn has Six different oxidation states from +2 to +7. Last element Cu has +1 and +2 oxidation states.
• Relative stability of different oxidation states of 3d-metals is correlated with the extra stability of half filled and fully filled electronic configurations.
• (Ex): Mn²⁺(3d⁵) is more stable than Mn⁴⁺(3d³)

Question 25.
Which metal in the 3d series exhibits +1 oxidation state most frequently and why?
Answer:
Copper has electronic configuration 3d¹⁰4s¹. It can easily lose 4s¹ electron to give the stable 3d¹⁰ configuration. Hence, it shows +1 oxidation state.

Question 26.
Why first ionization enthalpy of chromium is lower than that of Zinc.
Answer:

Question 27.
Transition metals show high melting points, why? (PTA -6)
Answer:
Transition metals show high melting points due to the presence of unpaired d-electrons available for metallic bonding.

III. Evaluate Yourself

Question 1.
Compare the stability of Ni⁴⁺ and Pt⁴⁺ from their ions enthalpy value.
Answer:

• The value of the ionisation enthalpies can be used is estimating the relative stability.
• Pl⁴⁺ relative to Ni⁴⁺ can be explained as follows:
  Ni⁴⁺ – IE₃ +IE₄ = (3395 +5297) = 8692 KJmol^-1
  Pt⁴⁺ – IE₃ +IE₄ = (2800 +4150) = 6950 KJmol^-1
• Pt⁴⁺ requires lesser energy as compared to the formation of Ni⁴⁺
• Therefore Pt⁴⁺ compounds are more stable than Ni⁴⁺ compounds.

Question 2.
Why ion is more stable in +3 oxidation state than in +2 and the reverse is true for manganese?
Answer:
(i) Atomic number of Fe=26
(ii) Electric configuration of Fe = [Ar] 3d⁶4S²
(iii) After removal of two electrons [Fe²⁺]=[Ar]3d⁶
(iv) Further, with removal of one more electron it becomes Fe³⁺ and d⁵ (half filled) that is most stable.
(v) Atomic number of Mn=25
(vi) Electronic configuration of Mn= [Ar] 3d⁵4S²
(vii) After removal of two outer electrons it will become Mn²⁺ and remaining d will be half filled d⁵ and that is most stable.
(viii) Removal of one electron Mn³⁺ having d⁴ configuration which is incompletely filled.

12th Chemistry Guide Transition and Inner Transition Elements Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

1. Transition elements are good conductors because
a) They are metals
b) They are all solids
c) They have free electrons in outer energy orbits
d) All of these
Answer:
d) All of these

2. Transition elements are
a) All metals
b) All non metals
c) Metals and non metals
d) Gases
Answer:
a) All metals

3. Transition elements form complexes very readily because of
a) Small cation size
b) Vacant – d – orbitals
c) Large ionic charge
d) All are correct
Answer:
d) All are correct

4. The transition metal present in vitamin B12 is
a) Fe
b) Co
c) Ni
d) Na
Answer:
b) Co

5. Transition elements are frequently used as catalysts, because of
a) Large ionic charge
b) Large surface area for the reactions to be absorbed
c) Unpaired d electrons
d) Both (b) and (c) are correct
Answer:
d) Both (b) and (c) are correct

6. d – Block elements are arranged in
a) Three series
b) Six series
c) Two series
d) Four series
Ans : d) Four series

7. d – Block element generally for
a) Covalent hydrides
b) Metallic hydrides
c) Interstitial hydrides
d) Salt like hydrides
Answer:
c) Interstitial hydrides

8. Metals which are hard and lustrous substances with high melting points form highly coloured compounds are known as
a) Alkali metals
b) Alkaline earth metals
c) Transition metals
d) None of these
Answer:
c) Transition metals

9. In the first transition series, the incoming electrons enters
a) 5 d orbitals
b) 4 d orbitals
c) 3 d orbitals
d) 2 d orbitals
Answer:
c) 3 d orbitals

10. Transition elements form alloys easily because they have
a) Same atomic number
b) Same electronic configuration
c) Nearly same atomic size
d) None
Answer:
c) Nearly same atomic size

11. Transition elements that show anaomalous electronic configuration in first series are.
a) Cr and Ni
b) Cu and Co
c) Fe and Ni
d) Cr and Cu
Answer:
d) Cr and Cu

12. Lightest transition element is
a) Fe
b) Sc
c) Os
d) None
Answer:
b) Sc

13. Densest transition element is
a) Fe
b) Sc
c) Os
d) Mn
Answer:
b) Sc

14. Variable valencies of transition elements is due to
a) Different energies of (n -1) d electrons
b) Different energies of ns electrons
c) Similar energies of (n – 1)d electrons
d) Similar energies of (n – 1)d and ns electrons
Answer:
d) Similar energies of (n-l)d and ns electrons

15. Which of the following ions has minimum ionic radius
a) Ni²⁺
b) Co²⁺
c) Cr²⁺
d) V²⁺
Answer:
a) Ni²⁺

16. Ionic radii of zirconium and hofminium become almost identical because
a) They are d block elements
b) They belong to the same group
c) Of increased nuclear charge
d) Of lanthanoids contraction
Answer:
d) Of lanthanoids contraction

17. The colour of Fe ions is
a) Blue
b) Light green
c) Very dark green
d) Yellow
Answer:
b) Light green

18. Magnetic property of transition metal is due to
a) Spin of electron
b) Orbital moment
c) Both
d) Neither of the two
Answer:
a) Spin of electron

19. All ferromagnetic substances
a) Can be magnetised
b) Can be electrolysed
c) Have completely filled d – orbitals
d) None
Answer:
a) Can be magnetised

20. Maxmium paramagnetism in 3d series shown by
a) Mn
b) Co
c) Ni
d) Fe
Answer:
a) Mn

21. Which of the following has the minimum magnetic moment
a) Mn²⁺
b) Fe²⁺
c) Cr²⁺
d) V³⁺
Answer:
d) V³⁺

22. Paramagnetism in the substance increases as
a) The number of paired electrons increases
b) The number of unpaired electrons increases
c) The number of unpaired electrons decreases
d) The number of paired electrons decreases
Answer:
b) The number of unpaired electrons increases

23. The number of unpaired electrons present in Cr³⁺ ion is
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3

24. Which of the following has more unpaired d electrons
a) Zn²⁺
b) Fe²⁺
c) N^3-
d) Cu⁺
Answer:
b) Fe²⁺

25. The lanthanoids contraction is responsible for the fact that
a) Zr and Y have the same radius
b) Zr and Nb have similar oxidation state
c) Zr and Hr have almost the same radius
d) Zr and Zn have same oxidation state
Answer:
c) Zr and Hr have almost the same radius

26. Which of the following is coloured
a) Cu⁺
b) Cu²⁺
c) Ti⁴⁺
d) V⁵⁺
Answer:
Cu⁺

27. In the four successive transition elements (Cr, Mn, Fe & Co) the stability of +2 oxidation state will be are in which of the following order? (PTA – 5)
a) Fe>Mn>Co>Cr
b) Co>Mn>Fe>Cr
c) Cr>Mn>Co>Fe
d) Mn>Fe>Cr>Co
Answer:
d) Mn>Fe>Cr>Co

28. Chromyl chloride when dissolved in NaOH solution gives yellow solution. The yellow solution contains (PTA – 3)
a) Cr₂O₇^2-
b) CrO₄^2-
c) CrO₅
d) Cr₂O₃
Answer:
b) CrO₄^2-

29. In the dichromate anion (Cr₂O₇)^2- (PTA – 1)
a) 4Cr – O bonds are equivalent
b) 6 Cr – O bonds are equivalent
c) All Cr – O bonds are equivalent
d) All Cr-O bonds are non-equivalent
Answer:
b) 6 Cr – O bonds are equivalent

30. Which of the following forms colourless compound?
a) Sc³⁺
b) V³⁺
c) Ti³⁺
d) Cr³⁺
Answer:
a) Sc³⁺

31. The basic character of the transition metal monoxides follows the order (Atomic number, Ti = 22; V = 23; Cr = 24)
a) VO > CrO > TiO > FeO
b) Cro > VO > FeO > TiO
c) TiO > FeO > VO > CrO
d) TiO > VO > CrO > FeO
Answer:
d) TiO > VO > CrO > FeO

32. The correct order of ionic radii of Y³⁺, La³⁺, EU³⁺, and Lu³⁺ is (Atomic number Y = 39; La = 57; Eu = 60; Lu = 71)
a) Y³⁺ < La³⁺ < Eu³⁺ < Lu³⁺
b) Y³⁺ < Lu³⁺ < Eu³⁺ < La³⁺
c) Lu³⁺ < Eu³⁺ < La³⁺ < Y³⁺
d) La³⁺ < Eu³⁺ < Lu³⁺ < Y³⁺
Answer:
b) Y³⁺ < Lu³⁺ < Eu³⁺ < La³⁺

33. During the smelting process silica is added to roasted copper to remove
a) Copper sulphide
b) Ferrous sulphide
c) Ferrous oxide
d) Ferrous Chloride
Answer:
c) Ferrous oxide

34. The ore which contains copper and iron both
a) Malachite
b) Chalcopyrite
c) Chalocite
d) Azurite
Answer:
b) Chalcopyrite

35. According to Ellingham diagram, the oxidation reaction of carbon to carbon monoxide may be used to reduce which one of the following oxides at lowest temperature?
a) Al₂O₃
b) Cu₂O
c) MgO
d) ZnO
Answer:
a) Al₂O₃

36. In electro chemical process (Electrolysis of fused salt) is used to extract
a) Iron
b) Lead
c) Sodium
d) Silver
Answer:
c) Sodium

37. In metallurgy, flux is a substance used to convert
a) Mineral into silicate
b) Fusible impurities to soluble impurities
c) Infusible impurities to soluble impurities
d) none of these
Answer:
c) Influsible impurities to soluble impurities

38. Heating of Iron pyrites in air to remove sulphur is called
a) Fusion
b) Calcination
c) Smelting
d) Roasting
Answer:
d) Roasting

39. Which of the following metal is leached by cyanide process?
a) Silver
b) Sodium
c) Aluminium
d) Copper
Answer:
a) Silver

40. The elements in which extra electrons enter into (n – 2) orbitals are called
a) f – block elements
b) p – block elements
c) d – block elements
d) s – block elements
Answer:
a) f – block elements

41. The most common oxidation state of lanthanoides
a) +4
b) +2
c) +6
d) +3
Answer:
d) +3

42. ………………. form complexes.
a) lanthanoides
b) actinides
c) Thorium
d) Cerium
Answer:
b) Actnides

43. ……………. form oxocations.
a) Actnides
b) lanthanoides
c) s – block
d) p – block
Answer:
a) Actnides

44. The correct electronic configuration of Gd³⁺ is
a) [Xe]4f¹⁴
b) [Xe]4f⁷
c) [Xe]4f⁰
d) [Xe]4f⁷
Answer:
b) [Xe]4f⁷

45. The general electronic configuration of lanthanoides
a) [Xe]4f^0-14
b) [Xe]5d^0-1
c) [Xe]4f^2-14 5d^0-1 bs²
d) [Xe]4f^0-14 5d^1-10 6s²
Answer:
c) [Xe]4f^2-14 5d^0-1 bs²

46. As we move from lanthanum to Lutetium, basic character of hydroxides.
a) decreases
b) increases
c) decrease and then increases
d) none of these
Answer:
a) decreases

47. The colour of U³⁺ is
a) red
b) green
c) yellow
d) pink
Answer:
a) red

II. Assertion and Reason

Question 1.
Assertion : For chromium, the ground state electronic configuration is 3d⁵4s¹ rather than 3d⁴4s².
Reason : The energy of the Cr atom is lower, when the six valence electrons are in different atomic orbitals with parallel spins
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason not is the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
A) Both Assertion and Reason are true and Reason is the correct explanation of A.

Question 2.
Assertion : The energies of the 5s and 4d orbitals are very close.
Reason: The relative energies of the 4d and 5s orbitals vary with the nuclear charge and the electronic distribution.
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 3.
Assertion : The densities, melting and boiling points of the transition elements are generally very high
Reason: Zn, Hg and Cd have low melting and boiling points as the d – block is complete.
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false bu t Reason is true
Answer:
B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 4.
Assertion: The transition metals more similar to one another than are representative metals of group 1 and group 2
Reason: Inner d orbitals are being filled
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
A) Both Assertion and Reason are true and Reason is the correct explanation of A.

Question 5.
Assertion : All the transition elements are metals and good conductors of heat and electricity.
Reason : The penultimate shell of electrons of all these elements is expanding and they are, therefore, expected to have physical and chemical properties in common.
A) Both Assertion and Reason are true and Reason is the correct explanation of A.
B) Both Assertion and Reason are true but Reason is the correct explanation of Assertion.
C) Assertion is true but Reason is false
D) Assertion is false but Reason is true
Answer:
A) Both Assertion and Reason are true and Reason is the correct explanation of A.

III. Match the following

1. Match the catalysts given in column I with the processes given in column II

Answer:

2. Match the properties given in column I with the metals in colum II

Answer:

3. Match the statements given in column I with oxidation states given in column II

Answer:

4. Match the following column I with column II

Answer:

5. Match the following

Answer:

IV. Choose the correct statements

Question 1.
Consider the following statements which is/are correct.
I. Sc ion has no unpaired electrons and the expected magnetic moemnt value is zero B.M
II. The oxidation state of Cr in Cr₂O₇^2- is +4
III. The spin only magnetic moment is given by \(\sqrt{n(n+2)} B M\)
a) I, III
b) II, I
c) III, II
d) II
Answer:
a) I, III

Question 2.
Consider the following statements which is / are correct
I. All Lanthanoids displaces hydrogen from acidified water.
II. The colour of tripositive Lanthanides becomes darker as one goes from Cr³⁺ to LU³⁺
III. The ionic size of tripositive Lanthanoid ions increases with atomic number
a) I
b) II
c) II, III
d) I, III
Answer:
a) I

Question 3.
Consider the following statement which is / are correct.
I. The maximum oxidation state of Osmium is +8
II. The highest oxidation state of a transition metal is given by outermost ‘s’ plus ‘d’ electrons.
III. The maximum magnetic moment is shown by the ion having d orbital electronic configuration is 3d⁵
a) I
b) II
c) I, III
d) I, II, III
Answer:
d) I, II, III

Question 4.
Consider the following statements which is / are correct.
I. Compared to Cu²⁺ salts Cu⁺ salts are less stable.
II. Iron does not form interstitial compounds
III. Ag⁺ is isoelectronic with Cd²⁺
a) I, III
b) II, III
c) III
d) II
Answer:
a) I, III

V. Choose the incorrect statements

Question 1.
Consider the following statements which is / are incorrect.
I. The common oxidation state of lanthanides is +3
II. All lanthanides form oxocations
III. All lanthanides are non – radioactive
a) I
b) II
c) II, III
d) II, III
Answer:
d) II, III

Question 2.
Consider the following statements which is / are incorrect.
I. The common oxidation state of Lanthanum is +4
II. UO₂²⁺ is colourless.
III. Actindes forms oxocations.
a) I, II
b) II, III
c) III, I
d) I, II, III
Answer:
a) I, II

Question 3.
Consider the following statements which is incorrect regarding potassium dichromate.
a) It oxidises ferric salt to ferrous salts.
b) It oxidises KI to I₂
c) It oxidises H₂S to S
d) None of the above
Answer:
a) It oxidises ferric salt to ferrous salts

Question 4.
Consider the following statements which is / are correct.
I. Potassium permanganate is extracted from chromite iron ore.
II. Potassium dichromate is extracted from Pyrolusite ore.
III. Potassium permangate is mainly used in Tanning of leather.
a) III
b) II
c) I, II
d) I, III
Answer:
c) I, II

VI. Fill in the blanks

1. The electronic configuration of scandium is ……………….. .
Answer:
[Ar] 3d¹4s²

2. The metal with highest electrical conductivity at room temperature is ……………….. .
Answer:
Silver

3. ……………….. values are used to predict the thermodynamic stability of the compounds
Answer:
Ionisation energy

4. The common oxidation state of scandium is ………………… .(MARCH 2020)
Answer:
+3

5. Mn²⁺ is ………………… than Mn⁴⁺
Answer:
More stable than

6. The oxidation state of W in Wcl₆ is
Answer:
+6

7. ………………… metal is unique in 3d series
Answer:
Copper

8. The metal shows maximum ferromagnetic character is
Answer:
Iron

9. Paramagnetism is the property of ………………… .
Answer:
Unpaired electrons

1o. Paramagnetism is common in
Answer:
Transition elements

11. Hydroformylation catalyst is
Answer:
(Co₂(CO)₈)

12. Potassium dichromate is in colour
Answer:
Orange red

13. The oxidation state of chromium in potassium dichromate is
Answer:
+6

14. The colour of potassium permanganate is
Answer:
dark purple colour

15. The oxidation state uranium in UP6 is
Answer:
+6

VII. Choose the correct Pair

1. a) Lead – Lead pencil
b) Cerusite – Zinc
c) Zinc sulphate – antiseptic
d) Lead pipes – softwater
Answer:
a) Lead – Lead pencil

2. a) Ag + hot NaOH – Products
b) Zn + hot NaOH – Products
c) Au + NaOH – Products
d) Cr + NaOH – Products
Answer:
b) Zn + hot NaOH – Products

3. a) Cu²⁺ – Diamagnetic
b) Cu²⁺ – Colourless
c) Cu²⁺ – Zero magnetic moment
d) Cu²⁺ – One unpaired electron
Answer:
d) Cu²⁺ – One unpaired electron

4. a) TiCl₄ – Polymerisation catalyst
b) Ni – Hydrogenation catalyst
c) Fe – Haber’s process
d) All the above pairs are correct
Answer:
All the above pairs are correct

VIII. Choose the incorrect Pair

1. a) Sc – 3d series
b) Zn – 3d series
c) Cr – 3d series
d) Cu – 4d series
Answer:
d) Cu – 4d series

2. a) Mn⁴⁺ – 3d³
b) Mn³⁺ – 3d⁴
c) Mn⁵⁺ – 3d⁵
d) Mn⁵⁺ – 3d²
Answer:
c) Mn⁵⁺ – 3d⁵

3. a) Cu⁺, Zn²⁺ – Diamagnetic
b) Sc³⁺, Ti⁴⁺, V⁵⁺ – Paramagnetic
c) Co⁵⁺, Fe²⁺ – Paramagnetic
d) Cu²⁺ – Paramagnetic
Answer:
b) Sc³⁺, Ti⁴⁺, V⁵⁺ – Paramagnetic

4.a) Potassium dichromate – +6
b) Potassium dichromate – oxidising
c) Potassium dichromate – Chrome tanning
d) Potassium dichromate – Bayer’s reagent
Answer:
d) Potassium dichromate – Bayer’s reagent

IX. Choose the odd man outer

1. Scandium, Titanium, Vanadium, Cadmium
Answer:
Cadmium
Hint : Cadmium belongs to 4d series. All others are 3d series.

2. Zirconium, Niobium, Technicium,
Ruthenium
Answer:
Technicium
Hint: Technicium is radio active. All others are non radio active.

3. U³⁺, UO₂²⁺, U⁴⁺, Ce³⁺
Answer:
Ce⁴⁺
Hint: Ce⁴⁺ is colourless. All others are coloured.

X. Additional Questions – 2 Mark

Question 1.
Write the classfication of transition elements.
Answer:

• d – Block elements composed of 3d series (4th period) Scandium to Zinc (10 elements),
• 4d series (5th period) Yttrium to Cadmium (10 elements)
• 5d series (6th period) Lanthanum, Hafinium to mercury (10 elements).
• 6d series (7th period) Actinium, Ruther Fordium to copernicium (10 mercury)

Question 2.
Give the electronic configuration of copper and chromium.
Answer:
The electronic configurations of
Chromium Cr [Ar] 3d⁵ 4S¹
Copper, Cu [Ar] 3d¹⁰ 4S¹

Question 3.
What are the type of packing possible in transition metals?
Answer:

• Most of the transition elements are hexagonal close packed,
• cubic close packed
• body centrered cubic which are the characteristics of true metals.

Question 4.
Why transition elements show variable oxidation states? (PTA – 4)
Answer:
Transition elements exhibit variable oxidation states due to

• loosing electrons from (n-1) d orbital
• ns orbital as the energy difference between them is very small.

Question 5.
Explain why Mn²⁺ is more stable than Mn³⁺?
Answer:

• Mn²⁺ has outer electronic configuration as 3d⁵
• Mn³⁺ has outer electronic configuration as 3d⁴
• Half filled orbitals and completely filled orbital are more stable than the partially filled orbitals so Mn²⁺ with half filled d orbital is more stable than Mn³⁺

Question 6.
Explain about ferromagnetic materials.
Answer:

• Ferromagnetic materials have domain structure.
• In each domain the magnetic dipoles are arranged.
• But the spin dipoles of the adjacent domains are randomly oriented.
• Some transition elements or ions with unpaired d electrons show ferromagnetism.

Question 7.
Explain why the melting and boiling points of Cd, Hg and Zn are low?
Answer:
The elements of group 12 [Zn, Hg, Cd] are quite soft with low melting points. Mercury is a liquid and melts at – 38°C. These elements has no unpaired electrons available for metallic bonding and therefore their melting and boiling points are low.

Question 8.
Which of the following ions would from colourless complexes?
Answer:
Cu²⁺, Zn²⁺, Ti , Ti⁴⁺, Cd²⁺
Zn²⁺ – 3d¹⁰ > Ti⁴⁺ – 3d° > Cd²⁺ – 4d¹⁰

These elements from colourless complexes because they have fully filled d – orbitals and do not possess impaired electrons and so no d – d transition is possible.

Question 9.
State Hume – Rothery rule for alloy formation.
Answer:
According to Hume – Rothery rule to form a substitute alloy the difference between the atomic radii of solvent and solute is less than 15%.

Both the solvent and solute must have the same crystal structure and valence and their electronegativity difference must be dose to zero.

Question 10.
Explain why transition metals form alloys? Transition metals form more alloys themselves because of their
Answer:
a tomic sizes are similar

one metal atom can be easily replaced by another metal atom from its crystal lattice to form an alloy.

Question 11.
Silver atom has completely filled d-orbitals (4d¹⁰) in its ground state, How can you say that it is a transition element?
Answer:
The outer electronic configuration of Ag (Z = 47) is 4d¹⁰ 5S¹, In addition to +1, it shows an oxidation state of +2 also. In +2 oxidation state, the configuration is d⁹
ie. the d – subshell is incompletely filled. Hence, it is transition element.

Question 12.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer:
Oxygen and fluorine have small size high electro negativity. Hence, they can oxidise the metal to the highest oxidation state by prompting all the valance electrons to participate in bonding.

Question 13.
Calculate the ‘Spin only’ magnetic moment of M²⁺ (aq) ion (z=27)
Answer:
Electronic configuration of M with z = 27 is [Ar] 3d⁷4S²
Thus, electronic configuation of M²⁺ will be [Ar] 3d⁷4S°

Question 14.
In what way is the electronic configuration of the transition elements different from that of the non – transition elements?
Answer:
Transition elements contain incompletely filled d-sub shell i.e their electronic configuration is (n-l)d^”1( ns0’1, , whereas non – transition elements have either no d- subshell or their

d-subshell is completely filled and have ns^1,2 or n² np^1-6 in their outermost shell.

Question 15.
Name the oxometa! anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer:
Cr²O₇^2- and Cr²O₄^2-
(Group number = Oxidation state of Cr = 6 MnO₄^–
(Group = oxidation stale of Mn=7)

Question 16.
What is the effect of increasing pH of a solution of Potassium dichromate?
Answer:
On increasing pH of potassium dichromate solution (i.e. on adding alkali) it changes to potassium chromate.

Question 17.
Write chemical equations for the reactions involved in the manufacture of potassium permanganate from Pyrolusite. (PTA- 5)
Answer:
Relevant equations for the manufacture of KMn04 from pyrolusite ore are given below:

Question 18.
Classify the following elementsintod-block and f-block elements. (MARCH 2020)
Answer:
(i) Tungsten
(ii) Ruthenium
(iii) Promethium
(iv) Einstenium
d-block elements – Tungsten, Ruthenium
f-block elements- promethium, Einstenium

Question 19.
The halides of transition elements become more covalent with increasing oxidation state of the metal why?
Answer:
As the oxidation slate of the metal increases, its charge increases. According to Eajan’s Rules, as the charge of the metal ion increases covalent character increases because the positively charged cation attracts the electron could on the anion towards itself.

Question 20.
Although Cr³⁺ and Co²⁺ ions have the same number of unpaired electrons but the magnetic moment of Cr³⁺ is 3.87 B.M. and that of Co²⁺ is 4.87 B.M. why?
Answer:
Cr³⁺ ion has symmetrical electronic configuration in the outermost orbit i.e. 3d³. In such ions there is no orbital configuration to magnetic moment. However appreciable orbital contribution takes place in Co²⁺ with 3d⁷ configuration.

Question 21.
Why E° value for Mn, Ni and Zn are more megative than expected?
Answer:
Negative values for Mn²⁺ and Zn²⁺ are related to the stabilities of half – filled and fully filled configurations respectively.

Question 22.
Calculate the spin only magnetic moment of Mn²⁺
Answer:
E.C of Mn²⁺ = [Ar] 3d⁵

Question 23.
Why Ni^II complexes are thermodynamically more stable than Pt^II complexes?
Answer:
The ionisation energy of Ni²⁺ less than that of Pt²⁺. Hence Ni²⁺ complexes are more stable than Pt²⁺ complexes.

Question 24.
Why do transition elements and its compounds act as catalyst? (PTA – 5)
Answer:
Transition metals has energetically available d orbitals that can accept electrons from reactant molecule or metal can form bond with reactant molecule using it’s d electrons.

Question 25.
WhatisZeigler-Natta Polymerisation
Answer:
A mixture of TiCl₄ and Tri alkyl aluminium is used for polymerisation.

Question 26.
Actinoid atoms are generally coloured? Justify your answer.
Answer:
The actinoid ions are generally coloured. This can be explained interms of unpaired
electrons u nde rgoi ng f – f Transi t ion.

Question 27.
How many unpaired electrons are present in Mn²⁺ ion. (Z=25) How does it influences magnetic behaviour of Mn²⁺ ion.
Answer:
Mn²⁺ :3d⁵ has 5 unpaired electorns. It is highly paramagnetic and it is attracted by magnet.

Question 28.
Transition metal atoms/ions are usually coloured. Justify.
Answer:
Transition metal ions have unpaired electrons. They can undergo d-d transition by absorbing light from visible region and radiating complementary colour.

Question 29.
Give the disproportionation of manganese (VI) in acidic medium.
Answer:
3MnO₄^2- + 4H⁺ → 2MnO^– + MnO₂ + 2H₂O

Question 30.
Oxoanions of a metal show higher oxidation state. Give reason.
Answer:
Oxo anions of a metal show higher oxidation state due to the ability of oxygen to form multiple bonds and its high electronegativity.

XI. Additional Questions – 3 Mark

Question 1.
In the series Sc (Z = 21) to Zn (Z = 30) the enthalpy of atomisation of Zinc is the lowest i.e. 126 KJ mol¹. why?
Answer:
In the series Sc to Zn, all elements have one or more unpaired electrons except zinc which has no unpaired electrons, its outer electronic configuration being 3d¹⁰ 4s². Lower the number of unpaired electrons, lower is the metal – metal bonding. Hence, metal – metal bonding is weakest in zinc. Therefore, enthalpy of atomisation is lowest.

Question 2.
Explain why Cu⁺ ion is not stable in aqueous solutions.
Answer:
Cu²⁺ (aq) is much more stable than Cu⁺ (aq). Although, second ionisation enthalpy of copper is large but ∆H_hyd for Cu²⁺ (aq) is much more negative than thatfor Cu⁺ (aq) and hence it more than compensates for second ionisation enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproport¬ionation.

Question 3.
Write down the electronic configuration of :
Answer:
i) Cr³⁺
ii) Cu⁺
iii) Co²⁺
iv) Mn²⁺
v) Pm³⁺
vi) Ce⁴⁺
vii) Lu²⁺
viii) Th⁴⁺
Answer:

Question 4.
What are the different oxidation states exhibited by lanthanoids?
Answer:
The most common oxidation state of lanthanoids is +3. However, some lanthanoids also show an oxidation state of +2 and +4. For example, Eu shows an oxidation state of +2 and Ce shows on oxidation state of +4.

Question 5.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reactions with (i) iodide (ii) iron II solution and (iii) H₂S (PTA – 5)
Answer:
i) Cr₂O₇^2- + 14H⁺ + 6I^– → 2Cr³⁺ + 7H₂O + 3I₂
ii) Cr₂O₇^2- + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺+ 7H₂O
iii) Cr₂O₇^2- + 8H⁺ + 3H₂S → 2Cr³⁺ + 7H₂O + 3S

Question 6.
What is meant by disproportionations’? Give two examples of disporportionation reaction in aqueous solution.
Answer:
When the oxidation state of an element in a reactant increases in one of the products and decreases in other product, the phenomenon is called disproportionation. For example, Mn(VI) in MnO₄² changes to Mn (VII) in the product MnO₄ and Mn(IV) in the product MnO₂ as shown by the reaction.
3MnO₄^2- + 4H⁺ → 2MnO_4-+ MnO₂ + 2H₂O
similarly, Cr (V) undergoes disproportionation in acidic medium as follows.
CrQ₄^3-+ 8H⁺ → 2CrO₄^2- + Cr³⁺ + 4H₂O

Question 7.
Calculate the no. of unpaired electrons in the following gaseous ions :
Answer:
Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺ which one of these is the most stable in aqueous solution?
Mn³⁺ : 3d⁴ has 4 unpaired electrons,
Cr³⁺ : 3d³ has 3 unpaired electrons,
V³⁺ : 3d² has 2 unpaired electrons,
Ti³⁺ : 3d¹ has 1 unpaired electrons.
Cr³⁺ is most stable out of these in aqueous solution because it has half filled t₂g level (i.e.t³_2g)

Question 8.
Why is the +2 oxidation state of manganese quite stable, while the same is not true for iron?
Answer:
Half filled configuration is more stable than others. Mn²⁺ is more stable due to half-filled d – orbitals.
Fe²⁺ is not stable because it does not have half filled d – orbitals. Configurations of Mn²⁺ and Fe²⁺ are as follows :
Mn²⁺ : 3d⁵ 4s⁰
Fe²⁺ : 3d⁶ 4s⁰

Question 9.
Write two characteristics of the transition elements. (PTA – 3)
Answer:
All the transition elements are metals.

All the transition metals are good conductors of heat and electricity.

Question 10.
Explain about diamagnetic materials.
Answer:

• Materials with no elementary magnetic dipoles are diamagnetic or a species with all paired electrons exhibits diamagnetism.
• This kind of materials are repelled by the magnetic field.
• Because the presence of external magnetic field, a magnetic induction is introduced to the material which generates weak magnetic field that oppose the applied field.

Question 11.
Explain about paramagnetic materials.
Answer:

• Paramagnetic solids having unpaired electrons.
• In the absence of external magnetic field, the dipoles are arranged at random and hence the solid shows no net magnetism.
• In the presence of magnetic field, the dipoles are aligned parallel to the direction of the applied field and therefore, they are attracted by an external magnetic field.

Question 12.
A substance is found to have a magnetic moment of 3.9 BM. How many number of unpaired electrons does it certain?
Answer:
Magnetic moment µ = 3.9
\(\sqrt{\mathrm{n}(\mathrm{n}+2)}\) = 3.9B.M
n(n + 2) = (3.9)² = 15
n (n + 2 ) = 3(3+ 2)
n = 3
The number of unpaired electrons is 3.

Question 13.
Why Zn²⁺ salts are white while Ni²⁺ salts are coloured? (PTA – 3)
Answer:

• Zn²⁺ has the configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰
• It has completely filled d orbital and there is no unpaired electrons.
• Hence Zn²⁺ is colourless.
• Ni²⁺ has the configurations 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸
• It has two unpaired electrons and d – d transitions are possible.
• Hence Ni²⁺ is coloured.

Question 14.
[Ti(H₂O)₆]³⁺ is coloured while
[Sc(H₂O)₆]³⁺ is colourless. Explain.
Answer:

• In [Ti(H₂O)₅]¹⁺ ,Tr³⁺ has outer electronic configuration as 3d¹
• In this case d-d transition is possible by the absorption of energy from the visble light and produce purple colour.
• But is [Sc(H₂O)₆]³⁺, Sc³⁺ has outer electronic configuration as 3d°.
• In this case d-d transition is not possible and it is colourless.

Question 15.
What are the characteristics of interstitial compounds?
Answer:

• They are hard and show electrical and thermal conductivity
• They have high melting points higher than those of pure metals
• Transition metal hydrides are used as powerful reducing agents
• Metallic carbides are chemically inert.

Question 16.
Explain why d block elements form more complexes?
Answer:

• Transition metal ions are small and highly charged
• They have vacant low energy orbitals to accept an electron pair donated by other groups.
• Examples : [Fe(CN)₆]^4-, [Co(NH₃)₆]³⁺

Question 17.
Explain chromyl chloride test. (MARCH 2020)
Answer:
When potassium dichromate is heated with any chloride salt in the presence of Cone H₂SO₄, orange red vapours of chromyl chloride (CrO₂Cl₂) is evolved.

This reaction is used in the detection of chloride ions in qualitative analysis.

Question 18.
Write the uses of potassium dichromate.
Answer:

• It is used as a strong oxidizing agent.
• It is used in dyeing and printing.
• It used in leather tanneries for chrome tanning.
• It is used in quantitiative analysis for the estimation of iron compounds and iodides.

Question 19.
Write the uses of potassium permanganate.
Answer:

• It is used as a strong oxidizing agent.
• It is used for the treatment of various skin infections and fungal infections of the foot.
• It used in water treatment industries to remove iron and hydrogen sulphide from well water.
• It is used as a Bayer’s reagent for detecting unsaturation in an organic compound.
• it is used in quantitative analysis for the estimation of ferroussalts, oxalates, hydrogen peroxide and iodides.

Question 20.
What is the action of heat on K₂Cr₂O₇ ?
Answer:
Potassium dichromate on heating gives potassium chromate.

XII. Additional Questions – 5 Mark

Question 1.
For M²⁺ /M and M³⁺ / M systems, the E° values for some metals are as follows.
Cr²⁺ /Cr = 0.9 V ; Cr³⁺ /Cr2+ = – 0.4 V
Mn²⁺ /Mn = -1.2 V ; Mn³⁺/Mn²⁺= +1.5 V
Fe²⁺ /Fe = +0.4 V ;Fe³⁺/Fe²⁺ = +0.8 V
Use this data to comment upon
(a) the stability of Fe³⁺ in acid solution as composed to that of Cr³⁺ or Mn²⁺ and
(b) the ease with which iron can be oxidised as compared to the similar process for either chromium or manganese metals.
Answer:
(a) Cr³⁺ / Cr²⁺ has a negative reduction potential Hence, Cr³⁺ cannot be reduced to Cr²⁺ i.e. Cr³⁺ is most stable. Mn³⁺/ Mn²⁺ has large positive E° value. Hence, Mn³⁺ can be easily reduced to Mn²⁺, i.e. Mn³⁺ is least stable. Eo value for Fe³⁺ / Fe²⁺ is positive but small. Hence, Fe³⁺ is more stable than Mn³⁺ but less stable than Cr³⁺. Thus, the stability follows the order.
Cr³⁺>Fe³⁺>Mn³⁺

(b) Oxidation potentials for the given pairs will be +0.9V, +1.2V, and +0.4V. Thus the order of getting oxidised will be
Mn > Cr > Fe

Question 2.
(i) Why do transition elements show variable oxidation states?
(ii) Why do most transition metal ions exhibit para magnetism?
(iii) How is the magnetic moment of a species related to the number of unpaired electrons?
Answer:
(i) This is because ns and (n-1) d orbitals do not differ much in energy. Electrons from both may take part in bonding. Hence, they show variable valency.
(ii) This is because of presence of unpaired electrons in most of the transition metal ions.
(iii) \(\mu_{\mathrm{s}}=\sqrt{\mathrm{n}(\mathrm{n}+2)}\) BM, where n stands for the number of unpaired electrons.

Question 3.
(i) Why is copper (z=29) considered as transition element?
(ii) K₂PtCl₆ is well – known compound while corresponding Ni compound is not known.
(iii) Why is radius of Fe²⁺ less than that of Mn²⁺?
(iv) Why is electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴ not correct for the ground state of Cr (Z-24)?
Answer:
(i) This is because Cu²⁺ ion has incomplete d – orbital
(ii) This is because Pt⁴⁺ is more stable than Ni⁴⁺ Energy required to remove 4 electrons from pt is less and that in Ni than.
(iii) This is because effective nuclear charge in Fe²⁺ is more compared to that in Mn²⁺
(iv) Correct electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵ because half – filled configuration

Question 4.
(i) Which trivalent cation is the largest in lanthanoid series?
(ii) One unpaired electron in atom contributes a magnetic moment of 1.1 B.M. Calculate the magnetic moment of Cr. (Atomic number = 24)
(iii) In a paramagnet ion, all the bonds formed between Mn and O are covalent. Give reasons.
Answer:
(i) La³⁺ is the largest ion
(ii) As Cr has 6 unpaired electrons, its magnetic moment is 6 x 1.1 = 6.6 B.M.
(iii) Oxidation state of Mn is MnO₄ is +7. It is energitacally not possible to lose 7 electrons to give ionic species. It forms bonds by sharing of elctrons. Hence, covalent bonds are formed.

Question 5.
There is only a marginal difference in decrease in ionisation elthalpy from Aluminium to Thallium. Explain why? (MARCH 2020)
Answer:
This is due to the presence of inner d-electrons and f-electrons which has poor shielding effect compared to s and p electrons.

Question 6.
(i) Name the metal with tripositive charge represented by the following electronic configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³
(ii) Why is K₂Cr₂O₇ generally preferred over Na₂Cr₂O₇ in volumetric analysis though both are oxidising agents?
(iii) Why does V₂O₅, act as catalyst?
Answer:
(i) Cr³⁺ is represented by Ihe configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³
(ii) Na₂Cr₂O₇ absorbs moisture from, the atmosphere.
(iii) It can form unstable intermediates with the reactants which readilv change into products

Question 7.
Explain the following observations.
(a) The elements of the d – series exhibit a large number of oxidation states than the elements of f – series.
(b) The Cu⁺ salts are colourless while Cu²⁺ salts are coloured (Atomic number of [Cu=29]
Answer:
(a) In d – series there are large number of unpaired electrons which take part in bond formation due to less effective nuclear charge, therefore, number of oxidation states are more. In f – block, there is more effective nuclear charge due to poor shielding effect of f – orbitals. Therefore, less number of electrons take part in bond formation.

(b) Cu⁺ does not have unpaired electtron, therefore the electrons cannot undergo d-d transitions. That is why, Cu⁺ salts are colourless. Cu²⁺ salts are coloured due to presence of one unpaired electron, it can undergo d-d transition by absorbing light from visible region and radiating blue colour.

Question 8.
How do you account for the following?
(i) All Scandium salts are white [Atomic number of Sc=21]
(ii) The first ionisation energies of the 5d transition elements are higher than those of the 3d and 4d transition elements in respective groups.
Answer:
(i) In Scandium Salts, Scandium has +3 oxidation state. Sc³⁺ does not have unpaired electrons and has empty d – orbitals. Therefore, there is no d-d transition, Hence its salts are white.

(ii) Due to poor shielding effect of 5d and 4f – electrons, effective nuclear charge increases, Hence, ionisation energy of 5d transition elements is more than that of 3d and 4d transition elements in respective groups.

Question 9.
How would you account for the following Situations?
(i) The transition metals generally form coloured compounds.
(ii) With 3d⁴ configuration, Cr²⁺ acts as a reducing agent, actinoids exhibit a larger number of oxidation agent.
(iii) The actinoids exhibit a larger number of oxidation states than the corresponding lanthanoids.
Answer:
(i) Transition metals contain unpaired d- electrons. The d – electrons absorb light from the visible range and are excited to higher energy d – orbitals (from t₂g to e_g orbitals). Transmitted light is the colour shown by the transition metals.

(ii) Cr²⁺ has d⁴ configuration while Cr³⁺ has more stable d⁵ configuration. Thus Cr, has a tendency to change from Cr²⁺ to Cr³⁺ or Cr²⁺ acts as reducing agent. Mn³⁺ has a d⁴ configuration. It has a tendency to change into Mn²⁺ with more stable d⁵ configuation. Thus, Mn³⁺ acts as an oxidising agent.

(iii) Actinoids exhibit a larger number of oxidation states than the corresponding lanthanoids because 5f, 6d and 7s levels are comparable energies.

Question 10.
A Violet compound of manganese (A) decomposes on heating to liberate oxygen and compound (B) and (C) of manganese are formed. Compound (C) reacts with KOH in presence of potassium nitrate to give compound.
(B) On heating compound (C) with Cone. H₂SO₄ and NaCl, Chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds (A) to (D) and also explain the reactions involved.
Answer:
The compounds A,B,C and D are given as under : A = KMnO₄, B = K₂MnO₄, C = MnO₂, D = MnCl₂
The reactions are explained as under:

2MnO₂ + 4KOH+ O₂ → 2K₂MnO₄ + 2H₂O
MnO₂ + 4NaCl + 4H₂SO₄ → MnCl₂ + 4NaHSO₄ + Cl₂ + 2H₂O

Question 11.
When a chromite ore (A) is fused with Sodium Carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallized from the solution. When compound (C) is treated with KC1, orange crystals of compound (D) crystallises out. Identify (A) to (D) and also explain the reactions.
Answer:
The compounds A,B,C and Dare given as under
A = FeCr₂O₄
B = Na₂CrO₄
C = Na₂Cr₂O₇-2H₂O
D = K₂Cr₂O₇
The reactions are explained as under :

Question 12.
Explain the following facts:
(a) Transition metals acts as catalysts.
(b) Chromium group elements have the highest melting points in their respective series.
(c) Transition metals form coloured complexes.
Answer:
(a) Transition metals have incomplete d-orbitals. They combine with the reactants to form intermediate products which change into the final products.
(b) Chromium group elements have the greatest number of unpaired electrons in d-orbitals. They ae capable of forming maximum interatomic metallic bonds. This raises the melting point of the chromium group elements.
c) Transition metals have incompletely filled d-orbitals and have unpaired electrons which are excited to higher energy d-orbitals in the same sub-shell (from t2g to eg). These are called d-d transitions. During this process, the molecules absorb energy from the visible region. The transmitted light is the colour shown by the substance.

Question 13.
Discuss the structure of dichromate ion:
Answer:
(i) Both dichromate and chromate ions are oxo anions of chromium and they are oxidising agents.
(ii) In these ions Cr is in +6 oxidation state.
(iii) In aqueous solution these ions are interconvertible.
(iv) In alkaline solution chromate ion is predominant.
(v) In acidic solution dichromate ion is predominant.

Question 14.
Discuss the oxidising power of KMnO₄ in
a) Acidic medium
b) neutral medium
c) alkaline medium
Answer:
a) Acidic medium
It oxidises ferrous salt to ferric salt
2MnO₄^– + 10Fe²⁺ + 16H⁺ → 2Mn²⁺ + 10Fe³⁺ + 8H₂O

b) Neutral medium:
It oxidises H2S to sulphur
2MnO₄^– + 3H₂S → 2MnO + 3S + 2OH^– + 2H₂O

c) Alkaline medium:
In the presence alkali the permanganate ion is converted into manganate ion.
MnO₄^– + 2H₂O + 3e^– → MnO₂ + 40H^–

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 9 Ratio Analysis Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 9 Ratio Analysis

12th Accountancy Guide Ratio Analysis Text Book Back Questions and Answers

I Multiple Choice Questions

Choose the correct answer

Question 1.
The mathematical expression that provides a measure of the relationship between two figures is called
(a) Conclusion
(b) Ratio
(c) Model
(d) Decision
Answer:
(b) Ratio

Question 2.
Current ratio indicates
(a) Ability to meet short term obligations
(b) Efficiency of management
(c) Profitability
(d) Long term solvency
Answer:
(a) Ability to meet short term obligations

Question 3.
Current assets excluding inventory and prepaid expenses is called
(a) Reserves
(b) Tangible assets
(c) Funds
(d) Quick assets
Answer:
(d) Quick assets

Question 4.
Debt equity ratio is measure of
(a) Short term solvency
(b) Long term solvency
(c) Profitability
(d) Efficiency
Answer:
(b) Long term solvency

Question 5.
Which of the following is not a tool of financial statement analysis?

Answer:
(a) (i) – 1,(ii) – 4,(iii) – 3,(iv) – 2

Question 6.
To test the liquidity of a concern, which of the following ratios are useful?
(i) Quick ratio
(ii) Net Profit ratio
(iii) Debt – equity ratio
(d) Current ratio
Select the correct answer using the codes given below:
(a) (i) and (ii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and iv)
Answer:
(b) (i) and (iv)

Question 7.
Proportion of share holders’ funds to total assets is called
(a) Proprietary ratio
(b) Capital gearing ratio
(c) Debt equity ratio
(d) Current ratio
Answer:
(a) Proprietary ratio

Question 8.
Which one of the following is not correctly matched?
(a) Liquid ratio – Proportion
(b) Gross profit ratio – Percentage
(c) Fixed assets turnover ratio – Percentage
(d) Debt – equity ratio – Proportion
Answer:
(c) Fixed assets turnover ratio – Percentage

Question 9.
Current liabilities ₹ 40,000; Current assets ₹ 1,00,000; Inventory ₹ 20,000. Quick ratio is
(a) 1:1
(b) 2,5:1
(c) 2:1
(d) 1:2
Hint:
Quick ratio or Liquid ratio = \(\frac{\text { Liquid Assets }}{\text { Current liabilities }}\)
Liquid assets = Current Assets – Inventory
= 1,00,000 – 20,000
= 80,000
= \(\frac{80,000}{40,000}\)
= 2:1
= 110%
Answer:
(c) 2:1

Question 10.
Cost of revenue from operation 3,00,000; Inventory at the beginning of the year 60,000; Inventory at the close of the year’ 40,000. Inventory turnover ratio is.
(a) 2 times
(b) 3 times
(c) 6 times
(d) 8 times
Hint:

Answer:
(c) 6 times

II Very Short Answer Questions

Question 1.
What is meant by accounting ratios?
Answer:
The ratio is a mathematical expression of the relationship between two related or interdependent items. It is the numerical or quantitative relationship between two items. It is calculated by dividing one item by the other related item. When ratios are calculated on the basis of accounting information, these are called ‘accounting ratios’.

Question 2.
What is the quick ratio?
Answer:
The quick ratio gives the proportion of quick assets to current liabilities. It indicates whether the business concern is in a position to pay its current liabilities and when they become due, out of its quick assets.

Question 3.
What is meant by debt-equity ratio?
Answer:
It is calculated to assess the long-term solvency position of a business concern. The debt equity ratio expresses the relationship between long term debt and shareholder’s funds.
Debt equity ratio = \(\frac{\text { Long term debt }}{\text { Shareholders funds }}\)
Capital employed = Shareholder’s funds + Noncurrent liabilities
Greater the return on investment better is than the profitability of a business and vice versa.

Question 4.
What does the return on investment ratio indicate?
Answer:
Return on investment shows the proportion of net profit before interest and tax to capital employed (shareholders’ funds and long term debts). This ratio measures how efficiently the capital employed is used in the business. It is an overall measure of the profitability of a business concern.

Question 5.
Statement any two limitations of ratio analysis.
Answer:
Ratios are only means: Ratios are not ended in themselves but they are only means to achieve a particular purpose. Analysis of related items must be done by the management or experts with the help of ratios. Change in price level: Ratio analysis may not reflect price level changes and current values as they are calculated based on historical data given in the financial statement.

III Short Answer Questions

Question 1.
Explain the objectives of ratio analysis.
Answer:
Following are the objectives of ratio analysis:

• To simplify accounting figures
• To facilitate analysis of financial statements
• To analysis the operational efficiency of a business
• To help in budgeting and forecasting
• To facilitate intra firm and inter-firm comparison of performance

Question 2.
What is the inventory conversion period? How is it calculated?
Answer:
The inventory conversion period is the time taken to sell the inventory. A shorter inventory conversion period indicates more efficiency in the management of inventory. It is computed as follows:

Question 3.
How is operating profit ascertained?
Answer:
Operating profit = Revenue from operations – Operating cost
Cost of revenue from operations = Purchases of stock – in – trade + Change in inventories of stock in trade + Direct expenses.
Operating expenses = Administrative expenses + Selling and distribution expenses.
Operating cost = Cost of revenue from operations + Operating expenses.

Question 4.
State any three advantages of ratio analysis.
Answer:
Following are the advantage of ratio analysis:

• Measuring operational efficiency: Ratio analysis helps to know the operational efficiency of a business by finding the relationship between operating cost and revenues and also by comparison of present ratios with those of the past ratios.
• Intra firm comparison: Comparison of the efficiency of different divisions of an organization is possible by comparing the relevant ratios.
• Inter-firm comparison: Ratio analysis helps the firm to compare its performance with other firms.

Question 5.
Bring out the limitations of ratio analysis:
Answer:

• Consistency in preparation of financial statements: Inter firm comparisons with the help of ratio analysis will be meaningful only if the firms follow uniform accounting procedures consistently.
• Non-availability of standards or norms: Ratios will be meaningful only if they are compared with accepted standards or norms. Only few financial ratios have universally recognized standards. For other ratios, comparison with standards is not possible.
• Change in price level: Ratio analysis may not reflect price level changes and current values as they are calculated based on historical data given in financial statements.

IV Exercises

Liquidity ratios

Question 1.
Calculate the current ratio from the following information.

Solution:
Current ratio = \(\frac{\text { Current Assets }}{\text { Current liabilities }}\)
Current Assets = Current Investments + Inventories + Trade Dr’s + B/R+Cash & Cash equivalents
= 40,000 + 2,00,000 + 1,20,000 + 80,000 + 10,000
= Rs. 4,50,000

Current Liabilities
= Trade Cr’s + B/P + Exps. Payable.
= 80,000 + 50,000 + 20,000
= Rs. 1,50,000
Cur. Ratio = \(\frac{4,50,000}{1,50,000}\)
= 3:1
Answer:
Current ratio : 3:1

Question 2.
Calculate quick ratio: Total current liabilities ₹ 2,40,000; total current assets ₹ 4,50,000; Inventories ₹ 70,000; Prepaid Expenses ₹ 20,000
Solution:
Quick Ratio = \(\frac{\text { Quick assets }}{\text { Current liabilities }}\)
Quick assets = Current Assets – Inventories & Prepaid exps.
= 4,50,000 – (70,000 + 2000)
= Rs.3,60,000
Quick Ratio = \(\frac{3,60,000}{2,40,000}\)
=1:5:1
Answer:
Quick ration: 1:5:1

Question 3.
Following is the balance sheet of Lakshmi Ltd. as of 31st March 2019.

Calculate: (i) Current ratio (ii) Quick ratio
Solution:
Current ratio = \(\frac{\text { Current Assets }}{\text { Current liabilities }}\)
Current Assets = Inventories + Trade Dr’s + Cash & Cash equivalents + Prepaid Exps
= 1,60,000 +3,20,000 + 80,000 + 40,000
= Rs. 6,00,000
Current Liabilities = Short term borrowings + Trade Payable + Expenses payable + Short term provisions.
= 50,000 + 3,10,000 + 15,000 + 25,000
= Rs. 4,00,000

Answer:
(i) Current ratio: 1.5:1;
(ii) Quick ratio: 1:1

Question 4.
From the following information calculate debt equity ratio.

Solution:
Debt Equity Ratio = \(\frac{\text { Long term debt }}{\text { Shareholder’s Funds }}\)
Long term debt = Debenture = Rs. 6,00,000
Shareholder’s Fund = Equity share capital + Reserves & Surplus
= 6,00,000 + 2,00,00 = Rs. 8,00,000
Debt Equity Ratio = \(\frac{6,00,000}{8,00,000}\)
= 0.75:1
Answer:
Debt equity ratio: 0.75:1

Question 5.
From the following Balance Sheet of Sundaram Ltd. Calculate proprietary ratio:

Answer:
Proprietary ratio: 0.75:1

Question 6.
From the following information calculate the capital gearing ratio:

Answer:
Capital gearing ratio: 0.5:1

Question 7.
From the following Balance Sheet of James Ltd. as of 31.03.2019 calculate
(i) Debt- Equity ratio
(ii) Proprietary ratio
(iii) Capital gearing ratio

Solution:
Debt Equity Ratio =  \(\frac{\text { Long Term Debt }}{\text { Shareholder’s Fund }}\)|
Shareholder’s Fund
Long term debt = Debentures = Rs. 3,00,000
Shareholder’s Fund = Eq. share capital +Pref. Shares capital + Reserves & surplus
= 2,50,000 ÷ 2,00,000 + 1,50,000
= Rs. 6,00,000

(3) Capital Gearing ratio

= \(\frac{\text { Funds bearing fixed interest (or) Fixed divided }}{\text { Equity shareholder’s Fund }}\)
Funds bearing fixed interest (or) fixed dividend
= Pref. Share Cap + Debentures
= 2,00,000 + 3,00,000 = Rs. 5,00,000
Equity share holder’s Fund .
= Equity Share cap + Reserves & Surplus
= 2,50,000 + 1,50,000
= Rs. 4,00,000
Capital gearing ratio = \(\frac{5,00,000}{4,00,000}\)
= 1.25:1

Answer:

• Debt-equity ration; 0.5:1;
• Proprietary ration; 0.5:1;
• Capital gearing ratio:1.25:1

Question 8.
From the given information calculate the inventory turnover ratio and inventory conversion period (in months) of Devi Ltd.

Solution:
Inventory Turnover Ratio = \(\frac{\text { cost of revenue from operations }}{\text { Average Inventory }}\)
Cost of revenue from operations = Purchase of stock + change in inventories of finished goods operations + Direct Exps.
= Rs. 6,90,000
AverageInventory = \(\frac{\text { Opening Inventory + Closing inventory }}{2}\)
= \(\frac{1,70,000+1,30,000}{2}\)
= Rs. 1,50,000
Change in inventory = Opening inventory – Closing inventory
= 1,70,000 – 1,30,000
= Rs. 40,000
Cost of revenue from operation
= 6,90,000 + 40,000 + 20,000
= Rs. 7,50,000

Answer:
Inventory turnover ratio 5times; Inventory conversion period 2.4 months

Question 9.
The credit revenue from operations of Velavan Ltd, amounted to ₹ 10,00,000. Its debtors and bills receivables at the end of the accounting period amounted to ₹ 1,10,000 and ₹ 1,40,000 respectively. Calculate trade receivables turnover ratio and also.collection period in months.
Solution:
Trade receivable Turnover ratio = \(\frac{\text { Credit revenue from Operations }}{\text { Average trade receivables }}\)
Average trade receivables = \(\frac{\text { Opening trade receivables + Closing trade receivables }}{2}\)
Trade receivable = Trade Drs + B/R
Inventory Turnovers Ratio = \(\frac{10,00,000}{2,50,000}\)
= 4 times
Average Trade receivable
= 1,10,000 + 1,40,000
= Rs. 2,50,000
Debt collection period = \(\frac{\text { Number of months in a year }}{\text { Trade receivable turnover ratio }}\)
= \(\frac{12}{4}\)
= 3 months
Answer:
Trade receivables turnover ratio: 4 time; Debt collection period: 3 months

Question 10.
From the following figures obtained from Arjun Ltd, calculate the trade payable turnover ratio and credit payment period (in days)

Solution:


Answer:
Trade payable turnover ratio: 10 times; Credit payment period: 36.5 days

Question 11.
From the following information of Geetha Ltd., Calculate fixed assets turnover ratio
(i) Revenue from operations during the year was ₹ 55,00,000.
(ii) Fixed assets at the end of the year ₹ 5,00,000
Solution:

Answer:
Fixed assets turnover ratio: 11 times

Question 12.
Calculate

• Inventory turnover ratio
• Trade receivable turnover ratio
• Trade payables turnover ratio and
• Fixed assets turnover ratio from the following obtained from Aruna Ltd.
  

  Particulars	As of 31st March 2018 ₹	As of 31st March 2019 ₹
  Inventory	3,60,000	4,40,000
  Trade receivables	7,40,000	6,60,000
  Trade Payable	1,90,000	2,30,000
  Fixed assets	6,00,000	8,00,000

   

Additional information:

• Revenue from operations for the year ₹ 35,00,000
• Purchases for the year ₹ 21,00,000
• Cost of revenue from operation ₹ 16,00,000
  Assume that sales and purchases are for credit.

Solution




Answer:

• Inventory turnover ratio; 4 times;
• Trade receivable turnover ratio; 5 times;
• Trade payables turnover ratio: 10 times;
• Fixed assets turnover ratio: 5 times

Question 13.
Calculate gross profit ratio form the following: Revenue from operations ₹ 2,50,000, Cost of revenue from operation ₹ 2,10,000 and Purchases ₹ 1,80,000.
Solution:

Answer:
Gross Profit ratio 16%

Question 14.
Following is the statement of profit and loss of Padma Ltd. for the year ended 31st March, 2018. Calculate the operating cost ratio.


Notes to Accounts

Solution:
Operating cost Ratio = \(\frac{\text { Operating cost }}{\text { Revenue from operation }}\) × 100
Operating cost = Cost of revenue from operations + Operating expenses.
Cost of revenue from operations = Purchase + Change in inventory + Direct Expenses
= 8,60,000 + 40,000 + Nil
= Rs. 9,00,000
Operating Exps = Salaries + Office & Administration Exps + Selling+Distribution Exps
= 1,60,000 + 50,000 + 90,000
= Rs. 3,00,000
Operating cost = 9,00,000 + 3,00,000
= Rs, 12,00,000
Operating cost Ratio = \(\frac {12,00000}{15,00000}\) × 100
= 80%
Answer:
Operating cost ratio 80%

Question 15.
Calculate operating profit ratio under the following cases.
Case 1 : Revenue from operations ₹ 8,00,000 Operating Profit ₹ 2,00,000.
Case 2 : Revenue from operations ₹ 20,00,000 Operating Cost ₹ 14,00,000.
Case 3 : Revenue from operations ₹ 10,00,000 Gross profit 25% on revenue from operations, operating expenses ₹ 1,00,000.
Solution:


Answer:
Operating profit ratio – Case 1: 25%; Case 2:30%; Case 3:15%

Question 16.
From the following details of a business, concern calculates net profit ratio.

Solution
Net Profit Ratio = \(\frac{\text { Net Profit }}{\text { Revenue from operations }}\)× 100
Gross profit = Revenue from operations – Cost of revenue from operation
= 9,60,000-5,50,000 Rs. 4,10,000
Operating Profit = Gross Profit – Operating Exps
Operating Exps = Office & Administrative Exps + Selling & Distribution Exps
= 1,45,000 + 25,000 = Rs. 1,70,000
Operating Profit = 4,10,000 – 1,70,000
Operating Profit = Rs. 2,40,000
Net Profit ratio = \(\frac{2,40,000}{9,60,000}\)× 100 = 25%
Answer :
Net Profit ratio 25%

Question 17.
From the following statement of profit. and loss of Dericston Ltd. Calculate
(i) Gross Profit ratio
(ii) Net Profit ratio.

Solution
Gross Profit Ratio = \(\frac{\text { Gross Profit }}{\text { Revenue from operations }}\)×100
Gross profit = Revenue from operations – Cost of revenue from operation
Cost of revenue from operations = Purchase + Change in inventories
= 18,80,000 – 80,000 = Rs. 18,00,000
Gross Profit = 24,00,000 – 18,00, 000 = Rs. 6,00,000

Answer :

• Gross profit ratio of 25%
• net Profit ratio 10%

Question 18.
From the following trading activities of Jones Ltd. Calculate

• Gross profit ratio
• Net Profit ratio
• Operating cost ratio
• Operating profit ratio

Solution
(1) Gross Profit Ratio = \(\frac{\text { Gross Profit }}{\text { Revenue from operations }}\)× 100
cost of Revenue from operations = Purchase + Change in inventories
= 2,10,000 + 30,000
= Rs. 2,40,000
Gross profit = Revenue from operations – cost of revenue from operations
= 4,00,000 – 2,40,000
= Rs. 1,60,000
Gross Profit ratio = \(\frac{1,60,000}{4,00,000}\)× 100 = 40%


Answer :

• Gross profit ratio 40%
• Net Profit ratio 20%
• Operating cost ratio 75%
• Operating profit ratio 25%

Question 19.
Following is the extract of the balance sheet of Abdul Ltd., as of 31st March 2019

Net Profit before interest and tax for the year was ₹ 60,000. Calculate the return on capital employed for the year.
Solution:

Answer :
Return on capital employed: 15%

12th Accountancy Guide Ratio Analysis Additional Important Questions and Answers

Other Important question & Answers
Question 1.
All solvency ratios are express in terms of
(a) Proportion
(b) Times
(c) Percentage
Answer:
(b) Times

Question 2.
All activity ratios, (or) Turnover ratios in terms of
(a) Proportion
(b) Times
(c) Percentage
Answer:
(b) Times

Question 3.
All profitability ratios are expressed in terms of ………………
(a) Proportion
(b) Times
(c) Percentage
Answer:
(c) Percentage

Question 4.
Shareholders funds include
(a) Equity share capital, preference share capital reserves & Surplus
(b) Loans from banks & financial institutions.
(c) Equity share capital, preference share capital, reserves & surplus, and loans from banks & financial institutions.
Answer:
(a) Equity share capital, preference share capital reserves & Surplus

Question 5.
The current ratio is a
(a) Solvency ratio
(b) Profitability ratio
(c) Liquidity ratio
Answer:
(a) Solvency ratio

Question 6.
The ratio is expressed in ……………… way
(a) 2
(b) 4
(c) 3
Answer:
(c) 3

Question 7.
The cost of revenue from the operation is Rs. 4,00,000. Average inventories Rs. 8,00,000 Inventory turnover ratio is
(a) 5 times
(b) 4 times
(c) 7 times
Answer:
(a) 5 times

Question 8.
The operating ratio is equal to
(a) 100-operating profit ratio
(b) 100 +operating profit ratio
(c) Operating profit ratio
Answer:
(a) 100-operating profit ratio

Question 9.
Operating expenses include
(a) Selling & administration expenses
(b) Selling & administration expens
(c) a & b
Answer:
(c) a & b

Question 10.
Equity share capital Rs. 2,00,000 Reserves & Surplus Rs. 30,000 Debenture Rs. 40,000 and shareholders fund will be.
(a) Rs. 200,000
(b) 2,70,000
(c) Rs. 2,30,000
Hint:
Share holder fund = Equity share capital + Reserve and surplus
= 2,00,000 + 30,000
= ₹ 2,30,000
Answer:
(c) Rs. 2,30,000

III Short Answer Questions

Question 1.
Define Ratio Analysis
Answer:
According to Myers, “Ratio analysis is a study of the relationship among various financial factors in a business”.

Question 2.
What do you mean by ratio Analysis?
Answer:
Ratio analysis is a tool which involves analyzing the financial statements by calculating various. It is a tool of financial statement analysis, in which, inferences are drawn based on the computation and analysis of different ratios.

Question 3.
What are the two ways of classifying the ratios?
Answer:
Ratios may be classified in the following two ways:

• Traditional classification
• Functional classification

Question 4.
What do you mean by the traditional classification of ratio? Explain
Answer:
The traditional classification of ratios is done on the basis of the financial statements from which the ratios are calculated. Under the traditional classification, the ratio is classified as:

• Balance sheet ratio: If both items in a ratio are from the balance sheet, it is classified as a balance sheet ratio.
• Income statement ratio: If the two items in a ratio are from the income statement, it is classified as an income statement ratio.
• Inter – Statements ratio: If a ratio is computed with a tone item from the income statement and another item from the balance sheet, it is called an inter-statement ratio.

Question 5.
What is the functional classification of ratios?
Answer:
Under the functional classification, the rations are classified as follows:

• Liquidity ratios
• Long term solvency ratios
• Turnover ratios.
• Profitability ratios.

Question 6.
What do you mean by Liquidity ratios?
Answer:
Liquidity means the capability of being converted into cash with ease. Liquidity ratios help to assess the ability of a business concern to meet its short term financial obligations. Short term assets (current assets) are more liquid as compared to long term assets (fixed assets). Liquidity ratios are also called as short term solvency ratios.

Question 7.
What do you mean by the current ratio?
Answer:
Current ratio gives the proportion of current assets to current liabilities of a business concern. It is computed by dividing current assets by current liabilities. The current ratio indicates the ability of an entity to meet its current liabilities as and when they are due for payment.

Question 8.
What do you mean by Long-term solves ratios? what are its types?
Answer:
Long term solvency means the firm’s ability to meet its liabilities in the long run. Long term solvency ratios help to determine the ability of the business to repay its debts in the long run. The following ratios are normally computed for evaluating long term solvency of the business:

• Debt equity ratio
• proprietary ratio
• Capital gearing ratio

Question 9.
What do you mean by Turnover Ratios? What are its types?
Answer:
Turnover ratios show how efficiently assets or other items have been used to generate revenue from operations. They are also called as activity ratios or efficiency ratios. They how the speed of movement of various items. They are expressed as number of times in relation to the item compared.
The important turnover ratios are:

• Inventory turnover ratio
• Trade receivable turnover ratio
• Trade payable turnover ratio
• Fixed assets turnover ratio.

Question 10.
What do you mean by Trade Receivable Turnover ratio?
Answer:
Trade receivable turnover ratio is the comparison of credit revenue from operations with average trade receivables during an accounting period. It gives the velocity of the collection of cash from trade receivables.

Question 11.
What do you mean by Debt collection Period?
Answer:
Debt collection period is the average time taken to collect the amount due from trade receivables. Lesser the debt collection period, grater is the efficiency of management in the collection of cash from trade receivables. It is calculated as follows.

What do you mean by Trade payable Turnover ratio?
Trade payable turnover ratio is the comparison of net credit purchases with average trade payables during an accounting period. It gives the velocity to payment of cash towards trade payables.

Question 12.
What do you mean by credit payment period?
Answer:
It is the average time taken by the business for payment of accounts payable. Lesser the credit payment period, greater is the efficiency of the management in managing accounts payable as it indicated quicker settlement of trade payables. It is calculated as follows:

Question 13.
What do you mean by Fixed Assets Turnover
Answer:
The fixed assets turnover ratio gives the number of times the fixed assets are turned over during the year in relation to the revenue from operations. This ratio indicates the efficiency of utilization of fixed assets.

Question 14.
What do you mean by profitability ratios? What are its types?
Answer:
Profitability ratios help to assess the profitability of a business concern. These rations also help to analyze the earning capacity of the business in terms of utilization of resources employed in the business. Generally, these rations are expressed as a percentage.
The profitability ratios commonly used are:

1. Gross profit ratio
2. Operating cost ratio
3. Operating profit ratio
4. Net profit ratio
5. Return on investment.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 5 Admission of a Partner Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 5 Admission of a Partner

12th Accountancy Guide Admission of a Partner Text Book Back Questions and Answers

I. Multiple Choice Questions

Choose the correct answer

Question 1.
Revaluation A/c is a
(a) Real A/c
(b) Nominal A/c
(c) Personal A/c
(d) Impersonal A/c
Answer:
(b) Nominal A/c

Question 2.
On revluation, the increase in the value of assets leads to
(a) Gain
(b) Loss
(c) Expense
(d) None of these
Answer:
(a) Gain

Question 3.
The profit or loss on revaluation of assets and liabilities is transferred to the capital account of
(a) The old partners
(b) The new partner
(c) All the partners
(d) The Sacrificing partners
Answer:
(a) The old partners

Question 4.
If the old profit sharing ratio is more than the new profit sharing ratio of a partner, the difference is called
(a) Capital ratio
(b) Sacrificing ratio
(a) all the partners
(b) the old partners
(c) the new partner
(d) the sacrificing partners
Answer:
(b) Sacrificing ratio

Question 6.
Which of the following statements is not true in relation to the admission of a partner
(a) Generally mutual rights of the partners change
(b) The profits and losses of the previous years are distributed to the old partners
(c) The firm is reconstituted under a new agreement
(d) The existing agreement does not come to an end
Answer:
(d) The existing agreement does not come to an end

Question 7.
Match List-I with List-II and select the correct answer using the codes given below:

Answer:
b

Question 8.
Select the odd one out
(a) Revaluation profit
(b) Accumulated loss
(c) Goodwill brought by a new partner
(d) Investment fluctuation fund
Answer:
(c) Goodwill brought by a new partner

Question 9.
James and Kamalesh are partners sharing profits and losses in the ratio of 2:1. They admit Yogesh into partnership. The new profit sharing ratio between Balaji, Kamalesh, and Yogesh is agreed to 3:1:1. Find the sacrificing ratio.
(a) 1:3
(b) 3:1
(c) 5:3
(d) 3:5
Hint:

Answer:
(c) 5:3

Question 10.
Balaji and Kamalesh are partners sharing profits and losses in the ratio of 2:1. They admit Yogesh into partnership. The new profit sharing ratio between Balaji, Kamalesh and Yogesh is agreed to 3:1:1. Find the sacrificing ratio between Balaji and Kamalesh.
(a) 1:3
(b) 3:1
(c) 2:1
(d) 1:2
Hint:
Sacrifice ratio = old ratio – new ratio
Old partner’s = Balaji, Kamalesh

Answer:
(d) 1:2

II. Very Short Answer Questions

Question 1.
What is meant by the revaluation of assets and liabilities?
Answer:
When a partner is admitted into the partnership the assets and liabilities are revealed as the current value may differ from the book value. Determination of current values of assets and liabilities is called revaluation of assets and liabilities.

Question 2.
How are accumulated profits and losses distributed among the partner at the time of admission of a new partner?
Answer:
The following and the adjustment are necessary at the time of admission of a partner.

•  Distribution of accumulated profits, reserves, and losses
•  Revaluation of assets and liabilities
•  Determination of new profit – sharing ratio and sacrificing ratio
•  Adjustment for goodwill
•  Adjustment of capital on the basis of the new profit sharing ratio (if no agreed).

Question 3.
What is sacrificing ratio?
Answer:
The sacrificing ratio is the proportion of the profit which is sacrificed or foregone by the old partners in favour of the new partner. The purpose of finding the sacrificing ratio is to share the goodwill brought in by the new partner.
Sacrifice Ratio = Old share – New share

Question 4.
Give the journal entry for writing off existing goodwill at the time of admission of a new partner.
Answer:
Old partners capital /current A/c (in old ratio) Dr. To G/W A/c

Question 5.
State whether the following will be debited or credited in the revaluation account.
Answer:

1. Depreciation on assets – Debited
2. Unrecorded liability – Debited
3. Provision for outstanding expenses – Debited
4. Appreciation of assets – Credited

III Short Answer Questions

Question 1.
What are the adjustments required at the time of admission of a partner?
Answer:
The following adjustments are necessary at the time of admission of a partner.

1. Distribution of accumulated profits, reserves, and losses
2. Revaluation of assets and liabilities
3. Determination of new profit sharing ratio and sacrificing ratio
4. Adjustment for goodwill
5. Adjustment of capital on the basis of new profit sharing ratio

Question 6.
What are the journal entries to be passed on revaluation of ssets and liabilities?
Answer:
Following are the journal entries to be passed to record the revaluation of assets and liabilities:

Question 7.
Write a short note on the accounting treatment of goodwill.
Answer:
Accounting treatment for goodwill on the admission of a partner is discussed below:

Question 1.
When a new partner brings cash towards goodwill
Answer:
When the new brings cash towards goodwill in addition to the amount of capital, it is distributed to the existing partners in the sacrificing ratio.
If the new partner does not bring goodwill in cash or in kind, his share of goodwill must be adjusted through the capital accounts of the partners.

Sometimes the new partner may bring only a part of the goodwill in cash or assets. In such a case, for the cash or the assets brought, the respective account is debited and for the amount not brought in cash or kind, the new partner’s capital account is debited.

If goodwill already appears in the books of accounts, at the time of admission if the partners decide, it can be written off by transferring it to the existing partner s capital account / current account in the old profit sharing ratio.

IV Exercises

Distribution of accumulated profits, reserves, and losses

Question 1.
Arul and Anitha are partners sharing profits and losses in the ratio of 4:3. On 31.03.2018, Ajay was admitted as partner. On the date of admissions, the book of the firm showed a general reserve of ₹ 42,000. Pass the journal entry to distribute the general reserve.
Solution:

(General reserve transferred to old partners capital A/c in the old profit sharing ratio 4:3.)
Answer :
Arul: ₹ 24,000 (Cr.); Anitha: ₹ 18,000 (Cr.)

Question 2.
Anjali and Nithya are partners of firms sharing profits and losses in the ratio of 5:3. They admit Pramila on 01.01.2018. On that date, their balance sheet showed an accumulated loss of ₹ 40,000 on the asset side of the balance sheet. Give the journal entry to transfer the accumulated loss on admission.
Solution:

(Accumulated loss transferred to old partner s capital A/cs in the old ratio.)
Answer:
Profit and loss: ₹ 25,000 (Dr.); Nithya: ₹ 15,000(Dr.)

Question 3.
Oviya and Kavya are partners in firm sharing profits and losses in the ratio of 5:3. They admit Agalya into the partnership. Their balance sheet as of 31st March 2019 is as follows:

The pass journal entry to transfer the accumulated profit and reserve on admission.
Solution:

Accumulated profits and reserve transferred to old partners capital A/c in the old ratio
Answer :
Oviya : ₹ 37,500; Kavya : ₹ 22,500

Question 4.
Hari, Madhavan, and Kesavan are partners, sharing profit and losses in the ratio of 5:32. As from 1st April 2017, Vanmathi is admitted into the partnership and the new profit sharing ratio is decided as 4:3:2:1. The folio 5 adjustments are to be made.
(a) Increase the value of premises by ₹ 60,000.
(b) Depreciate stock by ₹ 5,000, furniture by ₹ 2, 000 and machinery by ₹ 2,500.
(c) Provide for an outstanding liability of ₹ 500.
Pass journal entries and prepare revaluation account.
Solution:

Question 5.
Seenu and Siva are partners sharing profits and losses in the ratio of 5:3. In view of kowsalya admission, they decided.
(i) To increase the value of the building by ₹ 40,000.
(ii) To bring into record investment at ₹ 10,000, Which have not so far been brought into
(iii) To decrease the value of machinery by ₹ 14,000 and furniture by ₹ 12,000.
(iv) To write off sundry creditors by ₹ 16,000.
Pass journal entries and prepare revaluation account
Solution :

Answer:
Revolution Profit in Rs. 40.000

Question 6.
Sai and Shankar are partners, sharing profits and losses in the ratio of 5:3.The firm’s balance sheet as on 31st December, 2017, was as follows:


On 31st December 2017, Shanmugam was admitted into the partnership for 1/4 share of profit with ₹ 12,000 as capital subject to the following adjustments.
(a) Furniture is to be revalued at ₹ 5,000 and the building is to be revalued at ₹ 50,000.
(b) Provision for doubtful debts is to be increased to ₹ 5,500
(c) An unrecorded investment of ₹ 6,000 is to be brought into account.
(d) An unrecorded liability of ₹ 2,500 has to be recorded now.
Pass journal entries and prepare Revaluation Account and capital account of partners after admission.
Solution:


Answer:
Revaluation Profit: ₹16,000; Capital accounts; Sai: ₹ 58,000 (Cr.), Shankar: ₹ 46,000 (Cr.); Shanmugam: ₹ 12,000 (Cr.))

Question 7.
Amal and Vimal are partners in firm sharing profits and losses in the ratio of 7:5. Their valance sheet as of 31st March 2019, is as follows:

Nirmal is admitted as a new partner on 01.04.2018 by introducing a capital of 30,000 for 1/3 share in the future profit subject to the following adjustments.
(a) Stock to be depreciated by ₹ 5,000;
(b) Provision for doubtful debts to be created for ₹ 3,000
(c) Land to be appreciated by ₹ 20,000.
Prepare revaluation account and capital account of partners after admission.
Solution:

Answer:
Revaluation Profit: ₹ 12,000; Capital accounts: Amal ₹ 91,000 (Cr.), Vimal ₹ 65,000 (Cr.), Nirmal ₹ 30,000 (Cr.))

Question 8.
Praveena and Dhanya are partners sharing profits in the ration of 7:3 they admit Malini into the firm. The new ratio among Praveena, Dhanya, and Malini are 5:2:3. Calculate the sacrificing ratio.
Solution:
Sacrificing ratio = Old share – New share
OR = 7:3 NR = 5:2:3
Praveena
SR = OR – NR
\(\frac{7}{10}-\frac{5}{10}=\frac{2}{10}\)
Dhanya
= \(\frac{3}{10}-\frac{2}{10}=\frac{1}{10}\)
SR = OR – NR
SR = 2:1
Answer :
Sacrificing ratio = 2:1

Question 9.
Ananth and Suman are partners sharing profits and losses in the ratio of 3:2. They admit Saran for 1/5 share, which he acquires entirely from Ananth. Find out the new profit sharing ratio and sacrificing ratio.
Solution :

Answer:
New profit sharing ratio: 2:2:1 Sacrificing ratio 1:0

Question 10.
Raja and Ravi are partners, sharing profit in the ratio of 3:2. They admit Ram for 1/4 share of the Profit, he takes 1/20 share from Raja and 4/20 from Ravi. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:


Answer :
New Profit sharing ratio: 11:4:5; Sacrificing ratio 1: 4

Question 11.
Vimala and Kamala are partners, sharing profits in the ratio of 4:3. Vinitha enters into the partnership and she acquires 1/14 from Vimala and 1/14 from Kamala. find out the new profit sharing ratio and sacrificing ratio.
Solution:

Answer :
New Profit sharing ratio 7 : 5 : 2; sacrificing ratio 1 : 1

Question 12
Govinda and Gopal are partners is a firm sharing profit^ in the ratio of 5 : 4. They admit Rahim as a partner. Govind surrenders 2/9 of his share in favour of Rahim. Gopal surrenders 1 /9 of his share in favour of R thim. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:


Answer:
New profit sharing ratio 35:32:14; Sacrificing ratio 5:2

Question 13.
Prema and Chandra share profits in the ratio of 5:3. Hema has admitted a partner. Prema surrendered 1 /8 of her share and Chandra surrendered 1 /8 of her share in favour of Hema. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:
Prema : Chandra → 5:3 (OR)
Sacrificing Ratio

Answer:
New profit sharing ratio 35:21:8; Sacrificing ratio 5:3

Question 14.
Karthik and Kannan are equal partners. They admit Kailash with 1/4 share of the profit. Kailash acquired his share from old partners in the ratio of 7:3 Calculate the new profit sharing ratio and sacrificing ratio.
Solution :
Sacrificing Ratio → Share of New Parter 1/4 → Sacrificed → 7 : 3
karthicks SR

Answer:
New profit sharing ratio 13:17:10; Sacrificing ratio 7:3

Question 15.
Selvam and Senthil are partners sharing profit in the ratio of 2:3. Siva is admitted into the firm with 1/5 share of profit. Siva acquires equally from Selvam and Senthil. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:

Answer:
New profit sharing ratio 3:5:2; Sacrificing ratio 1:1

Question 16.
Mala and Anitha are partners, sharing profits and losses in the ratio of 3:2. Mercy is admitted into the partnership with 1/5 share in the profits. Calculate new profit sharing ratio and sacrificing ratio.
Solution:
Since share sacrificed proportion and new profit sharing ratio are not given, it is assumed that the existing partners sacrifice in their old profit sharing ratio that is 3:2
Sacrificing Ratio

Answer:
New profit sharing ratio 12:8:5; Sacrificing ratio 3:2

Question 17.
Ambika, Dharani and Padma are partners in a firm sharing profile in the ratio of 5:3:2. they admit Ramya for 25% profit. Calculate the new profit sharing ratio and sacrificing ratio.
Solution:

Answer:
New profit sharing ratio 15:9:6:10; Sacrificing ratio 5:3:2

Adjustment for goodwill

Question 18.
Aparna and Priya are partners who share profits and losses in the ratio of 3:2. Brindha joins the firm for 1/5 share of profits and brings in cash for her share of the goodwill of Rs. 10,000. Pass necessary journal entry for adjusting goodwill on the assumption that the fluctuating capital method is followed and the partners withdraw the entire amount of their share of goodwill.
Solution:


Answer:
Share of goodwill: Aparna: ₹ 6,000; Priya ₹ 4,000

Question 19.
Deepak, Senthil, and Santhosh are partners sharing profits and losses equally. They admit Jerald into a partnership for 1/3 share in future profits. The goodwill of the firm is valued at ₹ 45,000 and Jerald brought cash for his share of goodwill. The existing partners withdraw half of the amount of their share of goodwill. Pass necessary journal entries for adjusting goodwill on the assumption that the fluctuating capital method is followed.
Solution:
Jerald’s share of G/w = 45,000 x \(\frac{1}{3}\) = Rs. 15,000

Answer:
Share of goodwill: Deepak: ₹ 5,000; Senthil: ₹ 5,000; Santhosh: ₹ 5,000

Question 20.
Malathi and Shobana are partners sharing profits and losses in the ratio of 5:4. They admit Jayasri into a partnership for 1/3 share of profit. Jayasri pays cash ₹ 6,000 towards her share of goodwill. The new ratio is 3:2:1. Pass necessary journal entry for adjusting goodwill on the assumption that the fixed capital method is followed.
Solution:
= OR – NR
OR = 5:4
New Ratio = 3:2:1

Answer:
Share of goodwill: Malathi’s Current account: ₹ 2,000; Shobana’s Current account: ₹ 4,000;

Question 21.
Anu and Arul were partners in a firm sharing profits and losses in the ratio of 4:1. They have decided to admit Mano into the firm for 2/5 share of profits. The goodwill of the firm on the date f admission was valued at ₹ 25,000. Mano is not able to bring in cash for his share of goodwill. Pass necessary journal entry for goodwill on the assumption that the fluctuating Capital method is followed.
Solution:
Mano’s Share of G/w = 25,000 x 2/5 = Rs. 10,000

Answer:
Share of goodwill: Anu: ₹ 8,000; Arul: ₹ 2,000;

Question 22.
Varun and Barath are sharing profits and losses 5:4. They admit Dhamu into partnership. The new profit sharing ratio is agreed at 1:1:1. Dhamu’s share of goodwill is valued at ₹ 15,000 of which he pays ₹ 10,000 in cash. Pass necessary journal entries for adjustment of goodwill on the assumption that the fluctuating capital method is . followed. ‘
Answer:

Answer:
Share of goodwill: Varun : ₹ 10,000; Barath : ₹ 5,000;

Question 23.
Sam and Jose are partners in the firm sharing profits and losses in the ratio of 3:2. On 1st April 2018, they admitted Joel as a partner. On the date of Joel’s admission, goodwill appeared in the books of the firm at ₹ 30,000. By assuming the fluctuating capital method, pass the necessary journal entry if the partners decide
(a) Write off the entire amount of existing goodwill
(b) write off ₹ 20,000 of the existing goodwill.
Solution :
To write off the entire amount of goodwill

Answer:
Share of goodwill: (a) Sam: ₹ 18,000 (Dr); Jose : ₹ 12,000 (Dr) (b) Sam: ₹ 12,000 (Dr); Jose: ₹ 8,000 (Dr.))

Comprehensive problems:

Question 24.
Rajan and Selva are partners sharing profits and losses in the ratio of 3:1. Their balance sheet as on 31st March 2017 is as under.

On 01.04.2017, they admit Ganesan as a new partner on the following arrangements.
(i) Ganesan brings ₹ 10,000 as capital for 1/5 share of profit.
(ii) Stock and furniture are to be reduced by 10%, a reserve of 5% on debtors for doubtful debts is to be created.
(iii) Appreciate buildings by 20%
Prepare revaluation account, partner’s capital account, and the balance sheet of the firm after admission.
Solution :


Answer:
Revaluation profit:: ₹ 2,100; Capital accounts : Rajan : 27075; Selva; ₹ 15,025; Ganesan : ₹ 10,000 Balance sheet total: ₹ 89,600

Question 25.
Sundar and Suresh are partners sharing profit in the ratio of 3 : 2. Their balance sheet as on 1st January 2017 was as follows:

They decided to admit Sugumar into a partnership for 1/4 share in the profits on the following terms:
(a) Sugumar has to bring in ₹ 30,000 as capital. His share of goodwill is valued at Rs. 5,000. He could not bring cash towards goodwill.
(b) That the stock is valued at ₹ 20,000.
(c) That the furniture is depreciated by ₹ 2,000.
(d) That the value of building be depreciated by 20%
Prepare necessary ledger accounts and the balance sheet after admission.
Solution:



Answer:
Answer:
Revaluation loss : ₹ 15,000; Capital accounts: Sundar : ₹ 39,000; Suresh; ₹ 26,000; Sugumar : ₹ 25,000 Balance sheet total ₹ 1,40,000

Question 26.
The following is the balance sheet of James and Justina as on 1.1.2017. They share the profits and losses equally.

On the above date, Balan is admitted as a partner with a 1/5 share in future profits. Following are the terms for his admission:
(i) Balan brings ₹ 25,000 as capital.
(ii) His share of goodwill is ₹ 10,000 and he brings cash for it.
(iii) The assets are to be valued as under:
Building ₹ 80,000; Debtors ₹ 18,000; Stock ₹ 33,000
Prepare necessary ledger accounts and the balance sheet after admission.
Solution:


Answer:
Revaluation profit:: ₹ 11,000; Capital accounts : James : ₹ 58,000; Justina; ₹ 68,000; Balan : ₹ 25,000 Balance sheet total: ₹ 1,86,000

Question 27.
Anbu and Shankar are partners in a business sharing profits and losses in the ratio of 7:5. The balance sheet of the partners on 31.03.2018 is as follows:

Rajesh is admitted for 1/5 share on the following terms:
(i) Goodwill of the firm is valued at ₹ 80,000 and Rajesh brought cash ₹ 6,000 for his share of goodwill.
(ii) Rajesh is brought ₹ 1,50,000 as his capital.
(iii) Motor car is valued at ₹ 2,00,000; stock at ₹ 3,80,000 and debtors at ₹ 3,50,000.
(iv) Anticipated claim on workmen compensation fund is ^ 10,000
(v) Unrecorded investment of ₹ 5,000 has to be brought into account.
Prepare revaluation account, capital account, and balance sheet after Rajesh’s admission.
Solution:


Answer:
Revaluation profit:: ₹ 15,000; Capital accounts: Anbu: ₹ 5,11,417; Shankar; ₹ 3,79,583; Rajesh : ₹ 1,50,000 Balance sheet total: ₹ 11,71,000

12th Accountancy Guide Admission of a Partner Additional Important Questions and Answers

Question 1.
When A and B sharing profit and losses in the ratio of 3:2. They admit C as a partner giving him 1 /3 share of profits. This will be given by A 8t B
(a) Equally
(b) in the ration of their
(c) in the ratio of profits
Answer:
(c) in the ratio of profits

Question 2.
In order to maintain fair dealings at the time of admission, it is necessary to revalue assets 8t liabilities of the firm to their
(a) Cost Price
(b) Cost price less depreciation
(c) True value.
Answer:
(c) True value.

Question 3.
If the new share of the incoming partner is given without mentioning the details of the sacrifice made by the old partners then the presumption is that old partners sacrifice in the
(a) Old profit sharing ration
(b) Gaining ratio
(c) Capital ratio.
Answer:
(a) Old profit sharing ration

Question 4.
On admission of a new partner, the balance of general Reserve A/c should be transferred to the capital account of
(a) all partners in their new profit sharing ratio
(b) Old partners in their new old profit sharing ratio
(c) Old partners in their new profit sharing ratio.
Answer:
(b) Old partners in their new old profit sharing ratio

Question 5.
The old partners share all the accumulated profit & reserves in their
(a) new profit sharing ratio
(b) Old profit sharing ratio
(c) Capital ratio
Answer:
(b) Old profit sharing ratio

Question 6.
Hie reconstitution of the partnership requires a revision of the existing partners
(a) Profit sharing ratio
(b) Capital ratio
(c) Sacrificing ratio|
Answer:
(a) Profit sharing ratio

Question 7.
……………… ratio is computed at the time by the admission of a partner
(a) gaining ratio
(b) Capitalization
(c) Sacrificing ratio
Answer:
(c) Sacrificing ratio

Question 8.
When unrecorded liability is brought in to books of accounts, it results in
(a) profit
(b) loss
(c) Neither profit nor loss
Answer:
(b) loss

IV Additional Problems:

Question 1.
Anandan and Balaraman partners in a firm with a capital of Rs. 70,000 and Rs. 50,000 respectively. They decided to admit Chandran into the firm with a capital of Rs. 40,000. Give journal entry for Capital brought in by Chandran.
Solution :

Question 2.
Rathai and Kothai are partners sharing profits in the ratio of 3:2. They admit Kanmani for 1/5th share of future profits which she acquires 3/20th from Rathai and 1/20th from Kothai. Calculate new Pro iring ratio and. sacrifice s ratio of old partners.
Solution :

Question 3.
P and Q are partners sharing profits in the ratio of 3:2. They admit R for 1/5th Share which acquires equally from P and Q. Calculate new profit sharing ratio and sacrificing ratio of old partners.
Solution:

Question 4.
Sankar and Saleem are partner in a firm sharing profits and losses in the ration of 3:2 as on 31st March 2005. Their Balarr sheet was as under.

On 1st April 2005, they admit Solomon into partnership on the following condition: Solomon has brought Rs. 1,00,000 as capital.
The value of land and building was to be increased by Rs. 20,000.
Stock and furniture were to be depreciated by Rs. 10,000 and Rs. 5,000 respectively. Rs. 15,000 to be written off from sundry creditors as it is no longer liable.
Provision for doubtful debts is to be increased by Rs. 1,000.
Give journal entries, prepare Revaluation Account and the Balance Sheet.
Solution:

Question 5.
Amar and Akbar are partners in a firm sharing profits and losses in the ratio of 2:1  as 31st March 2005. Their Balan Sh» was as under.

On 1st April 2005, they admit Antony into partnership on the following conditions:
Antony has brought in the capital of Rs. 1,50,000 for 1/5th share of the future profits. Stock and machinery were to be depreciated by Rs. 6,000 and Rs. 15,000 respectively.
Investments Rs. 15,000 not recorded in the books brought into accounts.
Provision for doubtful debts is to be created at 5% on debtors.
A liability of Rs. 4,000 for outstanding repairs has been omitted to be recorded in the books. Give journal entries, prepare Revaluation Account, Capital Account, Bank Account, and the Balance Sheet.
Solution:

Question 6.
Sumathi and Sundari are partners of a firm sharing profit and loss in the ratio of 4 :3. Their Balance Sheet shows Rs. 14,000 as Profit and Loss A/c on the liabilities side. Pass entry.
Solution:

Question 7.
Mahalakshmi and Dhanalakshmi are partners sharing profit and loss in the ratio of 3:2. They admit Deepalakshmi on 1st January 2005. On that date, their Balance Sheet showed an amount of Rs. 25,000 as Profit and Loss A/c in the Asset side. Pass entry.
Solution:

Question 8.
Damodaran and Jagadeesan are partners sharing profits in the ratio of 3:2. They decided to admit Vijayan for 1/5th share of future profit. Goodwill of the firm is to be valued at Rs. 50,000.
Give Journal entries, if
There is no goodwill in the books of the firm.
The goodwill appears at Rs. 30,000
The goodwill appears at Rs. 60,000.
Solution:

Question 9.
Sankari and Sudha are partners sharing profit and loss in the ratio of 3:2. Their Balance Sheet as on 31st March 2005 is as under.

They decided to admit Santhi into the partnership with effect from 1st April 2005 on the following terms:
Santhi to bring in Rs. 60,000 as Capital for 1/3rd share of profits.
Goodwill was valued at Rs. 45,000
The land was valued at Rs. 1,50,000
The stock was to be written down by Rs. 8,000
The provision for doubtful debts was to be increased to Rs. 3,000
Creditors include Rs. 5,000 no longer payable and this sum was to be written off.
Investment of Rs. 10,000 be brought into books.
Prepare Revaluation A/c, Capital A/c, and Balance Sheet of the new firm.
Solution :

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.7 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.7

Question 1.
Evaluate the following
(i) \(\int_{0}^{∞}\) x⁵ e^-3x dx
Solution:

(ii) \(\int_{0}^{π/2}\) \(\frac{e^{-tanx}}{cos^6 x}\) dx
Solution:

Question 2.
\(\int_{0}^{∞}\) sin^αx² x³ dx = 32, α > 0, find α
Solution:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 1 Electrostatics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

12th Physics Guide Electrostatics Book Back Questions and Answers

Part – I:
I. Multiple choice questions:

Question 1.
Two identical point charges of magnitude – q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the displacement?

(a) A₁ and A₂
(b) B₁ and B₂
(c) both directions
(d) No stable
Answer:
(b) B₁ and B₂
Solution:
The potential due to an electric dipole along the equatorial line is zero.

Question 2.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) uniformly charged infinite line
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
c) uniformly charged infinite plane
Solution:

Uniform field lines are represented by equidistant parallel lines.

Question 3.
What is the ratio of the \(\left|\frac{\mathrm{q}_{1}}{\mathrm{q}_{2}}\right|\) charges for the following electric field line pattern?

(a) \(\frac{1}{5}\)
(b) \(\frac{25}{11}\)
(c) 5
(d) \(\frac{11}{25}\)
Answer:
(d) \(\frac{11}{25}\)
Solution:
Here q₁ is a negative charge and q₂ is a positive charge
Hint:
Count the number of lines on each charge.

Question 4.
An electric dipole is placed at an alignment angle of 30° with an electric field of 2 × 10⁵ NC^-1. It experiences a torque equal to 8 Nm. The charge on the dipole if the dipole length is lcm is
(a) 4 mC
(b) 8 mC
(c) 5 mC
(d) 7 mC
Answer:
(b) 8 mC
Solution:
T = PE sin θ
T = (q × 21)E sin30°
q = (q × 10^-2) × q = 8 × 10^-2c

Question 5.
Four Gaussian surfaces are given below with charges inside each Gaussian surface. Rank the electric flux through each Gaussian surface in increasing order:

(a) D < C < B < A
(b) A < B = C < D
(c) C < A – B < D
(d) D > C > B > A
Answer:
(a) D < C < B < A
Hint :
flux depends on charge

Question 6.
The total electric flux for the following closed surface which is kept inside water:

(a) \(\frac{80 \mathrm{q}}{\varepsilon_{\mathrm{o}}}\)
(b) \(\frac{\mathrm{q}}{40 \varepsilon_{0}}\)
(c) \(\frac{\mathrm{q}}{80 \varepsilon_{\mathrm{o}}}\)
(d) \(\frac{\mathrm{q}}{160 \varepsilon_{0}}\)
Answer:
(b) \(\frac{\mathrm{q}}{40 \varepsilon_{0}}\)
Solution:

Question 7.
Two identical conducting balls having positive charges q₁ and q₂ are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be
(a) less than before
(b) same as before
(c) more than before
(d) zero
Answer:
(c) more than before
Solution:

Question 8.
Rank the electrostatic potential energies for the given system of charges in increasing order:

(a) 1 = 4 < 2 < 3
(b) 2 = 4 < 3 < 1
(c) 2 = 3 < 1 < 4
(d) 3 < 1 < 2 < 4
Answer:
(a) 1 = 4 < 2 < 3
Solution:

Question 9.
An electric field \(\overrightarrow{\mathrm{E}}\)= 10 × Î exists in a certain region of space. Then the potential difference V = V_o – V_A, where V_o is the potential at the origin and V_A is the potential at x = 2 m is:
(a) 10 V
(b) -20 V
(c) +20 V
(d) -10 V
Answer:
(c) +20 V
Solution:
E = \(-\frac{d v}{d x}\)
dv = E.dx
= l0x
= 10 × 2
dv = 20 V

Question 10.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. The correct plot for electrostatic potential due to this spherical shell is
(a)

(b)

(c)

(d)
Answer:
(b)
Solution:
In a spherical shell, the electric field inside is zero. But the electric potential is constant.
\(\mathrm{V}=\frac{\mathrm{q}}{4 \pi \varepsilon_{\mathrm{o}} \mathrm{r}}\) as distance increases its potential decrease non-linearly.

Question 11.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is
(a) 8.80 × 10^-17 J
(b) -8.80 × 10^-17 J
(c) 4.40 × 10^-17 J
(d) 5.80 × 10^-17 J
Answer:
(a) 8.80 × 10^-17 J
Solution:
W_A→B = (V_A – V_B)q
=(7+4)ne
= 11 × 50 × 1.6 × 10^-19
=8.8 × 10^-17

Question 12.
If the voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains the same, Q doubled
(d) Both Q and C remain the same
Answer:
(c) C remains the same, Q doubled
Solution:

Question 13.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
(a) Capacitance
(b) Charge
(c) Voltage
(d) Energy density
Answer:
(d) Energy density

Question 14.
Three capacitors are connected in a triangle as shown in the figure. The equivalent capacitance between points A and C is

(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) \(\frac{1}{4}\)μF
Answer:
(b) 2 μF
Solution:

Question 15.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 × 10^-2 C and 5 × 10^-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
(a) 3 × 10^-2 C
(b) 4 × 10^-2 C
(c) 1 × 10^-2 C
(d) 2 × 10^-2 C
Answer:
(a) 3 × 10^-2 C
Solution:
Total charge Q = q₁ + q₂ = 4 × 10^-2 C
charge on bigger sphere,
q₂ = Q\(\left(\frac{r_{2}}{r_{1}+r_{2}}\right)\)
= 4 × 10^-2 × \(\frac{3}{4}\)
q₂ = 3 × 10^-2 C

II. Short Answer Questions:

Question 1.
What is meant by the quantization of charges?
Answer:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is an integer (0, ±1, ±2, ±3, ±4).
This is called quantisation of electric charge.

Question 2.
Write down Coulomb’s law in vector form and mention what each term represents.
Answer:
The force on the point charge q₂ exerted by another point charge q₁

where r̂₂ is the unit vector directed from charge q₁ to charge q₂, and ‘K’ is the proportionality constant K = \(\frac{1}{4 \pi \varepsilon_{0}}\); ε₀ is the permittivity of free space.

Question 3.
What are the differences between the Coulomb force and the gravitational force?
Answer:

• The gravitational force between two masses is always attractive but Coulomb’s force between two charges can be attractive or repulsive, depending on the nature of charges.
• The value of the gravitational constant G = 6.626 x 10^-11 N m² kg^-2. The value of the constant k in Coulomb law is k = 9 x 10⁹ N m² C².
• The gravitational force between two masses is independent of the medium. The electrostatic force between the two charges depends on the nature of the medium in which the two charges are kept at rest.
• The gravitational force between two point masses is the same whether two masses are at rest or in motion. If the charges are in motion, yet another force (Lorentz force) comes into play in addition to Coulomb force.

Question 4.
Write a short note on the superposition principle.
Answer:

1. The superposition principle explains the interaction of multiple charges.
2. The total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
3. The force on q1 exerted by the charge q<₂ is \(\overrightarrow{\mathrm{F}}_{12}=\mathrm{K} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{21}^{2}} \hat{\mathrm{r}}_{21}\)
4. The force on q1 exerted by the charge q<₃ is \(\overrightarrow{\mathrm{F}}_{13}=\mathrm{K} \frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{31}^{2}} \hat{\mathrm{r}}_{31}\)
5. \(\overrightarrow{\mathrm{F}}_{1}^{\text {tot }}=\mathrm{K}\left\{\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{21}^{2}} \hat{\mathrm{r}}_{21}+\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{31}^{2}} \hat{\mathrm{r}}_{31}+\ldots \frac{\mathrm{q}_{1} \mathrm{q}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}^{2}} \hat{\mathrm{r}}_{\mathrm{n}_{i}}\right\}\)
   

Question 5.
Define ‘electric field’.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by
\(\vec { E } \) = \(\frac { \vec { F } }{ { q }_{ 0 } } \)
The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC^-1).

Question 6.
What is mean by ‘electric field lines?
Answer:
Electric lines of force is an imaginary straight or curved path in which unit positive charge tends to move in the presence of an electric field.

Question 7.
The electric field lines never intersect. Justify.
Answer:
As a consequence, if some charge is placed at the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence, electric field lines do not intersect.

Question 8.
Define ‘electric dipole’. Give the expression for the magnitude of its electric dipole moment and the direction.
Answer:

1. Two equal and opposite charges separated by a very small distance constitute an electric dipole.
   Ex:- Carbondioxide, Water.
2. The magnitude of the dipole moment is given by the product of the magnitude of the one of the charges and the distance between them.
   \(|\overrightarrow{\mathrm{P}}|=2 \mathrm{qa}\)
   \(\overrightarrow{\mathrm{P}}=\mathrm{qr}_{+}+(-\mathrm{q}) \overrightarrow{\mathrm{r}}_{-}\)
   where r̂₊ is the position of vector of +1 from the origin and r̂_– is the position vector of -q from the origin
   \(\overrightarrow{\mathrm{P}}=\mathrm{qai}-\mathrm{qa}(-\hat{i})=2 \mathrm{qai}\)
3. It is a vector quantity having direction along the dipole axis -q to +q.
4. Unit: coulomb metre (Cm).

Question 9.
Write the general definition of electro dipole moment for a collection of point charge.
Answer:
The electric dipole moment vector lies along the line joining two charges and is directed from -q to + q. The SI unit of dipole moment is coulomb meter (Cm).
\(\vec { P } \) = qa\(\hat{i} \) -qa(\(\hat{-i} \)) = 2 qa\(\hat{i} \)

Question 10.
Define ‘electrostatic potential’.
Answer:
The electric potential at a point P is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to the point P in the region of the external electric field \(\overrightarrow{\mathrm{E}}\).

Question 11.
What is an equipotential surface?
Answer:
An equipotential surface is a surface on which all the points are at the same potential.

Question 12.
What are the properties of an equipotential surface?
Answer:

1. The work done to move a charge ‘q’ between two points lie on equipotential surface is zero.
2. The electric field is normal to an equipotential surface.

Question 13.
Give the relation between electric field and electric potential.
Answer:
Consider a positive charge q kept fixed at the origin. To move a unit positive charge by a small distance dx in the electric field E, the work done is given by dW = -E dx. The minus sign implies that work is done against the electric field. This work done is equal to an electric potential difference. Therefore,
dW = dV.
(or) dV = -Edx
Hence E = \(\frac { dV }{ dx }\)
The electric field is the negative gradient of the electric potential.

Question 14.
Define ‘electrostatic potential energy’.
Answer:
It is defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity.

Question 15.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 16.
What is meant by electrostatic energy density?
Answer:

1. The energy stored per unit volume of space is defined electrostatic energy density.
   \(\mathrm{U}_{\mathrm{E}}=\frac{\mathrm{U}}{\text { Volume }}=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^{2}\)
2. U – electrostatic potential energy
3. E – electric field
4. £0 – permittivity of free space

Question 17.
Write a short note on ‘electrostatic shielding’.
Answer:
Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

Question 18.
What is Polarisation?
Answer:
Polarisation is defined as the total dipole moment per unit volume of the dielectric.
\(\begin{array}{l}
\vec{p}=\chi_{\mathrm{e}} \overrightarrow{\mathrm{E}}_{\text {ext }} \\
\chi_{\mathrm{e}}=\text { electric susceptibility }
\end{array}\)

Question 19.
What is dielectric strength?
Answer:
The maximum electric field the dielectric can withstand before it breakdowns is called dielectric strength.

Question 20.
Define ‘capacitance’. Give its unit.
Answer:

• It is defined as the ratio of the magnitude of the charge on either of the conductor plates to the potential difference existing between the conductors.
• Unit: farad (or) C/V

Question 21.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge. This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

III. Long Answer Questions:

Question 1.
Discuss the basic properties of electric charges.
Answer:
Basic properties of charges
(i) Electric charge:
Most objects in the universe are made up of atoms, which in turn are made up of protons, neutrons and electrons. These particles have mass, an inherent property of particles. Similarly, the electric charge is another intrinsic and fundamental property of particles. The SI unit of charge is the coulomb.

(ii) Conservation of charges:
Benjamin Franklin argued that when one object is rubbed with another object, charges get transferred from one to the other. Before rubbing, both objects are electrically neutral and rubbing simply transfers the charges from one object to the other. (For example, when a glass rod is rubbed against silk cloth, some negative charges are transferred from glass to silk. As a result, the glass rod is positively charged and silk cloth becomes negatively charged).

From these observations, he concluded that charges are neither created or nor destroyed but can only be transferred from one object to other. This is called conservation of total charges and is one of the fundamental conservation laws in physics. It is stated more generally in the following way. The total electric charge in the universe is constant and charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

(iii) Quantisation of charges:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is any integer (0, ±1, ±2, ±3, ± ….). This is called quantisation of electric charge. Robert Millikan in his famous experiment found that the value of e = 1.6 x 10^-19C. The charge of an electron is -1.6 x 10^-19 C and the charge of the proton is +1.6 x 10^-19C. When a glass rod is rubbed with silk cloth, the number of charges transferred is usually very large, typically of the order of 10¹⁰. So the charge quantisation is not appreciable at the macroscopic level. Hence the charges are treated to be continuous (not discrete). But at the microscopic level, quantisation of charge plays a vital role.

Question 2.
Explain in detail Coulomb’s law and its various aspects.
Answer:
1. Coulomb’s law states that “force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of distance between them.
\(\overrightarrow{\mathrm{F}}=\mathrm{k} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{12}\)

2. The force on the charge q₂ exerted by the charge q&lt₁ always lies along the line joining the two charges.
r̂&lt₁2 is a unit vector pointing from q₁ to q₂.

3. In SI units, \(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\)= 9 x l0⁹Nm²C^-2
where £0 = 8.85 × 10^-12 C² N^-1m^-2 .

4. The magnitude of the electrostatic force between two charges each of one coulomb and separated by a distance of lm is
|F| = 9 x l0⁹N
This is a huge quantity, almost equivalent to weight of one million tons.

5. Coulombs law in vacuum takes the form
\(\overrightarrow{\mathrm{F}}_{21}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{12}\)
In a Medium,
\(\overrightarrow{\mathrm{F}}_{21}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{12}\)
£ > £0, the force between two point charges in a medium other than vacuum is always less than that in vacuum.

6. Relative permitivitv £r = \(\frac{\varepsilon}{\varepsilon_{o}}\)
for vacuum and air £r = 1
other media, £r > 1

Question 3.
Define ’electric field’ and discuss its various aspects.
Answer:
The electric field is defined as the force experienced by unit positive charge kept at that point.
\(\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{q}_{\mathrm{o}}}\)

Important aspects of the Electric field:

1. If the charge q is positive then the electric field points away from the source charge and if q is negative, the electric field points towards the source charge q.
2. If the electric field at a point P is E, then the force experienced by the test charge q0 placed at the point P is
   \(\overrightarrow{\mathrm{F}}=\mathrm{q}_{\mathrm{o}} \overrightarrow{\mathrm{E}}\)
3. Electric field is independent of the test charge q₀ and depends only on the source charge q.
   
4. Since the electric field is a vector quantity, at every point in space, this field has unique direction.
5. The test charge is made sufficiently small such that it will not modify the electric field of the source charge.
6. For continuous and finite size charge distributions, integration techniques must be used.
7. There are two kinds of electric field: uniform or constant electric field and non-uniform electric field.
8. Uniform electric field will have the same direction and constant magnitude at all points in space. Non-uniform electric field will have different directions or different magnitudes or both at different points in space.

Question 4.
Calculate the electric field due to a dipole on its axial line and the equatorial plane.
Answer:
Case (i):

Electric field due to an electric dipole at points on the axial line.

1. Consider an electric dipole placed on the x-axis A point C is located at a distance of r from the midpoint of the dipole along the axial line.
   

2. The electric field at a point C due to +q is
\(\overrightarrow{\mathrm{E}}_{+}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{(\mathrm{r}-\mathrm{a})^{2}}\) (along BC)

3. Since the electric dipole moment vector p is from -q to +q and is directed along BC, the above equation is rewritten as
\(\overrightarrow{\mathrm{E}}_{+}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{(\mathrm{r}-\mathrm{a})^{2}} \hat{\mathrm{p}}\) ………..(1)

4. The electric field at a point C due to -q is
\(\overrightarrow{\mathrm{E}}_{-}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{(\mathrm{r}+\mathrm{a})^{2}} \hat{\mathrm{p}}\) ……………(2)

5. Since +q is located closer to the point C than -q, \(\overrightarrow{\mathrm{E}}+\) is stronger than \(\overrightarrow{\mathrm{E}}_{-}\).
Therefore, the length of the \(\overrightarrow{\mathrm{E}}+\) vector is drawn larger than that of \(\overrightarrow{\mathrm{E}}_{-}\) vector.
The total electric field at point C is calculated using the superposition principle of the electric field.

6. The direction of is shown in Figure

Case (ii)
1. Electric field due to an electric dipole at a point on the equatorial plane

2. Consider a point C at a distance r from the midpoint 0 of the dipole on the equatorial plane.

3. Since point C is equidistant from +q and -q and are the same.

4. The direction of \(\overrightarrow{\mathrm{E}}+\) is along BC and the direction of \(\overrightarrow{\mathrm{E}}_{-}\) is along CA.

\(\overrightarrow{\mathrm{E}}+\) and \(\overrightarrow{\mathrm{E}}_{-}\) are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it.

5. The perpendicular components \(|\overrightarrow{\mathrm{E}}|\) sinθ and \(\left|\overrightarrow{\mathrm{E}}_{-}\right|\) sinθ are oppositely directed and cancel each other.
\(\mathrm{E}_{\text {tot }}=-\left|\mathrm{E}_{+}\right| \cos \theta \hat{\mathrm{p}}-\left|\overrightarrow{\mathrm{E}}_{-}\right| \cos \theta \hat{\mathrm{p}} \cdots \ldots .(6)\)

6. The magnitudes \(\overrightarrow{\mathrm{E}}+\) and \(\overrightarrow{\mathrm{E}}_{-}\) are the same and are given by

7. By substituting equation (7) into equation (6) we get
\(\overrightarrow{\mathrm{E}}_{\mathrm{tot}}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{2 \mathrm{q} \cos \theta}{\left(\mathrm{r}^{2}+\mathrm{a}^{2}\right)} \hat{\mathrm{p}}\)
since \(\cos \theta=\frac{a}{\sqrt{r^{2}+a^{2}}}\)

8. At very large distances (r»a), the equation (8) becomes
\(\overrightarrow{\mathrm{E}}_{\mathrm{tot}}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\overrightarrow{\mathrm{p}}}{\mathrm{r}^{3}}(\mathrm{r}>>\mathrm{a})\)

Question 5.
Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
1. Consider an electric dipole of dipole moment \(\overrightarrow{\mathrm{p}}\) placed in a uniform electric field . The charge+q will experience a force q\(\overrightarrow{\mathrm{E}}\) in the direction of the field and charge-q\(\overrightarrow{\mathrm{E}}\) will experience a force -qE in a direction opposite to the field. Since the external field \(\overrightarrow{\mathrm{E}}\) is uniform, the total force acting on the dipole is zero.

2. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole.

3. The total torque on the dipole about the point 0, is given by
\(\vec{\tau}=\overrightarrow{\mathrm{0A}} \times(-\mathrm{q} \overrightarrow{\mathrm{E}})+\overrightarrow{\mathrm{0B}} \times \mathrm{q} \overrightarrow{\mathrm{E}}\)
Torque on dipole.

4. Total torque is perpendicular to the plane of the paper and is directed into it. The magnitude of the total torque
\(\vec{\tau}=|\overrightarrow{\mathrm{0A}}|(-\mathrm{q} \overrightarrow{\mathrm{E}})|\sin \theta+| \overrightarrow{\mathrm{0B}}|\mathrm{q} \overrightarrow{\mathrm{E}}| \sin \theta\)

\(\tau\) = qE.2a sin θ

5. Where θ is the angle made by \(\overrightarrow{\mathrm{p}}\) with \(\overrightarrow{\mathrm{E}}\). Since p = 2aq, the torque is written in terms of the vector product as
\(\vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}}\)

6. The magnitude of thectrque is \(\tau=p E \sin \theta\) and is maximum when θ = 90^o.
This torque tends to rotate the dipole and align it with the electric field \(\overrightarrow{\mathrm{E}}\). once \(\overrightarrow{\mathrm{p}}\) is aligned with, the total torque on the dipole becomes zero.

7. If the electric field is nonuniform, then the force experienced by +q is different from that experienced by -q. In addition to the torque, there will be a net force acting on the dipole.

Question 6.
Derive an expression for electrostatic potential due to a point charge.
Answer:
1. Consider a positive charge ‘q’ kept fixed at the origin. Let P be a point at distance r from the charge ‘q’.

2. The electric potential at the point P is
\(\mathrm{V}=\int_{\infty}^{\mathrm{r}}(-\overrightarrow{\mathrm{E}}) \cdot \mathrm{d} \overrightarrow{\mathrm{r}}=-\int_{\infty}^{\mathrm{r}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dr}}\)

3. Electric field due to positive point charge is
\(\begin{array}{l}
\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \\
\mathrm{V}=\frac{-1}{4 \pi \varepsilon_{\mathrm{o}}} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}
\end{array}\)

4. The infinitesimal displacement vector,
d\(\overrightarrow{\mathrm{r}}\) = drr̂ and using r̂.r̂ = 1, we have
\(\mathrm{V}=\frac{-1}{4 \pi \varepsilon_{\mathrm{o}}} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \cdot \mathrm{dr} \hat{\mathrm{r}}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \int_{\infty}^{\mathrm{r}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \mathrm{dr}\)

5. After the integration,
\(\mathrm{V}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \mathrm{q}\left\{-\frac{1}{\mathrm{r}}\right\}_{\infty}^{\mathrm{r}}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}}\)

6. Hence the electric potential due to a point charge q at a distance r is
\(\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}}\)

Question 7.
Derive an expression for electrostatic potential due to an electric dipole.
Answer:
Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let θ be the angle between the line OP and dipole axis AB.

Potential due to electric dipole:
1. Let r₁ be the distance of point P from +q and r₂ be the distance of point P from -q.

2. Potential at P due to charge,
\(+q=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{1}}\)
Potential at P due to charge
\(-q=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{2}}\)
Total potential at the point P,
\(\mathrm{V}=\frac{\mathrm{q}}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right)\) (1)

3. By the cosine law for triangle BOP,
r₁² =r² + a² – 2ra cos θ

4. Similarly applying the cosine law for triangle A0P,
r₂² = r² + a²  – 2racos(180-θ)
since cos(180-θ) = – cos θ
r₂² = r² + a² +2racos θ

Using Binomial theorem, we get

\(\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\overrightarrow{\mathrm{p}} \cdot \hat{\mathrm{r}}}{\mathrm{r}^{2}}\)
Special Cases:

Question 8.
0btain an expression for potential energy due to a collection of three-point charges which are separated by finite
distances.
Answer:

1. The electric potential at a point at a distance r from point charge q&lt₁ is given by
   \(\mathrm{V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}_{1}}{\mathrm{r}}\)
2. The potential V is the work done to bring a unit positive charge front infinity to the point. If the charge q, is brought from infinity to that point at a distance r from q₁.
3. The work done is stored as the electrostatic potential energy W = q₂V
   \(\mathrm{U}=\mathrm{q}_{2} \mathrm{~V}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \mathrm{q}_{1} \mathrm{q}_{2}\) ………….(1)
4. The electrostatic potential energy depends only on the distance between the two point charges.

Three charges are arranged in the following configuration.

1. Bringing a charge q₁ from infinity to point A requires no work because there are no other charges already present in the vicinity of charge q₁.

2. To bring the second charge q₂ to point B, work must be done against the electric field at B created by the charge q₁. So the work done on the charge q₂ is Wq₂V_1B Here V_1B is electrostatic potential due to the charge q₁ at point B.
\(\mathrm{U}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{12}}\) (2)

3. Similarly to bring the charge q₂ to point C, work has to be done against the total electric field due to both the charges q₁ and q₂. So the work done to bring the charge q₃ is the electrostatic potential due to charge q₁ at point C and V₂ is the electrostatic potential due to charge q₂ at point C. The electrostatic potential is
\(\mathrm{U}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{13}}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{\mathrm{r}_{23}}\right)\) ………….(3)

4. Adding equations (2) and (3), the total electrostatic potential energy for the system of three charges q₁, q₂ and q₃ is
\(\mathrm{U}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{12}}+\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\mathrm{r}_{13}}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{\mathrm{r}_{23}}\right)\) ……………. (4)

Question 9.
Derive an expression for the electrostatic potential energy of the_dipole in a uniform electric field.
Answer:
1. Consider a dipole placed in the uniform electric field \(\overrightarrow{\mathrm{E}}\). A dipole experiences a torque when kept in an uniform electric field \(\overrightarrow{\mathrm{E}}\). This torque rotates the dipole to align it with the direction of the electric field.

2. To rotate the dipole an equal and opposite external torque must be applied on the dipole.

3. The work done by the external torque to rotate the dipole from angle θ<super>1 to θ at constant angular velocity is

4. The work done is equal to the potentiaL energy difference between the angular positions θ’ and θ.
U(θ) – U(θ’)= ΔU = – pE cos θ+pE cos θ’

5. If the initial angle is θ’ = 90° and is taken as reference point, then U(θ’)= pE cos 90°=0.

6. The potential energy stored in the system of dipole kept in the uniform electric field is given by
U= – pE cos θ = – \(\overrightarrow{\mathrm{p}}\).\(\overrightarrow{\mathrm{E}}\) ……….(3)

7. In addition to p and E, the potential energy also depends on the orientation 0 of the electric dipole with respect to the external electric field.

8. The potential energy is maximum when the dipole is aligned anti-parallel (θ = π) to the external electric field and minimum when the dipole is aligned parallel (θ = 0)

Question 10.
Obtain Gauss law from Coulomb’s law.
Answer:
1. Consider a positive point charge Q is surrounded by an imaginary sphere of radius r.

2. The total electric flux for the entire area is given by,
\(\Phi_{\mathrm{E}}=\oint \mathrm{E} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\oint \mathrm{EdA} \cos \theta \ldots \ldots\) (1)

3. The electric field of the point charge is directed radially outward at all points on the surface of the sphere. Therefore the direction of the area element d\(\overrightarrow{\mathrm{A}}\) is along the electric field \(\overrightarrow{\mathrm{E}}\) and θ = 0°.
\(\Phi_{\mathrm{E}}=\oint \mathrm{E.dA}\) (since cos 0° = 1)
E is a uniform on the surface of the sphere

4. This equation is Gauss Law.

Question 11.
Obtain the expression for electric field due to an infinitely long charged wire.
Answer:

Electric field due to infinite long charged wire:

1. Consider an infinitely long straight wire having uniform linear charge density λ.

2. Let P be a point located at a perpendicular distance r for the wire.

3. The charged wire possesses a cylindrical symmetry.

4. Let us choose a cylindrical Gaussian surface of radius r and length L,

5. The total electric flux in this closed surface is calculated as follows.

for the curved surface. \(\overrightarrow{\mathrm{E}}\) is parallel to \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{E}}\) .dA = EdA.
For the top and bottom surfaces,
\(\overrightarrow{\mathrm{E}}\) is perpendicular to \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{E}}\)
\(\begin{aligned}
&\Phi_{\mathrm{E}}=\int_{\text {Curved }} \mathrm{EdA}=\frac{\mathrm{Q}_{\text {encl }}}{\varepsilon_{0}} \ldots \ldots . \text { (2) }\\
&\text { surface }
\end{aligned}\)

Since the electric field is constant, E is taken out of the integration and
Q_encl is given by Q_encl=λL.
E \(\int_{\begin{array}{l}
\text { Curved } \\
\text { surface }
\end{array}} \mathrm{d} \mathrm{A}=\frac{\lambda \mathrm{L}}{\varepsilon_{\mathrm{o}}}\) ……..(3)
Here,
Φ_E=∫_{curved surface} dA= total area of the curved surface = 2πrL
E.2πrL=\(=\frac{\lambda \mathrm{L}}{\varepsilon_{\mathrm{o}}}\)

6. The electric field due to the infinite charged wired depends on \(\frac{1}{\mathrm{r}}\)rather than \(\frac{1}{r^{2}}\) for a point charge.

7. Equation (5) indicates that the electric field is always along the perpendicular direction (r̂) to wire. If λ > 0 then \(\overrightarrow{\mathrm{E}}\) points perpendicular outward (r̂) from the wire and if λ < 0, then points perpendicular inward (-r̂).

Question 12.
Obtain the expression for electric field due to a charged infinite plane sheet.
Answer:

1. Consider an infinite plane sheet of charges with uniform surface charge density σ.
2. Let P be a point at a distance of r from the sheet.

Electric field due to changed infinite Planar sheet:

1. A cylindrical shaped Gaussian surface of length 2r and area A of the flat surfaces are chosen such that the infinite plane sheet passes perpendicularly through the middle part of the Gaussian surface.
2. Applying Gauss law for this cylindrical surface.
   
3. n̂ is the outward unit vector normal to the plane. The electric field depends on the surface charge density and is independent of the distance r.
4. If σ > 0, the electric field at any point P is outward perpendicular to the plane.
5. If σ < 0, the electric field points inward perpendicularly to the plane.

Question 13.
Obtain the expression for electric field due to a uniformly charged spherical shell.
Answer:
Consider a uniformly charged spherical shell of radius R and total charge Q.
Case (a) At a point outside the shell (r > R)
1. Let us choose a point P outside the shell at a distance r from the center.

2. The charge is uniformly distributed on the surface of the sphere.

3. The electric field must point radially outward if Q > 0 and point radially inward if Q < 0.

4. A spherical Gaussian surface of radius r is chosen and the total charge enclosed by this Gaussian surface is Q.

5. By Gauss law.
\(\oint_{\text {Gaussian }} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\) ………..(1)

6. The magnitude of is also the same at all points due to the spherical symmetry of the charge distribution.
\(\mathrm{E} \oint_{\text {Gaussian }} \mathrm{d} \mathrm{A}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\) ………….(2)

7. The total area of the Gaussian surface,
∫dA = 4πr²
E.4πr²=\(\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\)
For points outside the sphere, a large spherical Gaussian surface is drawn concentric with the sphere.

8. For points outside the sphere, a spherical gaussian surface smaller than the sphere is drawn

Case (c):
The electric field due to a charged spherical cell

• The electric field is radially outward if Q > 0 and radially inward if Q < 0.

Case (b):
At a point on the surface of the spherical shell (r=R)

• The electrical field at points on the spherical shell (r=R) is given by
  
  Electric fields versus distance Gauss spherical shell of radius R

Case (c):
At point inside the spherical shell (r < R)

1. Consider a point P inside the shell at a distance r from the center.
2. A Gaussian sphere of radius r is constructed.
3. Applying Gauss law,
   \(\begin{aligned}
   &\oint_{\text {Gaussian }} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}\\
   &\mathrm{E} .4 \pi \mathrm{r}^{2}=\frac{\mathrm{Q}}{\varepsilon_{\mathrm{o}}}
   \end{aligned}\) …………..(5)
4. Since Gaussian surface encloses no charge, Q = 0.
   E = 0 ……………(6)
   Electric field versus distance for a spherical shell of radius R
5. The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.
6. A graph can be plotted between the electric field and radial distance.

Question 14.
Discuss the various properties of conductors in electrostatic equilibrium,
i) The electric field is zero everywhere, inside the conductor. This is true regardless of whether the conductor is solid or hollow.
Answer:

1. When we apply a uniform electric field, the free electrons accelerated.
2. It causes the left side to be negatively charged and the right side to be positively charged.
3. Due to this realignment of free electrons, there will be an internal electric field created inside the conductor.
4. It increases until it nullifies the external electric field and is said to be in electrostatic equilibrium in the order of
   10^-16 s.
   

ii) There is no net charge inside the conductors. The charges must reside only on the surface of the conductors.

1. Consider an arbitrarily shaped conductor and a Gaussian surface is drawn inside the conductor.
2. It is very close to the surface of the conductor.
3. Since the electric field is zero inside the conductor, the net electric flux is zero.
4. So, there is no net charge inside the conductor.
5. No net charge inside the conductor

iii) The electric field outside the conductor is perpendicular to the surface of the conductor and has a magnitude of σ/ε. Where σ is the surface charge density at that point.

1. If the electric field has components parallel to the surface of the conductor, then free electrons of the surface of the conductor ‘would experience acceleration.
   
   (a) Electric field is along the surface
   (b) Electric field is perpendicular to the surface of the conductor
2. At electrostatic equilibrium, the electric field must be perpendicular to the surface of the conductor.
3. Consider a small cylindrical gaussian surface. 0ne half of this cylinder is embedded inside the conductor.
   
4. Since the electric field is normal to the surface of the conductor, the curved part of the cylinder has zero electric flux.
5. Inside the conductor, the electric field is zero.
6. The bottom flat part of the Gaussian surface has no electric flux.
7. The top flat part of the Gaussian surface contributes to the electric flux.
8. The electric field is parallel to the area vector and the total charge inside the surface is σA.
9. Applying Gauss law, EA = \(\frac{\sigma \mathrm{A}}{\varepsilon_{\mathrm{o}}}\)
10. In Vector form, \(\overrightarrow{\mathrm{E}}\)=\(\frac{\sigma}{\varepsilon_{\mathrm{o}}} \hat{\mathrm{n}}\)

iv) The electrostatic potential has the same value on the surface and inside of the conductor.

1. This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface.
2. Since the electric field is zero inside the conductor, the potential is the same as the surface of the conductor.
3. At electrostatic equilibrium, the conductor is always at equipotential.

Question 15.
Explain the process of electrostatic induction.
Answer:
Definition: Charging without actual contact is called electrostatic induction.

Consider an uncharged (neutral) conducting sphere at rest on an insulating stand. A negatively charged rod is brought near the conductor without touching. (Fig. a)

1. The negative charge of the rod repels the electrons in the conductor to the opposite side. As a result, positive charges are induced near the region of the charged rod while negative charges on the farther side.
2. Before introducing the charged rod, the free electrons were distributed uniformly on the surface of the conductor and the net charge is zero. 0nce the charged rod is brought near the conductor, the distribution is no longer uniform with more electrons located on the farther side of the rod and positive charges are located closer to the rod. But the total charge is zero.

Now the conducting sphere is connected to the ground through a conducting wire, This is called grounding. Since the ground can always receive any amount of electrons, grounding removes the electron from the conducting sphere.
(Fig. b)

3. When the grounding wire is removed from the conductor, the positive charges remain near the charged rod.
(Fig. c)

4. Now the charged rod is taken away from the conductor. As soon as the charged rod is removed, the positive charge gets distributed uniformly on the surface of the conductor. By this process, the neutral conducting sphere becomes positively charged. (Fig. d)

Question 16.
Explain the dielectrics in detail and how an electric field is induced inside a dielectric.
Answer:

1. When an external electric field is applied to a conductor, the charges are aligned in such a way that an internal electric field is created which cancels the external electric field.
2. But in the case of a dielectric, which has no free electrons, the external electric field only realigns the charges so that an internal electric field is produced.
3. The magnitude of the internal electric field is smaller than that of the external electric field.
4. Therefore the net electric field inside the dielectric is not zero but is parallel to an external electric field with a magnitude less than that of the external electric field.
5. For example, let us consider a rectangular dielectric slab placed between two oppositely charged plates (capacitor).
   a)

1. The uniform electric field between the plates acts as an external electric field ext which polarizes the dielectric placed between plates.
2. The positive charges are induced on one side surface and negative charges are induced on the other side of the surface.
3. But inside the dielectric, the net charge is zero even in a small volume.
4. So the dielectric in the external field is equivalent to two oppositely charged sheets with the surface charge densities
   +σ_b and -σ_b.
5. These charges are called bound charges. They are not free to move like free electrons in conductors.

Question 17.
0btain the expression for capacitance for a parallel plate capacitor.
Answer:

1. Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d.
2. The electric field between two infinite parallel plates is uniform and is given by
   \(\mathrm{E}=\frac{\sigma}{\varepsilon_{\mathrm{o}}}\)
   \(\sigma=\frac{Q}{A}\)
   σ = surface charge density on the plates
   
   
3. Capacitance is directly proportional to the area of cross-section and is inversely proportional to the distance between the plates.
4. If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.
5. If the distance between the two plates is reduced, the potential difference between the plate decreases with the E constant.
6. The voltage difference between the terminals of the battery increases, leads to an additional flow of charge to the plates from the battery, till the voltage on the capacitor equals the battery’s terminal voltage.
7. If the distance is increased, the capacitor voltage increases and becomes greater than the battery voltage.
8. The charges flow from capacitor plates to the battery till both voltages become equal.

Question 18.
0btain the expression for energy stored in the parallel plate capacitor.
Answer:

1. Capacitor not only stores the charge but also stores energy.
2. When a battery is connected to the capacitor, electrons of total charge -Q are transferred from one plate to the other plate.
3. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor.
4. To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
   dW = V dQ ………….(1)
   
5. This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor.
   \(\mathrm{U}_{\mathrm{E}}=\frac{1}{2}\left(\frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}\right)(\mathrm{Ed})^{2}=\frac{1}{2} \varepsilon_{\mathrm{o}}(\mathrm{Ad}) \mathrm{E}^{2}\) …………(4) where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined \(\mathrm{U}_{\mathrm{E}}=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^{2}\) ……….(5)
6. Energy is stored in the electric field existing between the plates of the capacitor. 0nce the capacitor is allowed to discharge, the energy is retrieved.
7. The energy density depends only on the electric field and not on the size of the plates of the capacitor.

Question 19.
Explain in detail the effect of a dielectric placed in a parallel plate capacitor.
Answer:

1. Consider a capacitor with two parallel plates each of cross-sectional area A and are separated by distance d.
2. The capacitor is charged by a battery of voltage V₀ and the charge stored is Q₀
3. The capacitance of the capacitor without the dielectric is
   \(\mathrm{C}_{\mathrm{0}}=\frac{\mathrm{Q}_{\mathrm{0}}}{\mathrm{V}}\)

1. The modified electric field is,
   \(\mathrm{E}=\frac{\mathrm{E}_{\mathrm{o}}}{\varepsilon_{\mathrm{r}}} \ldots \ldots \ldots .(1)\)
2. Here E₀ is the electric field inside the capacitors when there is no dielectric and
   Er is the dielectric constant. Since εr >1, the electric E < E₀.
3. As a result, the electrostatic potential difference between the plates (V = Ed) is also reduced. But at the same time, the charge Q₀ will remain constant once the battery is disconnected.
4. Hence the new potential difference is,
   \(\mathrm{V}=\mathrm{Ed}=\frac{\mathrm{E}_{0}}{\varepsilon_{\mathrm{r}}} \mathrm{d}=\frac{\mathrm{V}_{\mathrm{o}}}{\varepsilon_{\mathrm{r}}}\) …………(2)
5. Thus new capacitance in the presence of a dielectric is,
   \(\mathrm{C}=\frac{\mathrm{Q}_{\mathrm{o}}}{\mathrm{V}}=\varepsilon_{\mathrm{r}} \frac{\mathrm{Q}_{\mathrm{o}}}{\mathrm{V}_{\mathrm{o}}}=\varepsilon_{\mathrm{r}} \mathrm{C}_{\mathrm{o}}\) ……………..(3)
   Since ε_r > 1, we have C > C₀ . Thus insertion of the dielectric constant ε_r increases the capacitance.
   \(\mathrm{C}=\frac{\varepsilon_{\mathrm{r}} \varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}=\frac{\varepsilon \mathrm{A}}{\mathrm{d}}\) ……………(4)
   where ε = ε_r ε₀ is the permititivity of the dielectric medium.
6. The energy stored in the capacitor before the insertion of a dielectric is given by
   \(\mathrm{U}_{\mathrm{o}}=\frac{1}{2} \frac{\mathrm{Q}_{0}^{2}}{\mathrm{C}_{0}}\) ………(5)
7. After the dielectric is inserted, Q₀ remains constant but the capacitance is increased.
8. The stored energy is decreased.
9. \(\mathrm{U}_{\mathrm{o}}=\frac{1}{2} \frac{\mathrm{Q}_{0}^{2}}{\mathrm{C}_{0}}=\frac{1}{2} \frac{\mathrm{Q}_{0}^{2}}{\varepsilon_{\mathrm{r}} \mathrm{C}_{0}}=\frac{\mathrm{U}_{\mathrm{o}}}{\varepsilon_{\mathrm{r}}}\) ………………..(6)
   Since ε_r > 1, we get U > U₀.

ii) When the battery remains connected to the capacitor:

1. The battery of voltage V₀ remains connected to the capacitor when the dielectric is inserted into the capacitor.
2. The potential difference V₀ across the plates remains constant.
3. The charge stored in the capacitor is increased by a factor ε_r.
   
4. The energy stored in the capacitor before the insertion of a dielectric is given by \(\mathrm{U}_{\mathrm{o}}=\frac{1}{2} \mathrm{C}_{\mathrm{o}} \mathrm{V}_{\mathrm{o}}^{2}\) ………..(10)
5. After the dielectric is inserted, the capacitance is increased and the stored energy is also increased
6. \(\mathrm{U}=\frac{1}{2} \mathrm{CV}_{\mathrm{o}}^{2}=\frac{1}{2} \varepsilon_{\mathrm{r}} \mathrm{C}_{\mathrm{o}} \mathrm{V}_{\mathrm{o}}^{2}=\varepsilon_{\mathrm{r}} \mathrm{U}_{\mathrm{o}}\) …………(11)
   Since ε_r>1 we have U > U₀.
7. Since the voltage between capacitor V, is constant, the electric field between the plates also remains constant.
8. The energy density is given by
   \(\mathrm{U}=\frac{1}{2} \varepsilon \mathrm{E}_{0}^{2}\) (12)
   where ε is the permittivity of the given dielectric material.

Question 20.
Derive the expression for resultant capacitance, when capacitors are connected in series and in parallel.
Answer:
i) Capacitor in series

1. Consider three capacitors of capacitance C₁, C_2, and C₃
2. connected in series with a battery of voltage V.
   
3. By these processes, each capacitor stores the same amount of charge Q. The capacitances of the capacitors are in general different so that the voltage across each capacitor is also different and are denoted as V₁, V₂ and V₃ respectively.
4. The total voltage across each capacitor must be equal to the voltage of the battery.
   V = V₁ + V₂ + V₃
5. Since, Q = CV, we have \(\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}_{1}}+\frac{\mathrm{Q}}{\mathrm{C}_{2}}+\frac{\mathrm{Q}}{\mathrm{C}_{3}}\)
   \(\mathrm{V}=\mathrm{Q}\left(\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\right)\)
   \(\frac{\mathrm{Q}}{\mathrm{C}_{\mathrm{s}}}=\mathrm{Q}\left(\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\right), \frac{1}{\mathrm{C}_{\mathrm{s}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\)
6. The inverse of the equivalent capacitance C_s of three capacitors connected in series is equal to the sum of the inverses of each capacitance.
7. The equivalent capacitance C_s is always less than the smallest individual capacitance in the series.

ii) Capacitance in parallel:

1. Consider three capacitors of capacitance C₁, C_2, and C₃ connected in parallel with a battery of voltage V.
2. The potential difference across each capacitor is the same.
   Total charge Q = Q₁ + Q₂ + Q₃
   Q₁, Q2, Q3 – Charge stored in C₁, C₂, C₃
   Now, since Q = CV, we have Q = C₁V + C₂V + C₃V
   
3. If these three capacitors are considered to form a single capacitance Cp which stores the total charge Q.
   Q = C_pV.
   C_pV= C₁V + C₂V + C₃V
   C_p = C₁ + C₂ + C₃
4. The equivalent capacitance of capacitors connected in parallel is equal to the sum of the individual capacitances.
5. The equivalent capacitance C_p in a parallel connection is always greater than the largest individual capacitance.

Question 21.
Explain in detail how charges are distributed in a conductor, and the principle behind the lightning conductor.
Answer:
1. Consider two conducting spheres A and B of radii r₁ and r₂ respectively connected to each other by a thin conducting wire.

2. If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres such that the electrostatic potential is the same in both the spheres.

3. They are now uniformly charged and attain electrostatic equilibrium.

4. Let q₁ be the charge residing on the surface of sphere A and is the charge residing on the surface of sphere B such that, Q = q₁ + q₂.

5. The charges are distributed only on the surface and there is no net charge inside the conductor.

6. The electrostatic potential at the surface of the sphere A is given by
\(V_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r_{1}}\)

7. The electrostatic potential at the surface of the sphere B is given by 1 q₂
\(V_{B}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{r_{2}}\)
V_A = V_B

8. The charge density on the surface of sphere A is σ₁, and the charge density on the surface of sphere B is σ₂. This implies that q₁ = 4πr₁² σ₁ and

q₂ = 4πr₂² σ₂ ………..(3)
σ₁r₁ = σ₂r₂ ………..(4)
σr = constant ………..(5)

9. The surface charge density σ is inversely proportional to the radius of the sphere.
Corona discharge:

Consider a charged conductor of irregular shape.

10. The smaller the radius of curvature, the larger is the charge density. The end of the conductor which has larger curvature (smaller radius) has a large charge accumulation.

11. As a result, the electric field near this edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge.

12. This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Lightning conductor:
Uses to protect tall buildings from lightning strikes.

Principle:
Action at points or corona discharge.

Construction:

1. This device consists of a long thick copper rod passing from the top of the building to the ground.
2. The upper end of the rod has a sharp spike or a sharp needle.
3. The lower end of the rod is connected to the copper plate which is buried deep into the ground.

Working:

1. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike.
2. Since the induced charge density on a thin sharp spike is large, it results in a corona discharge.
3. This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud.
4. The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the earth.
5. The lightning arrester does not stop the lightning; rather it diverts the lightning to the ground safely.

Question 22.
Explain in detail the construction and working of a Van-de-Graaff generator.
Answer:
Principle:
Electrostatic induction and action at points.

Construction:

1. Hollow metallic sphere A is mounted on an insulating pillar.
2. Pulley B is mounted at the center of the sphere.
3. Pulley C is mounted near the bottom.
4. A silk belt moves over the pulleys.
5. Pulley C is driven continuously by an electric motor.
6. D and E are the comb-shaped conductors mounted near the pulleys.
   

Working:

1. 1. 10⁴ V is given to comb D.
   2. Near the comb, D air gets ionized due to the action of points.
   3. Negative charges move towards the needle and
   4. Positive charges stick to the belt, moves up, and reach the comb E. Due to the electrostatic induction.
   5. Comb E gets a negative charge and the sphere gets a positive charge.
   6. Comb E ionize the air.
   7. Hence descending belt will be. left uncharged.
   8. The machine transfers the positive charge to the sphere.
   9. Leakage of charges can be reduced by enclosing it in a gas-filled steel chamber at very high pressure.
   10. It produces a large potential 10⁷v.

Uses:
To accelerate positive ions (protons, deuteron) for the purpose of nuclear disintegration.

IV.Exercises:

Question 1.
When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Answer:

Question 2.
The total number of electrons in the human body is typically in the order of 10²⁸. Suppose, due to some reason, you and your friend lost 1% of this number of electrons. Calculate the electrostatic force between you and your friend separated at a distance of lm. Compare this with your weight. Assume the mass of each person is 60kg and use point charge approximation.
Answer:

Question 3.
Five identical charges Q are placed equidistant on a semicircle as shown in the figure. Another point charge q is kept at the center of the circle of radius R. Calculate the electrostatic force experienced by the charge q.

Answer:


\(\mathrm{F}_{\text {resultent }}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{Qq}}{\mathrm{R}^{2}}(1+\sqrt{2}) \hat{\mathrm{i}}\)

Question 4.
Suppose a charge +q on Earth’s surface and another +q charge is placed on the surface of the Moon.
(a) Calculate the value of q required to balance the gravitational attraction between Earth and the Moon
(b) Suppose the distance between the Moon and Earth is halved, would the charge q change?
(Take m_E = 5.9 × 10²⁴ kg, M_m = 7.9 × 10²² (kg)
Answer:

Question 5.
Draw the free body diagram for the following charges as shown in figure (a), (b) and (c).
a)
Answer:

b)
Answer:

c)
Answer:

Question 6.
Consider an electron travelling with a speed y0 and entering into a uniform electric field which is perpendicular to as shown in the Figure. Ignoring gravity, obtain the electron’s acceleration, velocity, and position as functions of time.

Answer:

Question 7.
A closed triangular box is kept in an electric field of magnitude E = 2 × 10^-3 NC^-1 as shown in the figure.

Calculate the electric flux through the
(a) vertical rectangular surface
Answer:

(b) slanted surface
Answer:

c) entire surface.
Answer:

Question 8.
The electrostatic potential is given as a function of x in figure (a) and (b). Calculate the corresponding electric fields in regions A, B. and D. Plot the electric field as a function of x for figure (b)

Answer:
Figure a:
\(\overrightarrow{\mathrm{E}}=-\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}} \hat{\mathrm{i}}\)
From 0 to 0.2 m,

Figure b:

Question 9.
A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.

To create the spark, an electric field of magnitude 3 × 10⁶ Vm^-1 is required.
Question (a)
What potential difference must be applied to produce the spark?
Answer:
V = E × d
= 3 × 10⁶ × 0.6 × 10^-3
= 1800 V

Question (b)
If the gap is increased, does the potential difference increase, decrease, or remains the same?
Answer:
If the distance between the plate increased, then its capacitance will decrease which gives rise to an increase in potential

Question (c)
Find the potential difference if the gap is 1 mm.
Answer:
V = E × d
= 3 × 10⁶ × 1 × 10^-3
= 3000 V

Question 10.
A point charge of +10 μC is placed at a distance of 20 cm from another identical point charge of +10 μC. A point charge of -2 μC is moved from point a to b as shown in the figure. Calculate the change in potential energy of the system? Interpret your result.

Answer:

Question 11.
Calculate the resultant capacitances for each of the following combinations of capacitors.

Question a.

Answer:

Question b.

Answer:

Question c.

Answer:

Question d.

Answer:

Question e.

Answer:

Question 12.
An electron and a proton are allowed to fall through the separation between the plates of a parallel plate capacitor of voltage 5 V and separation distance h = 1 mm as shown in the figure. (Take m_p =1.6 × 10^-27 kg, m_e =9.1 × 10^-31 kg and g= 10ms^-2)

Question (a)
Calculate the time of flight for both electron and proton
Answer:
E = \(\begin{array}{l}
\mathrm{V} \\
\hline \mathrm{d}
\end{array}[latex]
= [latex]\frac{5}{10^{-3}}\)
= 5 × 10³ Vm^-1

Question (b)
Suppose if a neutron is allowed to fall, what is the time of flight?
Answer:

Question (c)
Among the three, which one will reach the bottom first?
Answer:
Hence electron will reach the bottom first.

Question 13.
During a thunderstorm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 × 10⁶ Vm^-1), lightning will occur.

Question (a)
If the bottom part of the cloud is 1000m above the ground, determine the electric potential difference that exists between the cloud and ground.
Answer:
\(\mathrm{E}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}\)
V = E.x
= 3 ×10⁶ ×10³
= 3 × 10⁹ V

Question (b)
In a typical lightning phenomenon, around 25 C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
Answer:

Question 14.
For the given capacitor configuration
a) Find the charges on each capacitor
b) potential difference across them
c) energy stored in each capacitor.

Answer:

Question 15.
Capacitors P and Q have identical cross-sectional areas A and separation d. The space between the capacitors is filled with a dielectric of dielectric constant Er as shown in the figure. Calculate the capacitance of capacitors P and Q.

Answer:
1. The arrangement can be supposed to be a parallel combination of two capacitors each with plate area A/2 and separation. d. Total capacitance C_p = C_air + C_dielectric

2. The arrangement can be supposed to be a series combination of two capacitors, each with plate area A and separation d/2.

Part – II:

12th Physics Guide Electrostatics Additional Important Questions and Answers

I. Matching Type Questions:

Question 1.
Match Column I and Column II.

Answer:
(A) → (2)
(B) → (1)
(C) → (4)
(D) → (3)

Question 2.
Match Column I and Column II.

Answer:
(A) → (2)
(B) → (3)
(C) → (1)
(D) → (4)

Question 3.
Match the entries of Column I and Column II.

Answer:
(A) → (2)
(B) → (4)
(C) → (3)
(D) → (1)

Question 4.
When a dielectric slab is inserted between the plates of one of the two identical capacitors shown in the figure then match the following:

Answer:
(A) → (1)
(B) → (1)
(C) → (2)
(D) → (2)

II. Assertion – Reason Type Questions:

a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion.
c) Assertion is correct, reason is incorrect
d) Assertion is incorrect, reason is correct.

Question 1.
Assertion:
Four-point charges q_1, q₂, q_3, and q₄ are as shown in figure. The flux over the shown Gaussian surface depends only on charges q₁ and q₂.
Reason:
Electric field at all points on Gaussian surface depends only on charges q₁ and q₂,

Answer:
(d) Assertion is incorrect, reason is correct.

Question 2.
Assertion:
On disturbing an electric dipole in stable equilibrium in an electric field, it returns back to its stable equilibrium orientation.
Reason :
A restoring torque acts on the dipole on being disturbed from its stable equilibrium.
Answer:
(a) Assertion is correct, reason is correct; reason is a correct explanation for the assertion.

Question 3.
Assertion:
Work done in moving a charge between any two points in an electric field is independent of the path followed by the charge, between these points.
Reason :
Electrostatic force is a non-conservative force.
Answer:
(c) Assertion is correct, reason is incorrect

Question 4.
Assertion:
Dielectric polarisation means the formation of positive and negative charges inside the dielectric.
Reason:
Free electrons are formed in this process.
Answer:
(c) Assertion is correct, reason is incorrect

III. Statement Type Questions:

Question 1.
Which of the following about potential difference between any two points is true?
I. It depends only on the initial and final position.
II. It is the work done per unit positive charge in moving from one point to other.
III. It is more for a positive charge of two units as compared to a positive charge of one
Answer:
I and II

Question 2.
Select the correct statements from the following.
I. The electric field due to a charge outside the Gaussian surface contributes zero net flux through the surface.
II. Total flux linked with a closed body, not enclosing any charge will be zero.
III. Total electric flux, if a dipole is enclosed by a surface is zero.
Answer:
I, II, and III

Question 3.
An electric dipole of moment \(\overrightarrow{\mathrm{P}}\) is placed in a uniform electric field \(\overrightarrow{\mathrm{E}}\). Then
I. the torque on the dipole is \(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{E}}\).
II. the potential energy of the system is \(\overrightarrow{\mathrm{P}}\) x \(\overrightarrow{\mathrm{E}}\).
III. the resultant force on the dipole is zero.
Which of the above statements is/are correct.
Answer:
I and III

Question 4.
Select the incorrect statements from the following.
I. Polar molecules have permanent electric dipole moment.
II. CO₂ molecule is a polar molecule
III. H₂O is a non-polar molecules.
Answer:
II and III

IV. Choose the incorrect statement:

Question 1.
Which of the following statements is incorrect?
I. The charge q on a body is always given by q=ne, where n is any integer positive or negative.
II. By convention, the charge on an electron is taken to be negative charge.
III. The fact that electric charge is always an integral multiple of e is termed as quantisation of charge.
IV. The quantisation of charge was experimently demonstrated by Newton in 1912.
(a) only I
(b) only II
(c) only lV
(d) only III
Answer:
(c) only IV

Question 2.
The energy stored in a parallel plate capacitor is given by \(\mathrm{V}_{\mathrm{E}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}\). Now which of the following statements is not true?
I. The work done in charging a capacitor is stored in the form of electrostatic potential energy \(\mathrm{V}_{\mathrm{E}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}\)
II. The net charge on the capacitor is Q.
III. The magnitude of the net charge on one plate of a capacitor is Q.
(a) I only
(b) II only
(c) I and II
d) I, II and III
Answer:
(b) II only

Question 3.
Which of the following is not true ?
I. For a point charge, electrostatic potential varies as 1/r .
II. For a dipole, the potential depends on the magnitude of position vector, and dipole moment vector.
III. The electric potential varies as is at large distance.
IV. For a point charge, the electrostatic field varies as \( \frac{1}{\mathrm{r}^{2}}\).
(a) I only
(b) II only
(c) III only
(d) I, II and III
Answer:
(c) III only

V. Diagram – Type Question:

Question 1.
The figure shows a charge +q at point P held in equilibrium in air with the help of four + q charges situated at the vertices of a square. The net electrostatic force on q is given by

(a) Gauss’s law
(b) Coulomb’s law
(c) Principle of superposition
(d) net electric flux out the position of +q.
Answer:
(c) Principle of superposition
Solution:
The weight mg of the charge hole in air is in equilibrium with the net electrostatic force exerted by the four charges situated at the corners. The net electrostatic force is given by the charges at the corners. This is the principle of superposition.

Question 2.
Which of the following graphs shows the correct variation of force when the distance r between two charges varies?


Answer:

Solution:
From Coulombs’ law \(\mathrm{F}=\frac{\mathrm{Kq}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\) i.e., \(\mathrm{F} \propto \frac{1}{\mathrm{r}^{2}}\) which is correctly shown by graph (d).

Question 3.
Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that

(a) E_A > E_B > E_C
(b)E_A = E_B = E_C
(c) E_A > E_C > E_B
(d) E_A > E_C > E_B
Answer:
(c) ) E_A > E_C > E_B

VI. Choose the odd man out:

Question 1.
(a) lightning arrestor
(b) Van-de-Graf generator
(c) Photocopying
(d) AC generator
Answer:
(d) AC generator

Question 2.
(a) Gold
(b) Silver
(c) electric potential
(c) Aluminium
(d) ebonite
Answer:
(d) ebonite

Question 3.
(a) Farad
(b) \(\frac{J}{C} m^{-1}\)
(c) Vm^-1
(d) NC^-1
Answer:
(a) Farad

Question 4.
(a) electric field
(b) electric force
(c) electric potential
(d) electric dipole moment
Answer:
(c) electric potential

VII. Choose The Incorrect Pair:
Question 1.

Answer:
C) Electric flux – Vector product

Question 2.

Answer:
A) Coulomb’s law – Force is directly proportional to square of distance

VIII. Choose the correct pair:

Question 1.

Answer:
D) Electric potential due to a uniform charged solid non-conducting sphere – Varies inversely of radius

Question 2.

Answer:
A) Inside a conductor placed in an external electric field – Electric field = 0

IX. Fill In the blanks:

Question 1.
The charge acquired by 5 ×  10¹⁰ electrons _______.
Answer:
q = ne
q = 5 × 10¹⁰ × l.6 × 10^-19 = 8 × 10^-9 C

Question 2.
The unit of permittivity is…………………
Answer:
C²N^-1m^-2

Question 3.
At sharp points, the charge density is ________.
Answer:
maximum

Question 4.
The charges in an electrostatic field are analogous to ………………… in a gravitational field.
Answer:
mass

X. Choose the best answer:

Question 1.
If electric field in a region is radially outward with magnitude E = Ar, the charge contained in a sphere of radius r centred at the origin is
(a) \(\frac{1}{4 \pi \varepsilon_{o}} A r^{3}\)
(b) \(4 \pi \varepsilon_{\mathrm{o}} \mathrm{Ar}^{3}\)
(c) \(\frac{1}{4 \pi \varepsilon_{0}} \frac{A}{r^{3}}\)
(d) \(\frac{4 \pi \varepsilon_{0}}{\mathrm{r}^{3}}\)
Answer:
b) \(4 \pi \varepsilon_{\mathrm{o}} \mathrm{Ar}^{3}\)
Solution:

Question 2.
The electric field intensity just sufficient to balance the earth’s gravitational attraction on an electron will be: (given
(a) -5.6 × 10^-11N/C
(b) -4.8 × 10^-15N/C
(c) -1.6 × 10^-19N/C
(d) -3.2 × 10^-19N/C
Answer:
(a) -5.6 × 10^-11N/C
Solution:

Question 3.
The insulation property of air breaks down when the electric field is 3 × 10⁶ vm^-1 The maximum charge that can be given to a sphere of diameter 5m is approximately
(a) 2 × 10^-2 C
(b) 2 × 10^-3 C
(c) 2 × 10^-4 C
(d) 2 × 10^-5 C
Answer:
(b) 2 × 10^-3 C
Solution:

Question 4.
ABC is an equilateral triangle. Charges +q are placed at each corner as shown in fig. The electric intensity at centre 0 will be

(a) \(\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}}\)
(b) \(\frac{1}{4 \pi \epsilon_{0}} \frac{q}{r^{2}}\)
(c) \(\frac{1}{4 \pi \epsilon_{o}} \frac{3 q}{r^{2}}\)
(d) zero
Answer:
(d) zero
Solution:
(Unit positive charge at 0 will be repelled equally by three charges at the three corners of the triangle. By symmetry, the resultant at 0 would be zero.)

Question 5.
A hollow insulated conduction sphere is given a positive charge of 10μC. What will be the electric field at the centre of the sphere if its radius is 2m?
(a) Zero
(b) 5μCm^-2
(c) 20μCm^-2
(d) 8μCm^-2
Answer:
(a) zero
Solution:
(Charge resides on the outer surface of a conducting hollow sphere of radius R. We consider a spherical surface of radius r< R.)

Question 6.
Five balls marked 1, 2, 3, 4, and 5 are suspended by separate threads. The pairs (1, 2) (2, 4) and (4, 1) show mutual attraction and the pairs (2, 3) and (4, 5) show repulsion. The nature of ball marked as 1 is
a) positive
b) negative
c) neutral
d) can’t determine
Answer:
c) neutral

Question 7.
The resultant capacitance of four plates, each is having an area A, arranged, as shown above, will be (plate separation is d)

(a) \(\frac{A \varepsilon_{0}}{d}\)
(b) \(\frac{\mathrm{A} \varepsilon_{\mathrm{o}}}{2 \mathrm{~d}}\)
(c) \(\frac{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}\)
(d) \(\frac{3 A \varepsilon_{o}}{d}\)
Answer:
(c) \(\frac{2 \mathrm{~A} \varepsilon_{\mathrm{o}}}{\mathrm{d}}\)

Question 8.
At infinity, the electrostatic potential is
(a) Infinity
(b) maximum
(c) minimum
(d) zero
Answer:
(d) zero

Question 9.
The electric field at a point on the equatorial line of a dipole and direction of the dipole moment

(a) will be parallel
(b) will be in the opposite direction
(c) will be perpendicular
(d) are not related
Answer:
(b) will be in the opposite direction
Solution:
(The direction of the electric field at equatorial point A or B will be in opposite direction, like that of direction of dipole moment.)

Question 10.
Debye is the unit of
(a) electric flux
(b) electric dipole moment
(c) electric potential
(d) electric field intensity
Answer:
(b) electric dipole moment

Question 11.
In the given diagram a point charge +q is placed at the origin 0. Work done in taking another point charge -Q from point A to point B is:

(a) \(\frac{q Q}{4 \pi \epsilon_{0} a^{2}}\left(\frac{a}{\sqrt{2}}\right)\)
(b) zero
(c) \(\left[\frac{-\mathrm{q} \mathrm{Q}}{4 \pi \epsilon_{0}} \frac{1}{\mathrm{a}^{2}}\right] \sqrt{2} \mathrm{a}\)
(d) \(\left[\frac{\mathrm{qQ}}{4 \pi \epsilon_{0}} \frac{1}{\mathrm{a}^{2}}\right] \sqrt{2} \mathrm{a}\)
Answer:
(b) zero

Question 12.
The total electric flux emanating from a closed surface enclosing an α -particle is (e-electronic charge)
(a) \(\frac{2 e}{\varepsilon_{0}}\)
(b) \(\frac{e}{\varepsilon_{0}}\)
(c) \(\mathrm{e} \varepsilon_{\mathrm{o}}\)
(d) \(\frac{\varepsilon_{0} \mathrm{e}}{4}\)
Answer:
(a) \(\frac{2 e}{\varepsilon_{0}}\)
Solution:
(According to Gauss’s law total electric flux \(\frac{1}{\varepsilon_{0}}\) through a closed surface is time the total charge inside that surface.
Electric flux, Φ_E = \(\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}}\)
Charge on a-particle = 2e)
Φ_E = \(\frac{2 e}{\varepsilon_{o}}\)

Question 13.
A coil of the area of cross-section 0.5 m² with 10 turns is in a plane that is parallel to a uniform electric field of 100 N/C. The flux through the plane is?
(a) 100 V.m
(b) 500 V.m
(c) 20 V.m
(d) zero
Answer:
(b) 500 V.m

Question 14.
0n moving a charge of Q coulomb by χ cm, WJ of work is done, then the potential difference between the point is
(a) \(\frac{\mathrm{W}}{\mathrm{Q}} \mathrm{v}\)
(b) QWV
(c) \(\frac{Q}{W} V\)
(d) \(\frac{Q^{2}}{W} V\)
Answer:
(a) \(\frac{\mathrm{W}}{\mathrm{Q}} \mathrm{v}\)
Solution:
Potential difference between two points in an electric field is.
V_A – V_B = \(\frac{\mathrm{W}}{\mathrm{q}_{\mathrm{o}}}\)

Question 15.
The positive terminal of 12 V battery is connected to the ground. Then the negative terminal will be
(a) – 6 V
(b) +12 V
(c) zero
(d)- 12V
Answer:
(d) – 12V
Solution:
When negative terminal is grounded, the positive terminal of battery is at +12 V. When positive terminal is grounded, the negative terminal will be at -12 V.

Question 16.
The maximum electric field that a dielectric medium can withstand without break-down is called is
(a) permittivity
(b) dielectric constant
(c) electric susceptibility
(d) dielectric strength
Answer:
(d) dielectric strength

Question 17.
The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E is
(a) \(\varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}\)
(b) \(\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}\)
(c) \(\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} / \mathrm{Ad}\)
(d) \(\varepsilon_{0} \mathrm{E}^{2} / \mathrm{Ad}\)
Answer:
(a) \(\varepsilon_{0} \mathrm{E}^{2} \mathrm{Ad}\)
Solution:
(Energy required to charge the capacitor is W = u = QV)
∵ E = V/d

Question 18.
Energy is stored in a capacitor in the form of
(a) electrostatic energy
(b) magnetic energy
(c) light energy
(d) heat energy
Answer:
(a) electrostatic energy

Question 19.
Dimension and unit of Electric flux is
(a) ML²T^-3 A^-2, Nm² C^-1
(b) ML³ T^-3 A^-1, Nm² C^-1
(c) ML² T^-1 A^-2, Nm² C^-1
(d) ML^-4 T^-3 A^-2, Nm² C^-1
Answer:
b) ML³ T^-3 A^-1, Nm² C^-1

Question 20.
An air-core capacitor is charged by a battery. After disconnecting it from the battery, a dielectric slab is fully inserted in between its plates. Now, which of the following quantities remains constant?
(a) Energy
(b) voltage
(c) Electric field
(d) Charge
Answer:
(d) Charge

Question 21.
In a charged capacitor, the energy is stored in
(a) the negative charges
(b) the positive charges
(c) the field between the plates
(d) both (a) and (b)
Answer:
(c) the field between the plates

Question 22.
The potential gradient at which the dielectric of a condenser just gets punctured is called
(a) dielectric constant
(b) dielectric strength
(c) dielectric resistance
(d) dielectric number
Answer:
(b) dielectric strength

Question 23.
The unit of permittivity is
(a) C²N^-1m^-2
(b) Nm²C^-2
(c) Hm^-1
(d) NC^-2m^-2
Answer:
(a) C²N^-1

Question 24.
In the case of a Van de Graaff generator, the breakdown field of air is
(a) 2 × 10⁸ Vm^-1
(b) 3 × 10⁶ Vm^-1
(c) 2 × 10⁸ Vm^-1
(d) 2 × 10⁴ Vm^-1
Answer:
(b) 3 × 10⁶ Vm^-1

Question 25.
Van de Graaff generator is used to
(a) store electrical energy
(b) build up the high voltage of a few million volts
(c) decelerate charged particle-like electrons
(d) both (a) and (b)
Answer:
(b) build up the high voltage of a few million volts

XI. Two mark questions:

Question 1.
What is meant by triboelectric charging?
Answer:
Charging the objects through rubbing is called triboelectric charging.

Question 2.
When does an object is said to be electrically neutral?
Answer:
If the net charge is zero in an object, It is said to be electrically neutral.

Question 3.
State Gauss’s Law?
Answer:
Definition:
Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux OE through the closed surface is Φ_E = \(\oint { \vec { E } } \) .d\(\vec { A } \) = \(\frac {{ q }_{encl}}{{ ε }_{0}}\)

Question 4.
State coulomb’s law.
Answer:
Coulomb’s law states that the force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Question 5.
What is meant by dielectric?
Answer:
A dielectric is a non-conducting material and has no free electrons. The electrons in a dielectric are bound within the atoms. Ebonite, glass, and mica are some examples of dielectrics.

Question 6.
What is meant by the superposition principle?
Answer:
The total force on a given charge is the vector sum of the forces exerted on it due to all other charges. The force on q, due to charges q < ₂, q < ₃ ………… q_n

Question 7.
What are polar molecules? Give examples.
Answer:
In polar molecules, the centers of the positive and negative charges are separated even in the absence of an external electric field. They have a permanent dipole moment.
The net dipole moment is zero in the absence of an external electric field. Examples of polar molecules are H₂O, N₂O, HCl, NH₃.

Question 8.
Explain the working of a microwave oven?
Answer:

1. It works on the principle of torque acting on an electric dipole.
2. The food we consume has water molecules which are permanent electric dipoles.
3. Oven produces microwaves that are oscillating electromagnetic fields and produce a torque on the water molecules.
   
4. Due to this torque on each water molecule, the molecules rotate very fast and produce thermal energy.
5. Thus, the heat generated is used to cook the food.

Question 9.
What are permittivity and relative permittivity? How are they related?
Answer:

• Permittivity of a medium = permittivity of free space × relative permittivity, ε = ε₀ ε_r
  Its unit is C² N^-1 m^-2
• Relative permitivity = \(\frac{\text { permitivity of a medium }}{\text { permitivity of vacuum }}=\varepsilon_{\mathrm{r}}=\frac{\varepsilon}{\varepsilon_{\mathrm{o}}}\)
• It has no unit.

Question 8.
What is a capacitor?
Answer:
The capacitor is a device used to store electric charge and electrical energy. Capacitors are widely used in many electronic circuits and have applications in many areas of science and technology.

Question 11.
State Gauss law?
Answer:
Gauss law states that the total flux of the electric field E over any closed surface is equal to l/£0 times of the net charge enclosed by the surface,
\(\phi_{\mathrm{E}}=\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}}\)

Question 12.
What is meant by electrostatic shielding?
Answer:

• It is the process of isolating a certain region of space from external field.
• It is based on the fact that electric field inside a conductor is zero.

Question 13.
What is meant by electrostatic Induction?
Answer:

• The phenomenon of producing induced charges without any contact with another charge is known as electrostatic induction.
• Electrostatic induction is used in electrostatic machines like Van de Graaff generators and capacitors.

Question 14.
Define farad.
Answer:
A conductor has a capacitance of one farad; if a charge of 1 coulomb given to it, rises its potential by 1 volt.

Question 15.
What are dielectrics? Give examples.
Answer:
A dielectric is an insulating material in which all the electrons are tightly bound to the nucleus of the atom. There are no free electrons to carry current.
E.g:- Ebonite, mica, and oil.

Question 16.
Can two equipotential surfaces intersect? Give reason.
Answer:
Since the electric field is normal to the equipotential surface and also the potential difference between any two points on the surface is nullified, the intersection is not possible.

Question 17.
Distinguish between polar and non-polar molecule?
Answer:

Question 18.
What is meant by dielectric breakdown?
Answer:
When the external electric field applied to a dielectric is very large, it tears the atoms apart so that the bound charges become free charges. Then the dielectric starts to conduct electricity. This is called dielectric breakdown.

Question 19.
The electric field outside a conductor is perpendicular to its surface. Justify.
Answer:

• The electric field has components parallel to the surface of the conductor.
• The free electrons in the conductor would experience acceleration.
• The conductor is not in equilibrium. At electrostatic equilibrium, the electric field must be perpendicular to its surface.

Question 20.
What is called ‘fringing field’ is a capacitor? when does it ignore?
Answer:
For finite-sized plates, the electric field is not strictly uniform between the plates. At both edges, the electric field is bent outward. Under the condition (d²<< A), this effect can be ignored.

Question 21.
How does a capacitor is used in a Computer keyboard?
Answer:
When the key is pressed, the separation between the plates decreases leading to an increase in the capacitance. This is turn triggers the electronic circuits in the computer to identify which key is pressed.

Question 22.
What will happen to the charge, voltage, electric field, the capacitance of a dielectric placed capacitor when it is connected with a battery and then disconnected?
Answer:

Question 23.
Calculate the number of electrons in one Coulomb of negative charge.
Solution:
According to the quantisation of charge q = ne
Here q = 1C. So the number of electrons in 1 coulomb of charge is
\(\mathrm{N}=\frac{q}{e}=\frac{1 \mathrm{C}}{1.6 \mathrm{X} 10^{-19}}=6.25 \times 10^{18} \text { electrons }\)

Question 24.
Define electric potential energy of two point charges?
Answer:
It is equal to work done to assemble the charges or work done in bringing each charge or work done in bringing a charge from an infinite distance.

Question 25.
Why is it safer to be inside a car than standing under a tree during lightning?
Answer:

• The metal body of the car provides electrostatic shielding.
• The electric field inside the car is zero.
• During lightning, the electric discharge passes through the body of the car. So during lightning, it is safer to sit inside a car than on open ground or under a tree.

Question 26.
Define the electric potential energy of an electric dipole placed in an electric field.
Answer:
Electric potential energy of an electric dipole in an electrostatic field is the work done in rotating the dipole to the desired position in the field.

Question 27.
What happens if a polar molecule is placed in an electric field?
Answer:

1. When a polar molecule is placed in an electric field, the dipoles orient themselves in the direction of the electric field. Hence a net dipole moment is produced.
2. The alignment of dipole moments in the direction of the applied electric field is called polarisation of electric polarisation.
3. The magnitude of the induced dipole moment p is directly proportional to the electric field E. The dipole moment, p = αE when α is called molecular polarisability.

Question 28.
Define volt.
Answer:

1. The unit of potential difference is volt.
2. The potential difference between the two points is I volt if 1 joule of work is done in moving 1 coulomb of charge from one point to another against the electric force.

Question 29.
What does an electric dipole experience when kept in a uniform electric field and non-uniform electric field?
Answer:
Uniform electric field:  When a dipole is kept in a uniform electric field at an angle θ, the net force is zero. It experiences a torque \(\vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}} \quad \tau=p \mathrm{E} \sin \theta\)

Non-uniform electric field: If the dipole is placed in a non-uniform electric field at an angle θ in addition to torque, it also experiences a force.

Question 30.
What is the Effective Capacitance of capacitors connected in series?
Answer:
When a number of capacitors are connected in series, the reciprocal of the effective capacitance is equal to the sum of reciprocal of the capacitances of the individual capacitors.

Question 31.
What is the effective capacitance of capacitors connected in parallel?
Answer:
The effective capacitance of the capacitors connected in parallel is equal to the sum of the capacitances of the individual capacitors.
C_p = C₁ + C₂ + ……….. + C_n

Question 32.
What are conductors and insulators? Give examples.
Answer:

1. Bodies which allow the charges to pass through are called conductors.
   e.g. Metals, human body, Earth, etc.
2. Bodies that do not allow the charge to pass through are called insulators.
   (e.g) Glass, mica, ebonite, plastic, etc.

Question 33.
What is meant by point charge?
Answer:
Any charge which occupies a space with dimensions less than its distance away from an observation point considered a point charge.

Question 34.
What is a capacitor?
Answer:
A capacitor is a device used to store electric charge and electrical energy.

Question 35.
Why does a balloon after rubbing stick to a wall?
Answer:
It is due to the polarisation of the wall due to the electric field due to a balloon.

Question 36.
How does the lightning conductor prevent the lightning stroke from the damage of the building?
Answer:

1. When a negatively charged cloud passes over the building, a positive charge will be induced on the pointed conductor.
2. The positively charged sharp points will ionize the air in the vicinity.
3. This will partly neutralise the negative charge of the cloud, thereby lowering the potential of the cloud.
4. The negative charges that are attracted to the conductor travel down to the earth. Thereby preventing the lightning stroke from the damage of the building.

Question 37.
Define the physical quantity whose unit is Vm, and state whether it is scalar or Vector.
Answer:

1. Electric flux has the unit Vm.
2. The number of electric field lines passing through a given area.
3. q = EA cosθ
4. It is a scalar quantity.
5. The other unit is Nm²c^-1.

XII. Three Mark Questions:

Question 1.
Give some important inference, over the expression of the electric field due to an electric dipole on its axial line and equational line.
Answer:
Important inferences
i) The magnitude of the electric field at points on the dipole axis is twice the magnitude of the electric field at points on the equatorial plane. The direction of the electric field at points on the dipole axis is directed along the direction of dipole moment vector \(\overrightarrow{\mathrm{p}}\) but at points on the equatorial plane it is directed opposite to the dipole moment vector, that is along \(\overrightarrow{\mathrm{p}}{-}\).

ii) At very large distances, the electric field due to a dipole varies as \(\frac{1}{\mathrm{r}^{3}}\) electric field due to a dipole at very large distances goes to zero faster than the electric field due to a point charge.

iii) The distance 2a approaches zero and q approaches infinity such that the product 2aq = p is finite, then the dipole is called a point dipole.

Question 2.
Mention some important points on an expression of electric potential due to point charge.
Answer:
Important points :
i) If the source charge q is positive, V > 0. If q is negative, then V is negative and equal to
\(\mathrm{V}=-\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{q}}{\mathrm{r}}\)
ii) The potential due to positive charge decreases as the distance increases, but for a negative charge the potential increases as the distance is increased. At infinity (r = 0) electrostatic potential is zero (V=0).
iii) The electric potential at a point P due to a collection of charges q₁, q₂, q₃,……..q_n is equal to sum of the electric potentials due to individual charges.

Question 3.
Mention some important points over the derivation of electric potential due to an electric dipole.
Answer:
Important points:
i) The potential due to an electric dipole falls as \(\frac{1}{r^{2}}\) and the potential due to a single point
charge falls as \(\frac{1}{\mathrm{r}}\). Thus the potential due to the dipole falls faster than that due to a monopole. As the distance increases from electric dipole, the effects of positive and negative charges nullify each other.

ii) The potential due to a point charge is spherically symmetric since it depends only on the distance r. But the potential due to a dipole is not spherically symmetric because the potential depends on the angle between \(\overrightarrow{\mathrm{p}}\) and position vector \(\overrightarrow{\mathrm{r}}\) of the point.

iii) However the dipole potential is axially symmetric. If the position vector \(\overrightarrow{\mathrm{r}}\) is rotated about \(\overrightarrow{\mathrm{p}}\) by keeping θ fixed, then all points on the cone at the same distance r will have the same potential

Question 4.
Small mercury drops of the same size are charged to the same potential V. If n such drops coalesce to form a single large drop, then calculate its potential.
Solution:
Let r be the radius of a small drop and R that of the large drop. Then, since the volume remains conserved,
\(\frac { 1 }{ 2 }\) πR² = \(\frac { 4 }{ 3 }\) πR³n
⇒ R³ = r³n
R = r³(n)^1/3
Further, since the total charge remains conserved, we have, using Q = CV
C_large V = n C_small v
Where V is the potential of the large drop.
4πε₀ RV = n (4πε₀r)v
V = \(\frac { nrv }{ R }\) = \(\frac { nrv }{{ r(n) }^{1/3}}\)
V = vn^2/3

Question 5.
Derive an expression for electric flux in a non-uniform electric field and an arbitrarily shaped area.
Answer:
1. Suppose the electric field is not uniform and the area is not flat, then the entire area is divided into n small area segments such that each area element is almost flat and the electric field through each area element is considered to be uniform.

2. The electric flux for the entire area A is approximately written as
\(\begin{array}{l}
\Delta \overrightarrow{\mathrm{A}}_{1}, \Delta \overrightarrow{\mathrm{A}}_{2}, \Delta \overrightarrow{\mathrm{A}}_{3} \ldots \ldots \Delta \overrightarrow{\mathrm{A}}_{\mathrm{n}} \\
\Phi_{\mathrm{E}}=\overrightarrow{\mathrm{E}}_{1} \cdot \Delta \overrightarrow{\mathrm{A}_{1}}+\overrightarrow{\mathrm{E}}_{2} \cdot \Delta \overrightarrow{\mathrm{A}}_{2}+\overrightarrow{\mathrm{E}}_{3} \cdot \Delta \overrightarrow{\mathrm{A}}_{3} \ldots . \overrightarrow{\mathrm{E}}_{\mathrm{n}} \cdot \Delta \overrightarrow{\mathrm{A}_{\mathrm{n}}} \\
=\sum_{\mathrm{i}=1}^{\mathrm{n}} \overrightarrow{\mathrm{E}}_{\mathrm{i}} \cdot \Delta \overrightarrow{\mathrm{A}_{\mathrm{i}}}
\end{array}\)

3. By taking the limit \(\Delta \overrightarrow{\mathrm{A}}_{i} \rightarrow 0\) (for all i) the summation in equation becomes integration. The total electric flux for the entire area is given by \(\Phi_{\mathrm{E}}=\int \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}\)

Electric flux for non-uniform electric Field:

1. The electric flux for a given surface depends on both the electric field pattern on the surface area and orientation of the surface with respect to the electric field.

Question 6.
The electric field is zero everywhere inside the conductor. This is true regardless of whether the conductor is solid or hollow. Justify.
Answer:

• This is an experimental fact. Suppose the electric field is not zero inside the metal, then there will be a force on the mobile charge carriers due to this electric field.
•  As a result, there will be a net motion of the mobile charges, which contradicts the conductors being in electrostatic equilibrium. Thus the electric field is zero everywhere inside the conductor.
• We can also understand this fact by applying an external uniform electric field on the conductor.

Question 7.
The electrostatic potential has the same value on the surface and inside the conductor. Justify.
Answer:

• We know that the conductor has no parallel electric component on the surface without doing any work.
• This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface.
• Since the electric field inside the conductor is zero inside the conductor, the potential is same as the surface of the conductor.
• Thus at electrostatic equilibrium, the conductor is always at equipotential.

Question 8.
Explain Faraday Cage’s experiment? Faraday cage is an instrument to demonstrate the effect of electrostatic shielding.
Answer:

• It is made up of metal bars configured in the instrument.
•  If an artificial lightning bolt is created outside the person inside is not affected.
• This is because the metal bar provides electrostatic shielding.
• The Electric field inside becomes zero.
• The charges flow through the metal bar to the ground with no effect on the person inside.

Question 9.
What do you understand from the expression of capacitance in a parallel plate capacitor?
Answer:
Capacitance is directly proportional to the area of cross-section and is inversely proportional to the distance between the plates.
1. If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.

2. If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with E constant. As a result, the voltage difference between the terminals of the battery increases which in turn leads to an additional flow of charge to the plates from the battery, till the voltage on the capacitor equals to the battery’s terminal voltage. Suppose the distance is increased, the capacitor voltage increases and becomes greater than the battery voltage. Then, the charges flow from capacitor plates to the battery till both voltages become equal.

Question 10.
Mention the applications of capacitors?
Answer:

1. The flash which comes from the digital camera when we take photographs is due to the energy released from the capacitor, called a flash capacitor.
2. During cardiac arrest, a device called heart defibrillator is used to give a sudden surge of a large amount of electrical energy to the patient’s chest to retrieve the normal heart function.
3. Capacitors are used in the ignition system of automobile engines to eliminate sparking.
4. Capacitors are used to reduce power fluctuations in power supplies and to increase the efficiency of power transmission.

XIII – Five mark questions:

Question 1.
Explain the historical background of electric charges?
Answer:

• Two millenniums ago, Greeks noticed that amber after rubbing with animal fur attracted small pieces of leaves and dust.
• The amber possessing this property is said to be charged.
• A glass rod rubbed with a silk cloth, attracts a piece of paper. So glass rod also becomes ‘charged’ when rubbed with a suitable material.
• Consider a charged rubbed rod hanging from a thread.
• Suppose another charged rubber rod is brought near the first rubber rod; the rods repel each other.
• Now if we bring a charged glass rod close to the charged rubber rod, they attract each other.
• At the same time, if a charged glass rod is brought near another charged glass rod both the rods repel each other.

Inferences:
i) The charging of the rubber rod and that of the glass rod are different from one another.
ii) The charged rubber rod repels another charged rubber rod, which implies that Tike charges repel each other’.
iii) The charged amber rod attracts the charged glass rod, implying that the charge in the glass rod is not the same kind of charge present in the rubber. Thus unlike charges attract each other.
Therefore, two kinds of charges exist in the universe. Benjamin Franklin called one type of charge as positive (+) and another type of charge as negative (-). Based on Franklin’s convention, rubber and amber rods are negatively charged while the glass rod is positively charged. If the net charge is zero in the object, it is said to be electrically neutral.

iv) The atom is electrically neutral and is made up of negatively charged electrons, positively charged protons, and neutrons which have zero charges. When an object is rubbed with another object, some amount of charge is transferred from one object to another due to the friction between them, and the object is then said to be ‘electrically charged’. Charging the object through rubbing is called ‘triboelectric charging’.

Question 2.
Derive an expression for electric field due to a system of point charges.
Answer:
Suppose a number of point charges are distributed in space, to find the electric field at some point P due to this collection of point charges, superposition principle is used.

The electric field due to a collection of point charges at an arbitrary point is simply equal to the vector sum of the electric fields created by the individual point charges. This is called the superposition of electric fields.

Question 3.
Discuss the electric flux of a uniform electric field?
Answer:
Electric flux for uniform Electric field:

1. Consider a uniform electric field in a region of space. Let us choose an area A normal to the electric field lines.
Φ_E = EA ……….(1)

2. Suppose the same area A is kept parallel to the uniform electric field, then no electric field lines pass through area A. The electric flux for this case is zero.
Φ_E > E = 0 ………..(2)

3. If the area is inclined at an angle e with the field, then the component of the electric field perpendicular to the area alone contributes to the electric flux.

4. The electric field component parallel to the surface area will not contribute to the electric flux. For this case, the electric flux  Φ_E = (E cosθ) A …………..(3)

5. Further, e is also the angle between the electric field and the direction normal to the area, Hence in general, for a uniform electric field, the electric flux is defined as
Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\) = EA cosθ ………………(4)

6. \(\overrightarrow{\mathrm{A}}\) is the area vector \(\overrightarrow{\mathrm{A}}\) = A n̂. Its magnitude is simply the area A and the direction are along the unit vector n̂ perpendicular to the area.

7. Using this definition for flux,
Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\)
equations (2) and (3) can be obtained as special cases.
In Figure (a), θ=0° Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\) = EA
In Figure (b), θ = 90° Φ_E = \(\overrightarrow{\mathrm{E}}\) . \(\overrightarrow{\mathrm{A}}\) = 0
Here \(\overrightarrow{\mathrm{A}}\) = An̂

Question 4.
Discuss some salient points about Gauss law?
Answer:
1. The total electric flux through the closed surface depends on the charges enclosed by the surface and the charges present outside the surface will not contribute to the flux and the shape of the closed surface which can be chosen arbitrarily.

2. The total electric flux is independent of the location of the charges inside the closed surface.

3. Chosen imaginary surface is called a Gaussian surface. The shape of the Gaussian surface to be chosen depends on the type of charge configuration and the kind of symmetry existing in that charge configuration.

4. The electric field E is due to charges present inside and outside the Gaussian surface but the charge Qencl denotes the charges which lie only inside the Gaussian surface.

5. The Gaussian surface cannot pass through any discrete charge but it can pass through continuous charge distributions.

6. Gauss law is another form of Coulomb’s law and it is also applicable to the charges in motion. Because of this reason, Gauss law is treated as much more general law than Coulomb’s law.

Question 5.
Derive an expression for electric field due to two parallel charged infinite sheet.
Answer:

1. Consider two infinitely large charged plane sheets with equal and opposite charge densities +σ and -σ which are placed parallel to each other.
2. The electric field between the plate and outside the plates is found using Gauss law.
3. The magnitude of the electric field due to an infinite charged plane sheet is \(\frac{\sigma}{2 \varepsilon}\) and it points perpendicularly outward if σ>0 and points inward if σ<0.
   
4. At the points P₁ and P₂, the electric field due to both plates are equal in magnitude and opposite in direction. As a result, electric field at a point outside the plate is zero. But inside the plate, electric fields are in the same direction
5. The direction of the electric field inside the plates is directed from positively charged plate to the negatively charged plate and is uniform everywhere inside the plate.
   \(\mathrm{E}_{\text {inside }}=\frac{\sigma}{2 \varepsilon_{0}}+\frac{\sigma}{2 \varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}}\)

Question 6.
Explain the principle of a capacitor?
Answer:

1. A capacitor is a device used to store electric charge and electric energy.
2. It consists of two conducting objects separated by some distance.
3. A simple capacitor consists of two parallel metal plates separated by a small distance.
4. When a capacitor is connected to a battery of potential difference V, the electrons are transferred from the battery to one plate and the other plate to the battery so that one plate becomes negatively charged with a charge of – Q and the other plate becomes positively charged with + Q.
5. The potential difference between the plates is equivalent to the battery’s terminal voltage. If the battery voltage is increased, the amount of charges stored in the plates also increase.
6. In general, the charge stored in the capacitor is proportional to the potential difference between the plates.
   Q α V, So that Q = CV, where C is the proportionality constant called capacitance.
   
7. The capacitance of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between conductors.
8. Unit:- Farad or Coulomb per volt. Farad is a very large unit of capacitance.
9. In Practice, we use micro & picofarad.
10. Total charge stored in the capacitor is zero when we say the capacitor stores charges, it means that amount of charge that can be stores charge, it means that amount of charge that can be stored in any one of the plates.
11. Nowadays it is available in different shapes and types. It is used in various kinds of electronic circuits.

Question 7.
Write the properties of lines of force:
Answer:

1. It starts from positive charge and ends at negative charge .
2. For a positive charge, the electric field lines point radially outward and for a negative charge, electric field lines point radially inward.
3. Tangent drawn at any point gives the direction of the electric field at the point.
4. The electric field lines are denser in a region where the electric field has a larger magnitude and less dense in a region of smaller magnitude.
5. They do not intersect
6. Field lines emanating from the positive charge are proportional to the magnitude of the charge.

XlV. Additional Problems (Two Marks):

Question 1.
An ele1tric dipole of charges 2 × l0^-10C separated by a distance 5 mm, is placed at an angle of 60 to a uniform field of 10 Vm^-1. Find the
i) magnitude and direction of the force acting on each charge and
ii) torque exerted by the field.
Data:
+q = 2 × 10^-10 C
-q = – 2 × 10^-10 C
θ = 60°
E = 10Vm^-1
2d = 5mm = 5 × 10^-3 m
i) Magnitude and directions of the force acts on each charge =?
ii) Torque exerted by the field =?
Solution:

1. Positive charge experience a force in the direction of the field.
2. Negative charge experiences a force opposite to the direction of the field.
   i) To find the force on each charge:
3. Force acts on positive charge: F = qE
   F = 2 × 10^-10 × 10
   F = 2 × l0^-9 × N in the direction of E
   Force acts on negative charge:
   F = qE = 2 × 10^-10 × 10
   F= 2 × l0^-9 N in the direction of E

ii) To find Torque exerted bj the field:

Question 2.
An electric dipole of charges 2 × 10^-6 C, – 2 × 10^-6 C is separated by a distance of 1 cm. Calculate the electric field due to dipole at a points
i) axial line 1 m from its centre and
ii) equatorial line 1 m from its centre.
Answer:


i) Electric field at a point on axial line = 360NC^-1
ii) Electric field at a point on equatorial line = 180NC^-1

Question 3.
Two charges +q and -3q are separated by a distance of lm. At what point in between the charges on its axis is the potential zero?
Answer:
Data :
q₁ = +q; q₂ = -3q; r = 1m
Solution:
Let the potential be zero at point 0 at a d . 1 from +q charge.


The potential is zero at a distance of 0.25m from the charge +q.

Question 4.
A parallel plate capacitor is maintained at some potential difference. A 3 mm thick slab is introduced between the plates. To maintain the plates at the same potential difference, the distance between the plates is increased by 2.4mm. Find the dielectric constant of the slab.
Data:
t = 3 mm; d’- d = 2.4 mm
V is same;
ε_r = ?
Solution:
Capacitance of air-filled parallel plate capacitor:

Since the potential are made the same capacitance in the above two cases is also the same.

The dielectric constant of the slab ε_r = 5

Question 5.
Three capacitors each of capacitance 9 pF are connected in series.
i) What is the total capacitance of the combination?
ii) What is the potential difference across each capacitor if the combination is connected to 120 V supply?
Answer:
Data:
C₁ = C₂ = C₃ = 9pF
V = 120V
C_s = ?
V₁, V₂, V₃ =?
To find effective capacitance:
Three capacitors are in series

∵ \(\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\)
C_s = 3pF
C_s = 3 × 10^-12 F

V₁, V₂, V₃ are the potential differences across the three capacitors as shown in the figure.

i) C_s = 3 × 10^-12 F
ii) V₁, V₂, V₃ = 40 V

XV. Five Mark Problems:

Question 1.
Four-point charges +q, +q, and -q is to be arranged respectively at the four corners of a square PQRS of side r. Find the work needed to assemble this arrangement.
Solution:

Here PQ = QS = SR= RP = r
\(\mathrm{PS}^{\prime}=\mathrm{RQ}=\mathrm{r} \sqrt{2}\)
W = Potential Energy of the system of four charges.
\(=\frac{(-q)(+q)}{4 \pi \varepsilon_{o} r}+\frac{(+q)(+q)}{4 \pi \varepsilon_{o} r}+\frac{(+q)(-q)}{4 \pi \varepsilon_{o} r}+\frac{(-q)(-q)}{4 \pi \varepsilon_{o} r}\)

Question 2.
Two small charged spheres repel each other with a force of 2 × 10^-3 N. The charge on one sphere is twice that of the other. When one of the charges is moved 10 cm away from the other, the force is 5 × 10^-4 N. Calculate the charges and the initial distance between them.
Answer:
Data:
F₁= 2 × 10^-3N when distance = r
F₂ = 5 × 10^-4N
when distance (i.e.) r’= (r+ 0. 1)rn
r = r + 10cm
q₁ = ?, q₂ = ?, r =?
Solution:
According to Coulomb’s law
\(\mathrm{F}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \cdot \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{2}}\)
Let q₁ = q and q₂ = 2q and distance between the charges be r experience a repulsive force F₁.


We know,
q₁ = q = 33.3 × 10^-9 C
q₂ = 2q = 2 × 33.3 × 10^-9 C
q₂ =66.6 × 10^-9 C
Initial distance r = 0.1m
Charges:
q₁ = 33.3 × 10^-9 C
q₂ = 66.6 × 10^-9 C

Question 3.
Two capacitors of unknown capacitance are connected in series and parallel. If net capacitances in two combinations are 6µF and 25µF respectively. Find their capacitances.
Answer:
Given:
Cs = 6µF
Cp = 25µF
\(\frac{1}{C_{S}}=\frac{1}{C_{1}}+\frac{1}{C_{2}} \quad \cdots \cdots\)

Question 4.
Three charges + 1 µC, + 3 µC, and – 5 µC are kept at the vertices of an equilateral triangle of sides 60 cm. Find the electrostatic potential energy of the system of charges.
Answer:


The electrostatic potential energy of the system of charges = – 0.255 J

Question 5.
Calculate the force between electron and proton in a Hydrogen atom.
(e = 1.6 × 10^-9 and r₀ = 0.53Å)
Solution:
The electron and proton attract each other. The force between these two particles is given by

Question 6.
Four point charges are placed at the four corners of a square in two ways (a) and (b) as shown in figure. Will the (i) electric potential and (ii) electric field, at the centre of the square be the same or different in the two configurations, and why?

Answer:
a) Electric potential is a scalar quantity. 0pposite charges cancel each other So, in both the configurations electric potential will be zero and the same.
b) Electric field is a vector quantity. It depends on positions of charge.
In the first diagram Electricfield,
\(E=2 \sqrt{2} \frac{\mathrm{kq}}{(\mathrm{r} / \sqrt{2})^{2}}\)
In the second diagram, E= 0
Because opposite charges are placed at the corner.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Accountancy Guide Pdf Chapter 4 Goodwill in Partnership Accounts Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 4 Goodwill in Partnership Accounts

12th Accountancy Guide Goodwill in Partnership Accounts Text Book Back Questions and Answers

I Multiple Choice Questions

Choose the correct answer

Question 1.
Which of the following statement is true?
(a) Goodwill is an intangible asset
(b) Goodwill is a current asset
(c) Goodwill is a fictitious asset
(d) Goodwill cannot be acquired
Answer:
(a) Goodwill is an intangible asset

Question 2.
Super profit is the difference between
(a) Capital employed and average profit
(b) Assets and liabilities
(c) Average profit and normal profit
(d) Current year’s profit and average profit
Answer:
(c) Average profit and normal profit

Question 3.
The average rate of return of similar concerns is considered as
(a) Average profit
(b) Normal rate of return
(c) Expected rate of return
(d) None of these
Answer:
(b) Normal rate of return

Question 4.
Which of the following is true?
(a) Super profit = Total profit / number of years
(b) Super profit = Weighted profit / number of years .
(c) Super profit = Average profit – Normal profit
(d) Super profit = Average profit × Years of purchase
Answer:
(c) Super profit = Average profit – Normal profit

Question 5.
Identify the incorrecte pair
(a) Good will under average profit method – Average profit × Number of years of purchase
(b) Goodwill under Super profit method – Super profit × Number of years of purchase
(c) Goodwill under Annuity method – Average Profit × Present value annuity factor
(d) Goodwill under Weighted average profit method – Weighted average profit x Number of years of purchase
Answer:
(c) Goodwill under Annuity method – Average Profit x Present value annuity factor

Question 6.
When the average profit is ₹ 25,000 and the normal profit ₹ 15,000 super profit is
(a) ₹ 25,000
(b) ₹ 5,000
(c) ₹ 10,000
(d) ₹ 15,000

Hints: Find out Super Profit
Super Profit = Average Profit – Normal Profit
= 25,000- 15,000
= ₹ 10,000
Answer:
(c) ₹ 10,000

Question 7.
Book profit of 2017 is ₹ 35,000; non -recurring income included in the profit is ₹ 1,000 and abnormal loss charged in the year 2017 was ₹ 2,000, then the adjusted profit is
(a) ₹ 36,000
(b) ₹ 35,000
(c) ₹ 38,000
(d) ₹ 34,000
Hints:

Answer:
(a) ₹ 36,000

Question 8.
The total capitalised value of a business is ₹ 1,00,000; asset are ₹ 1,50,000 and liabilities ₹ 80,000. the value of goodwill as per the capitalisation method will be
(a) ₹ 40,000
(b) ₹ 70,000
(c) ₹ 1,00,000
(d) ₹ 30,000
Hints:
Goodwill = Total capitalized value of the business – Actual capital employed
i) Capital employed = Fixed Assets + Current assets – Current Liabilities
= 1,50,000-80,000
= ₹ 70,000
ii) Capitalized Value = ₹ 1,00,000
= 1,00,000 – 70,000
= ₹ 30,000
Answer:
(d) ₹ 30,000

II. Very Short Answer Questions

Question 1.
What is goodwill?
Answer:
Goodwill is the good name or reputation of the business which brings benefit to the business. It enables the business to earn more profit. It is the present value of a firm’s future excess earnings.

Question 2.
What has acquired goodwill?
Answer:
Goodwill acquired by making payment in cash or kind is called acquired or purchased goodwill. When a firm purchases an existing business, the price paid of the purchase of such a business may exceed the net assets (Assets – Liabilities) of the business acquired. The excess of purchase consideration over the value of net assets acquired is treated as acquired goodwill.

Question 3.
What is super profit?
Answer:
Super profit is the excess of average profit over the normal profit of a business.
Super Profit = Average Profit – Normal Profit

Question 4.
What is normal rate of return?
Answer
:It is the rate at which profit is earned by similar business entities in the industry under normal circumstances.

Question 5.
State any two circumstances under which goodwill of a partnership firm is valued
Answer:

1. When there is a change in the profit-sharing ratio
2. When a new partner is admitted into a firm
3. When an existing partner retires from the firm or when a partner dies.
4. When a partnership firm is dissolved.

III Short Answer Questions

Question 1.
State any six-factor determining goodwill.
Answer:

1. Profitability of the firm.
2. Favourable location of the business enterprise.
3. Good quality of goods or services offered.
4. Tenure of the business enterprise.
5. Efficiency of management.
6. Degree of competition.
7. Other factors.

Question 2.
How is goodwill calculated under the super profits method?
Answer:
Under these methods, super profit is the base for calculation of the value of goodwill. Super profit is the excess of average profit over the normal profit of a business.
Super profit = Average profit – Normal profit

Average profit is calculated by dividing the total of adjusted actual profits of certain number of year by the total number of such years. Normal profit is the profit earned by the similar business firms under normal conditions.
Normal profit = Capital employed × Normal rate of return
Capital employed = Fixed assets + Currents assets – Current liabilities.

Question 3.
How is the value of goodwill calculated under the capitalisation method?
Answer:
Under these methods, super profit is the base for calculation of the value of goodwill. Super profit is the excess of average profit over the normal profit of a business.
Goodwill = Total capitalised value of the business – Actual capital employed
The total capitalised value of the business is calculation by capitalising the average profit on the basis of the normal rate of return.
Capitalised value of the business = \(\frac{\text { Average profit }}{\text { Normal rate of return }} \times 100\)
Actual capital employed = Fixed assets (excluding goodwill) + Current assets – Current liabilities

Question 4.
Computer average profit from the following information.
2016: ₹ 8,000; 2017: ₹ 10,000; 2018: ₹ 9,000
Solution:

Question 5.
Calculate the value of goodwill at 2 Years purchase of average profit when average profit is? 15,000.
Solution:
Goodwill = Average profit x No. of years of purchase
Goodwill = 15,000 x 2 Years
= Rs. 30,000

IV Exercises

Simple average profit method

Question 1.
The following are the profits of a firm in the last five years:
2014 : ₹ 10,000; 2015 : ₹ 11,000;
2016 :  ₹ 12,000; 2017 : ₹ 13,000; and 2018 : ₹ 14,000
Calculate the value of goodwill at 2 years purchase of average profit of five years.
Solution:
G/W = Average profit x No.of years of purchase

Answer :
Average profit: ₹ 12,000, Goodwill: ₹ 24,000

Question 2.
From, following information, calculate the value of goodwillOn the basis of 3 years purchase of average profits of last four years.

Solution:
G/W = Average profit x No.of years of purchase

Gs/W = 4,000 × 3 years = Rs. 12,000
Answer:
Average profit: ₹ 4,000; Goodwill: ₹ 12,000

Question 3.
From the following information relating to a partnership firm, find out the value of its goodwill based on 3 years purchas of average profits of the last 4 years.
(a) Profit of the years 2015, 2016, 2017 and 2018 are ₹ 10,000, ₹ 12,500, ₹ 12,000 and ₹ 11,500 respectively.
(b) The business was looked after by a partner and his fair remuneration amounts to ₹ 1,500, per year. This amount was not considered in the calculation of the above profits.
Solution:
G/W = Average profit = Average profit x No.of years of purchase

G/W Average profit × No.of years of purchase
= 10,000 × 3yrs
G/W = Rs. 30,000
Answer:
Average profit: ₹ 10,000; Goodwill: ₹ 30,000

Question 4.
From the following information relating to Sridevi enterprises, cal te the value of goodwill on the basis of 4 years pruchase of the average profits of 3 years.
(a) Profits for the years ending 31st December 2016, 2017 and 2018 were ₹ 1,75,000, ₹ 1,50,0 and ₹ 2,00,000 respectively.
(b) A non-recurring-income of ₹ 45,000 is include in the profits of the year 2016.
(c) The closing stock of the year 2017 was overvalued by ₹ 30,000.
Solution:
Calculation of Adjusted profit

Note : Overvaluationof cl. Stock in 2017 will results in over valuation of opening stock in 2018
Average Profit.
Average Profit. = \(\frac{1,30,000+1,20,000+2,30,00}{3}\)
Average Profit = \(\frac{4,80,000}{3}\)
= Rs. 1,60,000
= Average profit x No.of years of purchase
= 1,60,000 x 4 yrs
= Rs. 6,40,000
Answer :
Average profit: ₹ 1,60,000; Goodwill: ₹ 6,40,000

Question 5.
The following particulars are available in respect of the business carried on by a partnership firm:
(i) Profit earned : 2016 ₹ 25,000; 2017: ₹ 23,000 and 2018: ₹26,000.
(ii) Profit of 2016 includes a non-recurring income of ₹ 2,500.
(iii) Profit of 2017 is reduced by ₹ 3,500 due to stock destroyed by fire.
(iv) The stock was not insured. But, it is decided to insure the stock in future. The insurance premium is estimated to be ₹ 250 per annum.
you are required to calculate the value of goodwill of the firm on the basis of 2 years purchase of average profits of the last three years.
Solution:
Calculation of Adjusted profit

G/W = Average profit × No.of years of purchase
G/W = 24,750 × 2 yrs
= Rs. 49,500
Answer:
Average profit: ₹ 24,750; Goodwill: ₹ 49,500

Weighted average profit method.

Question 6.
Find out the value of goodwill at three years pruchase of weighted average profit of last four years.

Solution:
Calculation of Weighted Average profit

Weighted profts = \(\frac{\text { Weighted profts }}{\text { Total of weights }}\)
Av.pft = \(\frac{1,54,000}{10}\) = Rs. 15,400
G/w = Weighted Average profit × No. of yrs of purchase
G/w = 15,400 × 3yrs
= Rs. 46,200
Answer:
Average profit: ₹15,400; Goodwill: ₹ 46,200

Purchase of super profit method

Question 7.
From the following details, calculate the value o oodwill at 2 years purchase of super profit:
(a) Total assets of a firm are ₹ 5,00,000;
(b) The liabilities of the firm are ₹ 2,00,000
(c) Normal rate of return in this class of business is 12.5%.
(d) Average profit of the firm is ₹ 60,000.
Solution:
G/W = Super Profit x No.of years of purchase

Cap. Emp = Total Assets – Total liabilities
= 5,00,000 – 2,00,000
= Rs. 3,00,000
Normal Profit = \(\frac{3,00,000 \times 12.5}{100}\)
Rs. 37,500
Average Profit = Rs. 60,00
Super profit = Average Profit – Normal profit
= 60,000 – 37,500
= Rs. 22,500
G/w = 22,500 x 2 yrs
= Rs. 45,000
Answer:
Super profit: ₹ 22,500; Goodwill: ₹ 45,000

Question 8.
A partnership firm earned net profits during the last three years as follows:
2016 ₹ 20,000; 2017: ₹ 17,000 and 2018: ₹ 23,000.
The capital investement of the firm throghout the above mentioned period has been ₹ 80,000. Having regard to the risk involved, 15% is considered to the a fair return on capital employed in the business. Calculate the value of goodwill on the basis of 2 years purchase of super profit.
Solution:
G/W = Super Profit x No.of years of purchase

Super profit = Average Profit – Normal profit
= 20,000 – 12,000
= Rs. 8,000
G/w = 8,000 x 2 yrs
= Rs. 16, 000
Answer :
Super profit: ₹ 8,000; Goodwill: ₹ 16,000

Annuity Method:

Question 9.
From the following information, calculate the value of goodwill under annuity method:

Present value of ₹ 1 for 5 years at 15% per annum as per the annuity table is 3.352
Solution:
G/W = Super Profit x Value of A nnuity
SuPer profit = Average Profit – Normal Profit
SuPer profit = 14,000- 4,000 = Rs. 10,000
G/w = 10.0 x 3.352 = Rs. 33,520
Answer :
Super profit: ₹ 10,000; Goodwill: ₹ 33,520

Capitalisation of super profit method:

Question 10.
Find out the value of goodwill by caitalising super profits:
(a) Normal Rate of Return 10%
(b) Profits for the last four years are ₹ 30,000, ₹ 40,000, ₹ 50,000, ₹ 45,000,
(c) A non-recurring income of ₹ 3,000 is included in the above mentioned profit of ₹ 30,000,
(d) Average capital employed is ₹ 3,00,000.
Solution:

= Rs. 30,000
Super profit – Average Profit – Normal Profit
Super Profit – 40,500 – 30,000
= Rs. 10,500

Answer :
Super Profit: ₹ 10,500; Good will: ₹ 1,05,000

Question 11.
From the following information, find out the value of goodwill by capitalisatio method:
(i) Average profit ₹ 20,000
(ii) Normal rate of return 10%
(iii) Tangible assets of the firm ₹ 2,20,000
(iv) Liabilities of the firm ₹ 70,000.
Solution:
G/W = Total capitalaised Value of the business – Actual cap.Employee Average Profit

Actual cap employed = Tangible Assets – Liabilities of the firm.
= 2,20,000 – 70,000
= Rs. 1,50,000
G/w = 2,00,000 – 1,50,000
= Rs. 50,000
Answer:
Total Capitalised value: ₹ 2,00,000; Net tangible assets: ₹ 1,50,000; Goodwill: ₹ 50,000

12th Accountancy Guide Goodwill in Partnership Accounts Additional Important Questions and Answers

Question 1.
The excess of average profit over normal profit is
(a) Super profit
(b) Goodwill
(c) Undistributed profit
Answer:
(a) Super profit

Question 2.
The goodwill that can be recoreded in the books of accounts and is shown on the assets side of the B/S is
(a) Acquired goodwill
(b) Self-Generated goodwil
(c) None of these two
Answer:
(a) Acquired goodwill

Question 3.
How many methods are there for valuing G/W?
(a) 3
(b) 2
(c) 5
Answer:
(a) 3

III Short Answer Questions

Question 4.
Describe the nature of G/W.
Answer:
The nature of goodwill can be described as follows:

• Goodwill is an intangible fixed asset. It is intangible because it has no physical existence. It cannot be seen or touched.
• It has a definite value depending on the profitability of the business enterprise.
• It cannot be separated from the business.
• It help in earning more profit and attracts more customers.
• It can be purchased or sold only when the business is purchased or sold in full or in part.

Question 5.
Explain any 5 factors that determine the value of G/W
Answer:
Factors determining the value of goodwill of a partnership firm
(i) Profitability of the firm
The profit earning capacity of the firm determines the value of its goodwill. A firm earning higher profits and having potential to generate higher profits in future will have higher value of goodwill.

(ii) Favourable location of the business enterprise
If the firm is located in a prominent place which is easily accessible to the customers, it can attract more customers. Its sales and profit will be higher when compared to a firm which is not located in a prominent place. Hence, it will ahve high value of goodwill.

(iii) Good equality of goods or services offered
If a firm is located in a prominent place which is easily accessible to the customers, it can attract more customers. Its sales and profit will be higher when compared to a firm which is not located in a prominent place. Hence, it will have high value of goodwill.

(iv) Tenure of the business enterprise
A firm which has caried on business for several years will have higher reputation among its customers as it is better known to the customers. Such a firm will have higher earnings and higher value of goodwill when compared to a new firm.

(v) Efficiency of management
A firm having efficient management will earn more profitsa and the value of its goodwill will be higher compared to a firm with less efficient managerial personnel.
(iii) It cannot be separated from the business.

Question 6.
What do you mean by self generated G/W?
Answer:
It is the goodwill which is self generated by a firm based on features of the business such as favourable location, loya customers, etc. Such self-generated goodwill cannot be recorded in the books of accounts.

Question 7.
How do you calculated Adjusted Profit?
Answer:
Adjusted profit = Actual profit
+ Past expanse not required in the future.
– Past revenues not likely to be earned in the future
+ Additional income expected in the future
– Additional expenses expected to be paid in the future

Question 8.
How do you calculate G/W under weighted average profit method?
Answer:
Weighted average profit method
Under this method, goodwill is calculated by mulitplying the weighted average profit by a certain number of years of purchase.
Goodwill = Weighted average profit × Number of years of purchase

In this method, weights are assigned to each year’s profit. Weighted profit is ascertained by multiplying the weights assigned with the respective years profit. The sum of the wieghted profits is divided by the sum of weights assigned to determine the weighted averagge profit.

Weighted average profit = \(\frac{\text { Total of weighted profits }}{\text { Total of weights }}\)

Question 9.
What do you mean by Annuity Method of valuing G/w?
Answer:
Under this method, value of goodwill is calculated by multiplying the super profit with the present value of annuity.
Goodwill = Super profit x Present value annuity factor
Present value annuity factor is the present value of annuity of rupee one at a given time. It can be found out from annuity table or by using formula.
Annuity factor = \(\frac{\mathrm{i}(1+\mathrm{i})^{\mathrm{n}}}{\mathrm{i}(1+\mathrm{i})^{\mathrm{n}-1}}\)
Where i = interest rate
n = estimated number of years.

Question 10.
Write the formula for calcultaing G/w under capitalisation of Super profit method.
Answer: