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Students can download 5th Maths Term 1 Chapter 2 Numbers Ex 2.3 Questions and Answers, Notes, Samacheer Kalvi 5th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 5th Maths Solutions Term 1 Chapter 2 Numbers Ex 2.3

Question 1.
Read the following numbers by placing the commas at appropriate periods and write their number names.
a. 15731997
b. 341964
c. 29121972
d. 347810
Answer:
a. 15731997
15731997 = 1,57,31,997
Number Name: One crore fifty seven lakhs thirty one thousands nine hundred and ninety seven.

b. 341964
341964 = 3,41,964
Number Name: Three lakhs forty one thousands nine hundred and sixty four.

c. 21921972
21921972 = 2,19,21,972
Number Name; Two crores nineteen lakhs twenty one thousands nine hundred and seventy two.

d. 347810
347810 = 3,47,810
Number Name: Three lakhs forty seven thousands eight hundred and ten.

Question 2.
Write the place value of 5 in the following numbers.
a. 287500
b. 586012
c. 5869732
d. 5467859
Answer:
a. 287500
The place value of 5 is 5 × 100 = 500

b. 586012
The place value of 5 is 5 × 1,00,000 = 5,00,000

c. 5869732
The place value of 5 is 5 × 10,00,000 = 50,00,000

d. 5467859
The place value of first 5 is 5 × 10 = 50
The place value of Second 5 is 5 × 10,00,000 = 50 00,000

Question 3.
Write the following in standard notation.
a. 30000 + 3000 + 300 + 30 + 3
b. 200000 + 7000 + 7
c. 8000000 + 70000 + 3000 + 30 + 5
d. 4000000 + 400 + 4
Answer:
a. 30000 + 3000 + 300 + 30 + 3
= 33333 = 33,333

b. 200000 + 7000 + 7
= 207007 = 2,07,007

c. 8000000 + 70000 + 3000 + 30 + 5
= 8073035 = 80,73,035

d. 4000000 + 400 + 4
= 4000404 = 40,00,404

Question 4.
Write the following numbers in expanded form.
a. 63,570
b. 36,01,478
c. 1,45,70,004
d. 28,48,387
Answer:
a. 63,570
= 60000 + 3000 + 500 + 70 + 0

b. 36,01,478
= 3000000 + 600000 + 1000 + 400 + 70 + 8

c. 1,45,70,004
= 10000000 + 4000000 + 500000 + 70000 + 4

d. 28,48,387
= 2000000 + 800000 + 40000 + 8000 + 300 + 80 + 7

Students can Download Tamil Nadu 11th Physics Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 4 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

PART – 1

Answer all the questions: [15 × 1 = 15]

Question 1.
In some region, the gravitational field is zero. The gravitational potential in this region is …………………….
(a) A variable
(b) A constant
(c) Zero
(d) Can’t be zero
Answer:
(b) A constant

Question 2.
The angle between two vectors 2\(\hat { i } \) + 3\(\hat { j } \) + \(\hat { k } \) and -3\(\hat { j } \) + 6k is …………………..
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Hint:
The angle between the two vector is 90°.
[cos θ = \(\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\mathrm{AB}}\)]
Answer:
(d) 90°

Question 3.
A stretched rubber has
(a) Increased kinetic energy
(b) Increased potential energy
(c) Decreased kinetic energy
(d) The axis of rotation
Answer:
(b) Increased potential energy

Question 4.
The direction of the angular velocity vector is along ……………………..
(a) The tangent to the circular path
(b) The inward radius
(c) The outward radius
(d) The axis of rotation
Answer:
(d) The axis of rotation

Question 5.
For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is …………………….
(a) \(\frac{1}{2}\)
(b) \(\frac { 1 }{ \sqrt { 2 } } \)
(c) 2
(d) \(\sqrt{2}\)
Hint:

Answer:
(a) \(\frac{1}{2}\)

Question 6.
If the linear momentum of the object is increased by 0.1 %, then the kinetic energy is increased by ………………………..
(a) 0.1%
(b) 0.2%
(c) 0.4%
(d) 0.01%
Hint:
The relation b/w linear momentum and kinetic energy
P = \(\sqrt{2 \mathrm{mE}_{\mathrm{K}}} \Rightarrow \mathrm{E}_{\mathrm{K}}=\frac{\mathrm{P}^{2}}{2 \mathrm{m}}\)
If L is increased by 0.1% 1′ = P + \(\frac{0.1}{100}\)P

Answer:
(c) 0.4%

Question 7.
A block of mass 4 kg is suspended through two light spring balances A and B. Then A and B will read respectively ………………………
(a) 4 kg and 0 kg
(b) 0 kg and 4 kg
(c) 4 kg and 4 kg
(d) 2kg and 2 kg
Hint:
Tension is uniformly transmitted if the springs are massless.
Answer:
(c) 4 kg and 4 kg

Question 8.
At the same temperature, the mean kinetic energies of molecules of hydrogen and oxygen are in the ratio of ………………………
(a) 1 : 1
(b) 1 : 16
(c) 8 : 1
(d) 16 : 1
Hint:
Average kinetic energy of a molecule is proportional to the absolute temperature.
Answer:
(a) 1 : 1

Question 9.
A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. Which of the following plots shows the correct variation of speed v with height h from the lower end?
(a)
(b)
(c)
(d)
Answer:

(d)

Question 10.
In a simple hormonic oscillation, the acceleration against displacement for one complete oscillation will be ………………………
(a) An ellipse
(b) A circle
(c) A parabola
(d) A straight line
Answer:
(d) A straight line

Question 11.
A shell is fired from a canon with velocity v m/s at an angle 0 with the horizontal direction. At the highest point in the path it explodes into two pieces of equal mass. One of the pieces retraces it path of the cannon and the speed in m/s of the other piece immediately after the. explosion is ………………………..
(a) 3v cos θ
(b) 2v cos θ
(c) \(\frac{3}{2}\)v cos θ
(d) \(\frac { \sqrt { 3 } }{ 2 } \) cos θ
Hint:
Velocity at the highest point = horizontal component of velocity = V cos θ
Momentum of shell before explosion = mv cos θ
Momentum of two pieces after explosion = \(\frac{m}{2}\) (- V cos θ) + \(\frac{m}{2}\) v
Law of conservation of momentum mv cos θ = –\(\frac{mv}{2}\)cos θ + \(\frac{m}{2}\) v
∴ V = 3V cos θ
Answer:
(a) 3v cos θ

Question 12.
In which process, the p – v indicator diagram is a straight line parallel to volume axis?
(a) Isothermal
(b) Adiabatic
(c) Isobaric
(d) Irreversible
Answer:
(c) Isobaric

Question 13.
For a liquid to rise in capillary tube, the angle of contact should be ………………………….
(a) Acute
(b) Obtuse
(c) Right
(d) None of these
Answer:
(a) Acute

Question 14.
The increase in internal energy of a system is equal to the work done on the system which process does the system undergo?
(a) Isochoric
(b) Adiabatic
(c) Isobaric
(d) Isothermal
Answer:
(b) Adiabatic

Question 15.
The change in frequency due to Doppler effect does not depend on …………………………
(a) The speed of the source
(b) The speed of the observer
(c) The frequency of the source
(d) Separation between the source and the observer
Answer:
(d) Separation between the source and the observer

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
What are the advantages of SI system?
Answer:

1. This system makes use of only one unit for one physical quantity, which means a rational system of units
2. In this system, all the derived units can be easily obtained from basic and supplementary units, which means it is a coherent system of units.
3. It is a metric system which means that multiples and submultiples can be expressed as powers of 10.

Question 17.
Using the principle of homogunity of dimensions, check dimensionally the given equations are correct,
(a) \(\mathbf{T}^{2}=\frac{4 \pi^{2} r^{3}}{\mathbf{G}}\)
(b) \(T^{2}=\frac{4 \pi^{2} r^{3}}{G M}\)?
Answer:
Here G-gravitational constant
r – radius of orbit M – mass
Dimensional formula for T = T
Dimensional formula for r = L
Dimensional formula for G = M^-1L³T^-2
Dimensional formula for M = M
(a) T² M° L° = [L³] [M L^-3T^-2]
T² M° L° = L° MT² – Not correct

(b) T² = [L³][ML^-3T²][M^-1]
T² = T² – Dimensionally correct

Question 18.
Find out the workdone required to extract water from the well of depth 20 m. Weight of water and backet is 2.8 kg wt?
Answer:
Workdone, W = mgh
W = (Weight) × depth
= 2.8 × 20
W = 56 J

Question 19.
Is a single isolated force possible in nature?
Answer:
No. According to Newton’s third law, for every action, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is always a reaction. So it is impossible.

Question 20.
State the factors on which the moment of inertia of a body depends?
Answer:

1. Mass of body
2. Size and shape of body
3. Mass distribution w.r.t. axis of rotation
4. Position and orientation of rotational axis

Question 21.
If a drop of water falls on a very hot iron, it takes long time to evaporate. Explain why?
Answer:
When a drop of water falls on a very hot iron it gets insulated from the hot iron due to a thin layer of water vapour which is a bad conductor of heat. It takes quite long to evaporate as heat is conducted from hot iron to the drop through the insulating layer of water vapour very slowly.

Question 22.
What is meant by gravitational field. Give its unit?
Answer:
The gravitational field intensity \(\vec { E } \)₁, at a point is defined as the gravitational force experienced by unit mass at that point. It’s unit N kg^-1.

Question 23.
Why dimensional methods are applicable only up to three quantities?
Answer:
Understanding dimensions is of utmost importance as it helps us in studying the nature of physical quantities mathematically. The basic concept of dimensions is that we can add or subtract only those quantities which have same dimensions. Also, two physical quantities are equal if they have same dimensions, these basic ideas help us in deriving the new relation between physical quantities, it is just like units.

Question 24.
Why does a porter bend forward while carrying a sack of rice on his back?
Answer:
When a porter carries a sack of rice, the line of action of his centre of gravity will go away from the body. It affects the balance, to avoid this he bends. By which centre of gravity will realign within the body again. So balance is maintained.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Explain variation of ‘g’ with latitude?
Answer:
When an object is on the surface for the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.
OP_z, cos λ = \(\frac{PZ}{OP}\) = \(\frac{R’}{R}\)
R’ = R cos λ

where λ is is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
a_PQ = ω²R cos λ = ω²R cos² λ
Since R’ = R cos λ
Therefore, g’ = g – ω’²R cos²λ
From the above expression, we can infer that at equator, λ = 0, g’ = g – ω²R. The acceleration due to gravity is minimum. At poles λ = 90; g1 = g. it is maximum. At the equator, g’ is minimum.

Question 26.
Cap you associate a vector with
(a) the length of a wire bent into a loop
(b) a plane area
(c) a sphere.
Answer:
(a) We cannot associate a vector with the length of a wire bent into a loop, this is cause the length of the loop does not have a definite direction.

(b) We can associate a vector with a plane area. Such a vector is called area vector and its direction is represented by a normal drawn outward to the area.

(c) The area of a sphere does not point in any difinite direction. However, we can associate a null vector with the area of the sphere. We cannot associate a vector with the volume of a sphere.

Question 27.
What is mean by inertia? Explain its types with example?
Answer:
The inability of objects to move on its own or change its state of motion is called inertia.
Inertia means resistance to change its state. There are three types of inertia:

1. Inertia of rest: The inability of an object to change its state of rest is called inertia of rest.
Example:

• When a stationary bus starts to move, the passengers experience a sudden backward push.
• A book lying on the table will remain at rest until it is moved by some external agencies.

2. Inertia of motion: The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:

• When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.
• An athlete running in a race will continue to run even after reaching the finishing point.

3. Inertia of direction: The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:

• When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
• When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.

Question 28.
A light body and body with greater mass both are having equal kinetic energy. Among these two which one will have greater linear momentum?
Answer:
Given Data:
E₁ – E₂

then P₂ > P₁
i.e; a heavier body has greater linear momemtum.

Question 29.
Three particles of masses m₁ = 1 kg, m₂ = 2 kg and m₃ = 3 kg are placed at the corners of an equilateral triangle of side lm as shown in Figure. Find the position of center of mass?
Answer:
The center of mass of an equilateral triangle lies at its geometrical center G.
The positions of the mass m₁, m₂ and m₃ are at positions A, B and C as shown in the Figure. From the given position of the masses, the coordinates of the masses m₁ and m₂ are easily marked as (0,0) and (1,0) respectively.

To find the position of m3 the Pythagoras theorem is applied. As the ∆DBC is a right angle triangle,
BC² = CD² + DB²
CD² = BC² – DB²
CD² = 1² – (\(\frac{1}{2}\))² = 1 – (\(\frac{1}{4}\)) = \(\frac{3}{4}\)
CD = \(\frac { \sqrt { 3 } }{ 2 } \)

The position of mass m₃ is or (0.5, 0.5\(\sqrt{3}\))
X coordìnate of center of mass,
y_CM = \(\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}}\)
y_CM = \(\frac { \sqrt { 3 } }{ 4 } \) m
∴ The coordinates of center of mass G (x_CM, y_CM) is (\(\frac{7}{12}\), \(\frac { \sqrt { 3 } }{ 4 } \))

Question 30.
Define precision and accuracy. Explain with one example?
Answer:
The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. Precision of a measurement is a closeness of two or more measured values to each other.

The true value of a certain length is near 5.678 cm. In one experiment, using a measuring instrument of resolution 0.1 cm, the measured value is found to be 5.5 cm. In another experiment using a measuring instrument of greater resolution, say 0.01 cm, the length is found to be 5.38 cm. We find that the first measurement is more accurate as it is closer to the true value, but it has lesser precision. On the contrary, the second measurement is less accurate, but it is more precise.

Question 31.
Derive an expression for total acceleration in the non uniform circular motion?
Answer:
If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time. Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration as shown in the figure.

The resultant acceleration is obtained by vector sum of centripetal and tangential acceleration. Since centripetal acceleration is \(\frac { v^{ 2 } }{ r } \), the magnitude of this resultant acceleration is given by
a_R = \(\sqrt{a_{t}^{2}+\left(\frac{v^{2}}{r}\right)^{2}}\)

This resultant acceleration makes an angle θ with the radius vector as shown in figure.
This angle is given by tan θ = \(\frac{a_{t}}{\left(v^{2} / r\right)}\)

Question 32.
Calculate the value of adiabatic exponent for monoatomic molecule?
Answer:
Monoatomic molecule:
Average kinetic energy of a molecule = [\(\frac{3}{2}\)kT]
Total energy of a mole of gas = \(\frac{3}{2}\) kT × N_A = \(\frac{3}{2}\)RT
For one mole, the molar specific heat at constant volume

Question 33.
At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s surface, (mass of oxygen molecule : 2.76 × 10^-26kg Boltzmann’s constant (k_B) = 1.38 × 10^-23 J mol^-1 k^-1
Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
State kegler’s laws of planetary motion?
Answer:
1. Law of Orbits:
Each planet moves around the Sun in an elliptical orbit with the Sun at one of the foci.

2. Law of area:
The radial vector (line joining the Sun to a planet) sweeps equal areas in equal intervals of time.

3. Law of period:
The square of the time period of revolution of a planet around the Sun in its elliptical orbit is directly proportional to the cube of the semi-major axis of the ellipse.
T² ∝ a³
\(\frac { T^{ 2 } }{ a^{ 3 } } \) = Constant

(b) The distance of planet Jupiter from the sun is 5.2 times that of the earth. Find the period of resolution of Jupiter around the sun?
Answer:
Here r₁ = 5.2 r_e; T_J = ?; T_e = 1 year

= 11.86 years

[OR]

(c) Explain the propagation of errors in multiplication?
Answer:
Error in the product of two quantities:
Let ∆A and ∆B be the absolute errors in the two quantities A, and B, respectively. Consider the product Z = AB,
The error AZ in Z is given by Z ± AZ = (A ± ∆A) (B ± ∆B)
= (AB) ± (A ∆ B) ± (B ∆ A) ± (∆A • ∆B)
Dividing L.H.S by Z and R.H.S by AB, we get,
1 ± \(\frac{∆Z}{Z}\) = 1 ± \(\frac{∆B}{B}\) ± \(\frac{∆A}{A}\) ± \(\frac{∆A}{A}\). \(\frac{∆B}{B}\)
As ∆A/A, ∆B/B are both small quantities, their product term \(\frac{∆A}{A}\).\(\frac{∆B}{B}\) can be neglected.
The maximum fractional error in Z is
\(\frac{∆Z}{Z}\) = ± (\(\frac{∆A}{A}\) + \(\frac{∆B}{B}\))

(d) The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56 s, 2.42s, 271s and 2.80s respectively. Calculate the average absolute error?
Answer:
Mean absolute error = \(\frac{\Sigma\left|\Delta \mathrm{T}_{i}\right|}{n}\)
∆T_m = \(\frac{0.01+0.06+0.20+0.09+0.18}{5}\)
∆T_in = \(\frac{0.54}{5}\) = 0.108s = 0.1 1s (Rounded of 2^nd decimal place).

Question 35 (a).
Explain in detail the triangle law of vector addition?
Answer:
Let us consider two vectors \(\vec { A } \) and \(\vec { B } \) as shown in figure.

To find the resultant of the two vectors we apply the triangular law of addition as follows:

Represent the vectors \(\vec { A } \) and \(\vec { B } \) by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.

To explain further, the head of the first vector \(\vec { A } \) is connected to the tail of the second vector \(\vec { B } \). Let θ be the angle between \(\vec { A } \) and \(\vec { B } \). Then \(\vec { R } \) is the resultant vector connecting the tail of the first vector A to the head of the second vector B.

The magnitude of R (resultant) is given geometrically by the length of \(\vec { R } \) (OQ) and the direction of the resultant vector is the angle between \(\vec { R } \) and \(\vec { A } \). Thus we write \(\vec { R } \) = \(\vec { A } \) + \(\vec { B } \).

1. Magnitude of resultant vector:
The magnitude and angle of the resultant vector are determined by using triangle law of vectors as follows. From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure, let R is the magnitude of the resultant of \(\vec { A } \) and \(\vec { B } \).
cos θ = \(\frac{AN}{B}\) ∴AN = B cos θ ans sin θ = \(\frac{BN}{B}\) ∴ BN = B sin θ
For ∆OBN, we have OB² = ON² + BN²

2. Direction of resultant vectors: If θ is the angle between \(\vec { A } \) and \(\vec { B } \), then

(b) State and prove Bernoulli’s theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid?
Answer:
Bernoulli’s theorem:
According to Bernoulli’s theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains a constant. Mathematically,
\(\frac { P }{ \rho } \) + \(\frac{1}{2}\)v² + gh = Constant
This is known as Bernoulli’s equation.

Proof: Let us consider a flow of liquid through a pipe AB. Let V be the volume of the liquid when it enters A in a time t which is equal to the volume of the liquid leaving B in the same time. Let a_A, v_A and P_A be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively.

Let the force exerted by the liquid at A is
F_A = P_Aa_A

Distance travelled by the liquid in time t is d = v_at

Therefore, the work done is W = F_Ad = P_Aa_Av_At
But a_Av_At = a_Ad = V, volume of the liquid entering at A.

Thus, the work done is the pressure energy (at A), W = P_Ad = P_AV
Pressure energy per unit volume at
A = \($\frac{\text { Pressure energy }}{\text { Volume }}$\) = \(\frac{P_{A} V}{V}\) = P_A

Pressure energy per unit mass
A = \($\frac{\text { Pressure energy }}{\text { Mass }}$\) = \(\frac{P_{A} V}{m}\) = \(\frac{P_{A}}{\frac{m}{V}}=\frac{P_{A}}{\rho}\)

Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
EP_A = P_AV = P_AV × (\(\frac{m}{m}\)) = m\(\frac { P_{ A } }{ \rho } \)

Potential energy of the liquid at A,
PE_A = mg h_A

Due to the flow of liquid, the kinetic energy of the liquid at A,
KE_A = \(\frac{1}{2} m v_{\mathrm{A}^{2}}\)

Therefore, the total energy due to the flow of liquid at A,
E_A = EP_A + KE_A + PE_A
E_A = m \(m \frac{P_{A}}{\rho}+\frac{1}{2} m v_{A}^{2}+m g h_{A}\)

Similarly, let a_B, v_B and P_B be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at B. Calculating the total energy at F_B, we get
E_B = m \(m \frac{P_{B}}{\rho}+\frac{1}{2} m v_{B}^{2}+m g h_{B}\)
From the law of conservation of energy,
E_A = E_B

Thus, the above equation can be written as
\(\frac{P}{\rho g}+\frac{1}{2} \frac{v^{2}}{g}\) + h = Constant

The above equation is the consequence of the conservation of energy which is true until there is no loss of energy due to friction. But in practice, some energy is lost due to friction.

This arises due to the fact that in a fluid flow, the layers flowing with different velocities exert frictional forces on each other. This loss of energy is generally converted into heat energy. Therefore, Bernoulli’s relation is strictly valid for fluids with zero viscosity or non-viscous liquids. Notice that when the liquid flows through a horizontal pipe, then
h = 0 ⇒ \(\frac{P}{\rho g}+\frac{1}{2} \frac{v^{2}}{g}\) = Constant

Question 36 (a).
What are the limitations of dimensional analysis?
Answer:
Limitations of Dimensional analysis

1. This method gives no information about the dimensionless constants in the formula like 1, 2, …………….. π, e, etc.
2. This method cannot decide whether the given quantity is a vector or a scalar.
3. This method is not suitable to derive relations involving trigonometric, exponential and logarithmic functions.
4. It cannot be applied to an equation involving more than three physical quantities.
5. It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation.

For example, using dimensional analysis, s = ut + \(\frac{1}{3}\)at² is dimensionally correct whereas the correct relation is s = ut + \(\frac{1}{2}\)at²

(b) The escape velocity v of a body depends on

1. The acceleration due to gravity ‘g’ of the planet
2. The radius R of the planet. Establish dimensionally the relation for the escape velocity?

Answer:
\(v \propto g^{a} \mathrm{R}^{b} \Rightarrow v=k g^{a} \mathrm{R}^{b}\), K → dimensionally proportionality constant.
[v] = [g]^a [R]^b
[M⁰L¹T^-1] = [M⁰L¹T^-2]^a [M⁰L¹T¹⁰]^b
equating powers
1 = a + b
-1 = -2a ⇒ a = \(\frac{1}{2}\)
b = 1 – a = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∴ v = k\(\sqrt{gR}\)

[OR]

(c) Discuss the law of transverse vibrations In stretched string?
Answer:
Laws of transverse vibrations in stretched strings:
There are three laws of transverse vibrations of stretched strings which are given as follows:

(I) The law of length:
For a given wire with tension T (which is fixed) and mass per unit length p (fixed) the frequency varies inversely with the vibrating length. Therefore,
f ∝\(\frac{1}{l}\) ⇒ f = \(\frac{C}{2}\)
⇒1 × f = C, where C is constant

(II) The law of tension:
For a given vibrating length I (fixed) and mass per unit length p (fixed) the frequency varies directly with the square root of the tension T,
f ∝\(\sqrt{T}\)
⇒f = A\(\sqrt{T}\), where A is constant

(III) The law of mass:
For a given vibrating length l (fixed) and tension T (fixed) the frequency varies inverely with the square root of the mass per unit length µ,
f = \(\frac{1}{\sqrt{\mu}}\)
⇒f = \(\frac{B}{\sqrt{\mu}}\), where B is constant

(d) Explain how to determine the frequency of tuning for k using sonometer?
Answer:
Working: A transverse stationary or standing wave is produced and hence, at the knife edges P and Q, nodes are formed. In between the knife edges, anti-nodes are formed. If the length of the vibrating element is l then
l = \(\frac{\lambda}{2}\) ⇒ λ = 2l
Let f be the frequency of the vibrating element, T the tension of in the string and µ the mass per unit length of the string. Then using equation ,we get
f = \(\frac{v}{\lambda}=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}\) in Hertz …………………. (1)

Let ρ be the density of the material of the string and d be the diameter of the string. Then the mass per unit length µ,

Question 37.
(a) To move an object, which one is easier, push or pull? Explain?
Answer:
When a body is pushed at an arbitrary angle θ (o to \(\frac{π}{2}\)), the applied force F can be resolved into two components as F sin e parallel to the surface and F cos θ perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is
no acceleration along the vertical direction the normal force N is equal to
N_push = mg + F cos θ ………………………. (1)
As a result the maximal static friction also increases and is equal to
\(f_{s}^{\max }=\mu_{r} N_{p u s h}=\mu_{s}(m g+F \cos \theta)\) ………………… (2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.
When an object is pulled at an angle θ, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is
N_pull = mg – F cos θ ……………….. (3)

Equation (3) shows that the normal force needs is less than N_push. From equations (1) and (3), it is easier to pull an object than to push to make it move.

[OR]

(b) Derive an expression for the velocities of two objects colliding elastically in one dimension?
Answer:
Consider two elastic bodies of masses m₁ and m₂ moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure.

In order to have collision, we assume that the mass m₁ moves, faster than mass m₂ i.e., u₁ > u₂. For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same.

From the law of conservation of linear momentum,
Total momentum before collision (ρ_i) = Total momentum after collision (ρ_f)
m₁u₁ + m₂u₂ = m₁v₁ + m₂ v₁ …………………. (1)
or m₁ (u₁ – v₁) = m₂(v₂ – u₂) …………………… (2)
Furthur,

For elastic collision,
Total kinetic energy before collision KE_i = Total kinetic energy after collision KE_f
\(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}\) ………………….. (3)
\(m_{1}\left(u_{1}^{2}-v_{1}^{2}\right)=m_{2}\left(v_{2}^{2}-u_{2}^{2}\right)\)
After simplifying and rearranging the terms,
Using the formula a² – b² = (a + b) (a – b), we can rewrite the above equation as
\(m_{1}\left(u_{1}+v_{1}\right)\left(u_{1}-v_{1}\right)=m_{2}\left(v_{2}+u_{2}\right)\left(v_{2}-u_{2}\right)\) ……………….. (4)
Dividing the equation (4) by (2) we get

Equation (5) can be written as
u₁ – u₂ = -(v₁ – v₂)
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for v₁ and v₂,
v₁ = v₂ + u₂ – u₁ …………………….. (6)
or v₂ = u₁ + v₁ – u₁ …………………….. (7)
To find the final velocities V₁ and v₂:
Substituting equation (5) in equation (2) gives the velocity of m₁ as
m₁ (u₁ – v₁) = m₂ (u₁ + v₁ – u₂ – u₂)
m (u₁ – v₁) = m₂ (u₁ + v₁ – 2u₂)
m₁u₁ – m₁v₁ = m₂u₁ + m₂v₁ + 2m₂u₂
m₁u₁ – m₂u₁ + 2m₂u₂ = m₁v₁ + m₂v₁
(m₁ – m₂)u₁ + 2m₂u₂ = (m₁ + m₂) v₁
or v₁ = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\) u₁  + \(\left(\frac{2 m_{2}}{m_{1}+m_{2}}\right)\) u₂
Similarly, by substituting (6) in equation (2) or substituting equation (8) in equation (7), we get the final m₂ as
v₂ = \(\left(\frac{2 m_{1}}{m_{1}+m_{2}}\right)\) u₁ + \(\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right)\) u₂ …………………. (9)

Question 38 (a).
Prove that at points near the surface of the Earth the gravitational potential energy of the object is v = mgh?
Answer:
When an object of mass m is raised to a height h, the potential energy stored in the object is mgh. This can be derived using the general expression for gravitational potential energy. Consider the Earth and mass system, with r, the distance between the mass m and the Earth’s centre. Then the gravitational potential energy.
U = –\(\frac{\mathrm{GM}_{e} m}{r}\) …………………… (1)

Here r = R_e + h, where Re is the radius of the Earth, h is the height above the Earth’s surface

By using Binomial expansion and neglecting the higher order

Replace this value and we get,
U = \(-\frac{\mathrm{GM}_{e} m}{\mathrm{R}_{e}}\left(1-\frac{h}{\mathrm{R}_{e}}\right)\) ………………… (4)

We know that, for a mass m on the Earth’s surface,
\(\mathrm{G} \frac{\mathrm{M}_{e} m}{\mathrm{R}_{e}}=m g \mathrm{R}_{e}\) …………………… (5)

Substituting equation (4) in (5) we get,
U = -mgR_e + mgh ………………….. (6)

It is clear that the first term in the above expression is independent of the height h. For example, if the object is taken from height h₁ to h₂, then the potential energy at h₁ is
U(h₁) = -mgR_e + mgh₁ …………………… (7)
and the potential energy at h₂ is
U(h₂) = -mgR_e + mgh₂ …………………… (8)

The potential energy differenqe between h₁ and h₂ is
U(h₂) – U(h₁) = mg(h₁-h₂) …………………. (9)

The term mgR_e in equation (7) and (8) plays no role in the result. Hence in the equation (6) the first term can be omitted or taken to zero. Thus it can be stated that the gravitational potential energy stored in the particle of mass m at a height h from the surface of the Earth is U = mgh. On the surface of the Earth, U = 0, since h is zero.

[OR]

(b) Derive an expression for Radius of gyration?
Answer:
For bulk objects of regular shape with uniform mass distribution, the expression for moment of inertia about an axis involves their total mass and geometrical features like radius, length, breadth, which take care of the shape and the size of the objects.

But, we need an expression for the moment of inertia which could take care of not only the mass, shape and size of objects, but also its orientation to the axis of rotation. Such an expression should be general so that it is applicable even for objects of irregular shape and non-uniform distribution of mass. The general expression for moment of inertia is given as,
I = MK²

where, M is the total mass of the object and K is called the radius of gyration. The radius of gyration of an object is the perpendicular distance from the axis of rotation to an equivalent point mass, which would have the same mass as well as the same moment of inertia of the object.

As the radius of gyration is distance, its unit is m. Its dimension is L. A rotating rigid body with respect to any axis, is considered to be made up of point masses m₁, m₂, m₃, . . . mn at perpendicular distances (or positions) r₁, r₂, r₃ . . . r_n respectively as shown in figure. The moment of inertia of that object can be written as,

I = \(\sum m_{1} r_{1}^{2}=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+m_{3} r_{3}^{2}+\ldots .+m_{n} r_{n}^{2}\)
If we take all the n number of individual masses to be equal
m = m₁ = m, = m₂ = m₃ = ……………… = m_n
then

I = MK²

where, nm is the total mass M of the body and K is the radius of gyration.

The expression for radius of gyration indicates that it is the root mean square (rms) distance of the particles of the body from the axis of rotation. In fact, the moment of inertia of any object could be expressed in the form, I = MK²

For example, let us take the moment of inertia of a uniform rod of mass M and length l. Its moment of inertia with respect to a perpendicular axis passing through the center of mass is,
I = \(\frac{1}{12}\)Ml²
In terms of radius of gyration, I = MK²
Hence, MK² = \(\frac{1}{12}\)Ml²
K² = \(\frac{1}{12}\)l²
K = \(\frac{1}{12}\) or K = \(\frac{1}{2 \sqrt{3}} l\) or K = (0.289)l

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 7 Mensuration Ex 7.4

Question 1.
An aluminium sphere of radius 12 cm is melted to make a cylinder of radius 8 cm. Find the height of the cylinder.
Answer:
Sphere – Radius r₁ = 12 cm
Cylinder – Radius r₂ = 8 cm
h₂ = ?
Volume of cylinder = Volume of sphere melted

∴ Height of the cylinder made = 36 cm.

Question 2.
Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tanks will rise by 21 cm.
Answer:
Length of the rectangular tank (l) = 50 m = 5000 cm
Width of the rectangular tank (b) = 44 m = 4400 cm
Level of water in the tank (h) = 21 cm
Volume of the tank = l × b × h cu. units = 5000 × 4400 × 21 cm³
Radius of the pipe (r) = 7 cm
Speed of the water = 15 km/hr.
(h) = 15000 × 100 cm / hr.
Volume of water flowing in one hour

Question 3.
A conical flask is full of water. The flask has base radius r units and height h units, the water poured into a cylindrical flask of base radius x r units. Find the height of water in the cylindrical flask.
Answer:
Radius of the conical flask = r units
Height of the conical flask = h units
Volume of the conical flask = \(\frac{1}{3} \pi r^{2} h\) cu.units
Radius of the cylindrical flask = x r units
Let the height of the cylindrical flask be “H” units
Volume of the cylindrical flask = Volume of the Conical flask

Height of the cylindrical flask = \(\frac{h}{3 x^{2}}\) units

Question 4.
A solid right circular cone of diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal diameter.
Answer:
Radius of a cone (V) = 7 cm
Height of a cone (h) = 8 cm
External radius of the hollow sphere (R) = 5 cm
Let the internal radius be “x”
Volume of the hollow sphere = Volume of the Cone


Internal diameter of the Hollowsphere = 2 × 3 = 6 cm.

Question 5.
Seenu’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank) which is in the shape of a cuboid. The sump has dimensions 2 m × 1.5 m × 1 m. The overhead tank has its radius of 60 cm and height 105 cm. Find the volume of the water left in the sump after the overhead tank has been completely filled with water from the sump which has been full, initially.
Answer:
Length of the cuboid tank (l) = 2 cm = 200 cm
Breadth of the cuboid tank (b) = 1.5 cm = 150 cm
Height of the tank (h) = 1 m = 100 cm
Volume of the cuboid = l × b × h cu. units
= 200 × 150 × 100 cm³
= 30,00,000 cm³
Radius of the tank (r) = 60 cm
Height of the tank (h) = 105 cm
Volume of the cylindrical tank = πr²h cu. units
= \(\frac{22}{7}\) × 60 × 60 × 105 cm³
= 22 × 60 × 60 × 15 cm³
= 1188000 cm³
Volume of water left in the sump = Volume of the sump – Volume of the tank
= 3000000 – 1188000 cm³
= 1812000 cm³

Question 6.
The internal and external diameter of a hollow hemispherical shell is 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, then find the height of the cylinder.
Answer:
Internal radius of the shell (r) = 3 cm
External radius of the shell (R) = 5 cm
Radius of the cylinder (r) = 7 cm
Let the height of the cylinder be “h”
Volume of the cylinder = Volume of the hemispherical shell

Height of the cylinder = 1.33 cm

Question 7.
A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, then find the thickness of the cylinder.
Answer:
Radius of a sphere (r) = 6 cm
External radius of the cylinder (R) = 5 cm
Height of the cylinder (h) = 32 cm
Let the internal radius of the cylinder be ‘x’
Volume of the hollow cylinder = Volume of a sphere
πh (R² – r²) = \(\frac{4}{3}\) πr³
π × 32 (5 + x) (5 – x) = \(\frac{4}{3}\) × π × 6 × 6 × 6
32 (25 – x²) = 4 × 2 × 6 × 6
25 – x² = 9
x² = 25 – 9 = 16
x = √16 = 4
Thickness of the cylinder = 5 – 4 = 1 cm.

Question 8.
A hemispherical bowl is filled to the brim with juice. The juice is poured into a cylindrical vessel whose radius is 50% more than its height. If the diameter is same for both the bowl and the cylinder then find the percentage of juice that can be transferred from the bowl into the cylindrical vessel.
Answer:
Let the height of the cylinder be “h”
radius is 50% more than the height


From (1) and (2) we get,
Volume of the cylinder = Volume of the hemisphere
It is possible to transfer the full quantity from the bowl into the cylindrical vessel.
100 % of the juice can be transferred.

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Tamilnadu Samacheer Kalvi 6th Social Science Economics Solutions Term 2 Chapter 1 Economics-An Introduction

Samacheer Kalvi 6th Social Science Economics-An Introduction Text Book Back Questions and Answers

I. Fill in the blanks

1. The producers of food grains are ……………..
2. Collection of honey is a …………….. occupation.
3. The conversion of raw materials into finished goods is called ……………..
4. According to Gandhiji the villages are …………….. of the nation.
5. The percentage of population in the cities of Tamil Nadu is ……………..

Answer:

1. Farmers
2. Primary
3. Secondary activities
4. backbone
5. 47%

II. Match the following

Answer:
1. – c
2. – d
3. – a
4. – b
5. – d

III. Match and find the odd pair

Answer:
1. – d
2. – c
3. – b
4. – a

IV. Choose the correct answer

1. Agriculture is a (primary / Secondary) occupation.
2. Economic activities are divided on the basis of (ownership / use)
3. Sugar Industries are (Primary / Secondary) activity.
4. Agro based industry (Cotton / Furniture)
5. Dairy farming is a (Public sector/Co-operative sector)

Answer:

1. Primary
2. use
3. Secondary
4. Cotton
5. Co-operative sector

V. Answer the following questions

Question 1.
Sandhai – Define
Answer:
In villages once in a week or month, all things are sold in a particular place at a specific time to meet the needs of the people. That is called Sandai.

Question 2.
What is called the barter system?
Answer:
A system of exchanging goods for other goods is called a barter system. Example: Exchange a bag or rice for enough clothes.

Question 3.
What is trade?
Answer:
Trade involves the transfer of goods or services from one person to another often in exchange of money.

Question 4.
What is Savings?
Answer:
The amount from the income which is left for future needs after consumption is called sayings.

Question 5.
What was the necessity for the invention of money?
Answer:

1. When traders exchange commodities there arises a difference in the value of the commodity.
2. To solve this problem people invented money.

Question 6.
What was the reason for the development of settlements near water bodies?
Answer:

1. Rivers act as the main source for the cultivation of crops.
2. So early man settled permanently near the rivers.

Question 7.
What are called secondary occupation?
Answer:
The raw materials obtained from the primary activities are converted into finished products is called a secondary occupation.

Question 8.
Name the city-centered industries.
Answer:
Cement, iron, and Aluminium industries, seafood processing are some of the city centered industries.

VI. Answer the following in detail

Question 1.
List out the important primary occupations of your district.
Answer:

1. Agriculture
2. Cattle rearing
3. Collection of fruits, nuts, honey, and medicinal herbs.

Question 2.
Mention the manufacturing industries found in your district.
Answer:

1. Cotton textiles
2. Spinning and weaving
3. Food processing industries
4. Beedi production
5. Wind power generations

Question 3.
How are the industries classified on the basis of raw materials?
Answer:
On the basis of raw materials industries are classified as

1. Agro-Based Industries – Cotton textiles, Sugar mills, and Food processing.
2. Forest-Based Industries – Paper mills, Furniture making, Building materials.
3. Mineral Based Industries – Cement, Iron, Aluminium Industries.
4. Marine Based Industries – Seafood processing.

Question 4.
Write down the occupations in the service sector.
Answer:
The service sector serves the people to fulfill their daily needs like:

1. Transport – Roadways, Railways, Waterways, Airways.
2. Communication – Post, Telephone, Information Technology.
3. Trade – Procurement of goods, selling.
4. Banking – Money Transaction, Banking Services.

Question 5.
What do you know about the features of cities?
Answer:

1. A city is a large human settlement.
2. The high density of population.
3. Four-way roads, flyovers, skyscrapers, parks.
4. Educational institution, hospital, Government offices.
5. Private and public industries and technological institutions.
6. Employment opportunities permanent monthly income, basic requirements are

VII. Fill in the tabular column given below

Activity:
Write the lyrics of Bharathiyar’s Analyze the lyrics and write down the commodities which were exchanged in yesteryears with the help of the teacher.

(1) Wheat
(2) Betel
(3) Tusk

VIII. Stick picture

Samacheer Kalvi 6th Social Science Economics-An Introduction Additional Important Questions and Answers

I. Fill in the blanks Answer

1. The permanent settlements near the rivers were called ……………
2. More than …………… percentage of the world’s population live in cities.
3. The …………… sector serves the people to fulfill their daily needs.
4. One who uses the products is called ……………
5. …………… are the real shadow of cities.

Answer:

1. Villages
2. 50
3. Service
4. Consumer
5. Villages

II. Choose the correct answer

Question 1.
Tertiary activities are also called as sector ……………
(a) Private
(b) Service
(c) Public
Answer:
(b) Service

Question 2.
Secondary and tertiary activities are …………… centred activities.
(a) City
(b) Town
(c) Village
Answer:
(a) City

Question 3.
…………… is the main occupation in villages.
(a) Mining
(b) Fishing
(c) Farming
Answer:
(c) Farming

III. Answer the following questions

Question 1.
Sandhai – Define
Answer:
In villages, once in a week or month, things are sold in a particular place at a specific time to meet the needs of the people. That is called Sandhai.

Question 2.
What is called the barter system?
Answer:

1. A system of exchanging goods for other goods is called a barter system.
2. Example: Exchange a bag or rice for enough clothes.

Question 3.
What are consumer goods?
Answer:
The finished goods which are brought from the market to fulfill the daily needs of the consumers are called consumer goods.

Question 4.
Who are cultivators?
Answer:
Persons who are involved in farming and grazing are called cultivators or farmers.

Question 5.
How are industries classified?
Answer:
Industries are classified on the basis of the:

1. Availability of raw materials
2. Capital
3. Ownership

Question 6.
Name the sectors that are helpful in the economic development of our country.
Answer:
Agriculture and Industries are helpful in the economic development of our country.

IV. Mind map

Students can Download Tamil Nadu 11th Physics Model Question Paper 5 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 5 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
Two protons are travelling. along the same straight path but in opposite directions. The relative velocity between the two is ……………………
(a) c
(b) \(\frac{c}{2}\)
(c) 2c
(d) 0
Hint:
One of the velocity V₁ = V; Other velocity V₂ = -V
Relative velocity (V_rel) = \(\frac{V_{1}-V_{2}}{\left[1+\frac{V^{2}}{C^{2}}\right]}\) = \(\frac{V-(-V)}{\left[1+\frac{V^{2}}{C^{2}}\right]}\) = \(\frac{2v}{2}\)
V = C
V_relative = C
Answer:
(a) c

Question 2.
If the Earth stops rotating about its own axis, g remains unchanged at …………………..
(a) Equator
(b) Poles
(c) Latitude of 45°
(d) No where
Answer:
(b) Poles

Question 3.
When train stops, the passenger moves forward. It is due to ……………………
(a) Inertia of passenger
(b) Inertia of train
(c) Gravitational pull by Earth
(d) None of the above
Answer:
(a) Inertia of passenger

Question 4.
A particle of mass m moves in the xy plane with a velocity v along the straight line AB. If the angular momention of the particle with respect to origin O is L_A when it is at A and L_B when it is at B, then …………………….
(a) L_A = L_B
(b) L_A < L_B
(c) L_A > L_B
(d) The relationship between L_A and L_B depends uopn the slope of the line AB

Hint:
Magnitude of L is, L = mvr sin ϕ = mvd
d = r sin ϕ is the distance of closest approach of the particle so the origin, as ‘d’ is same for both particles.
So, L _A = L_B
Answer:
(a) L_A = L_B

Question 5.
A couple produces ……………………….
(a) Pure rotation
(b) Pure translation
(c) Rotation and translation
(d) No motion
Answer:
(a) Pure rotation

Question 6.
A body starting from rest has an acceleration of 20 ms~2 the distance travelled by it in the sixth second is ……………………..
(a) 110 m
(b) 130m
(c) 90m
(d) 50 m
Hint:
Distance travelled in n^th second, u = 0
S_n = u + \(\frac{1}{2}\)a(2n – 1)
S₆ = 0 + \(\frac{1}{2}\) × 20 × (2 × 6 – 1); S₆ = 110 m
Answer:
(a) 110 m

Question 7.
A lift of mass 1000 kg, which is moving with an acceleration of 1m/s² in upward direction has tension has developed in its string is ………………………
(a) 9800 N
(b) 10800 N
(c) 11000 N
(d) 10000 N
Hint:
Tension, T = mg + ma = m(g + a) = 1000 (10 + 1)
T = 11000 N
Answer:
(c) 11000 N

Question 8.
The relation between acceleration and displacement of four particles are given below ………………………..
(a) a_x = 2x
(b) a_x = + 2x²
(c) a_x = -2x²
(d) a_x = -2x
Answer:
(d) a_x = -2x

Question 9.
A sonometer wire is vibrating in the second overtone. In the wire there are, ……………………..
(a) Two nodes and two antinodes
(b) One node and two antinodes
(c) Four nodes and three antinodes
(d) Three nodes and three antinodes
Answer:
(d) Three nodes and three antinodes

Question 10.
Which of the following is the graph between the light (h) of a projectile and time (t), when it is projected from the ground ………………………..
(a)
(b)
(c)
(d)
Answer:
(c)

Question 11.
According to kinetic theory of gases, the rms velocity of the gas molecules is directly proportional to ………………………
(a) \(\sqrt{T}\)
(b) T³
(c) T
(d) T⁴
Hint:
The rms velocity, V_rms = \(\sqrt{3KT/m}\) ⇒ V_rms ∝ \(\sqrt{T}\)
Answer:
(a) \(\sqrt{T}\)

Question 12.
A body of mass m moving with velocity v collides head on with another body of mass 2m which is initially at rest. The ratio of K.E of colliding body before and after collision will be ……………………..
(a) 1 : 1
(b) 2 : 1
(c) 4 : 1
(d) 9 : 1
Hint:
KE of colliding bodies before collision = \(\frac{1}{2}\) mv²
After collision the mass = m + 2m = 3m
velocity becomes V’ = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)v = \(\frac{mv}{3m}\) = \(\frac{v}{3}\)
KE after collision = \(\frac{1}{2}\)m (\(\frac{V}{3}\)² = \(\frac{1}{9}\) (\(\frac{1}{2}\) mv²)
\(\frac{\mathrm{KE}_{\text {before }}}{\mathrm{KE}_{\text {after }}}=\frac{\frac{1}{2} \mathrm{mv}^{2}}{\frac{1}{9}\left(\frac{1}{2} \mathrm{mv}^{2}\right)}=9: 1\)
Answer:
(d) 9 : 1

Question 13.
Four particles have velocity 1, 0, 2 and 3ms^-1. The root mean square velocity of the particles is ……………………..
(a) 3.5 ms^-1
(b) \(\sqrt{3.5}\) ms^-1
(c) 1.5ms^-1
(d) Zero
Hint:

Answer:
(b) \(\sqrt{3.5}\) ms^-1

Question 14.
Two vibrating tuning forks produce progressive waves given by y₁ = 4 sin 500 πt and y₂ = 2 sin 506 πt where t is in seconds Number of beat produced per minute is ………………………
(a) 360
(b) 180
(c) 3
(d) 60
Hint:

f₂ – f₁ = 3 = beats per sec and 3 × 60 = 180 beats per min

Answer:
(b) 180

Question 15.
Workdone by a simple pendulum in one complete oscillation is …………………………..
(a) Zero
(b) Jmg
(c) mg cos θ
(d) mg sin θ
Answer:
(a) Zero

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
A girl is swinging in the sitting position. How will the period of the swing be changed if she stands up?
Answer:
This can be explained using the concept of a simple pendulum. We know that the time period of a simple pendulum is given by
T = 2π\(\sqrt{l/g}\) i.e; T ∝\(\sqrt{l}\)
When the girl stands up, the distance between the point of suspension and the centre of mass of the swinging body decreases i.e., l decreases, so T will also decrease.

Question 17.
A car starts to move from rest with uniform acceleration 10 ms^-2 then after 2 sec, what is its velocity?
Answer:
a =10 ms^-2;
t = 2s;
w = 0;
v = ?
v = u + at
v = 0 + 10 × 2
= 20 ms^-1

Question 18.
State Lami’s theorem?
Answer:
If a system of three concurrent and coplanar forces is in equilibrium, then Lami’s theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces. The constant of proportionality is same for all three forces.

Question 19.
Due to the action of constant torque, a wheel from rest makes n rotations in t seconds? Show that the angular acceleration of a wheel as \(\frac{4 \pi n}{t^{2}}\) rad s^-2
Answer:
Initial angluar velocity ω₀ = 0
Number of rotations in t seconds = n
angular displacement θ = 2πn
but, θ = ω₀t + \(\frac{1}{2}\)αt²
2πn = \(\frac{1}{2}\)αt²
α = \(\frac{4 \pi n}{t^{2}}\)

Question 20.
Why a given sound is louder in a hall than in the open?
Answer:
In a hall, repeated reflections of sound take place from the walls and the ceiling. These reflected sounds mix with original sound which results in increase the intensity of sound. But in open, no such a repeated reflection is possible. .’. sound will not be louder as in hall.

Question 21.
What are the differences between connection and conduction?
Answer:
Conduction:
Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. The objects which allow heat to travel easily through them are called conductors.

Convection:
Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.

Question 22.
Why two holes are made to empty an oil tin?
Answer:
When oil comes out through a tin with one hole, the pressure inside the tin becomes less than the atmospheric pressure, soon the oil stops flowing out. When two holes are made in the tin, air keeps on entering the tin through the other hole and maintains pressure inside.

Question 23.
If the length of the simple pendulum is increased by 44% from its original length, calculate the percentage increase in time period of the pendulum?
Answer:
Since T ∝ \(\sqrt{l}\) = Constant \(\sqrt{l}\)

∴ T_f = 1.2 T_i = T_i + 20% T_i

Question 24.
When do the real gases obey more correctly the gas equation PV = nRT?
Answer:
An ideal gas is one whose molecules have zero volume and no mutual force between them. At low pressure, the volume of a gas is large and so the volume occupied by the molecules is negligible in comparison to the volume of the gas.

At high temperature, the molecules have large velocities and so the intermolecular force has no influence on their motion. Hence at low pressure and high temperature, the behaviour of real gases approach the ideal gas behaviour.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
A stone is thrown upwards with a speed v from the top of a tower. It reaches the ground with a velocity 3v. What is the height of the tower?
Answer:
From equation of motion,
v’ = u + at ……………….. (1)
h = ut + \(\frac{1}{2}\) at²
here, v’ = av; u = v; a = +g
Using equ. (1)
3v = v + gt ⇒ 3v – v = gt
t = \(\frac{2v}{g}\)

Substitute ‘t’ value in equ. (2)
h = v(\(\frac{2v}{g}\)) + \(\frac{1}{2}\)g(\(\frac{2v}{g}\))² = \(\frac { 2v^{ 2 } }{ g } \) + \(\frac{1}{2}\)g (\(\frac{2v}{g}\))²
h = \(\frac { 2v^{ 2 } }{ g } \) + \(\frac { 2v^{ 2 } }{ g } \)
= \(\frac { 4v^{ 2 } }{ g } \) g = 10 ms^-2
= \(\frac { 4v^{ 2 } }{ 10 } \); h = \(\frac { 2v^{ 2 } }{ 5 } \)

Question 26.
An object is projected at an angle such that the horizontal range is 4 times of the maximum height. What is the angle of projection of the object?
Answer:
Given:
Horizontal range = 4 H_max
Horizontal range = \(\frac{u^{2} \sin 2 \theta}{g}\) = \(\frac{2 u^{2} \sin \theta \cos \theta}{g}\)
Maximum height = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
as given, \(\frac{2 u^{2} \sin \theta \cos \theta}{g}\) = \(\frac{4 u^{2} \sin ^{2} \theta}{2 g}\)
2 cos θ = 2 sin θ
tan θ = 1
∴ θ = 45°

Question 27.
A room contains oxygen and hydrogen molecules in the ratio 3 : 1. The temperature of the room is 27°C. The molar mass of 0₂ is 32 g mol^-1 and for H₂, 2 g mol^-1. The value of gas constant R is 8.32 J mol^-11 k^-1. Calculate rms speed of oxygen and hydrogen molecule?
Answer:
(a) Absolute Temperature T = 27°C = 27 + 273 = 300 K.
Gas constant R = 8.32 J mol^-1 K^-1
For Oxygen molecule: Molar mass

M = 32 gm/mol = 32 × 10^-3 kg mol^-1
rms speed v_rms = \(\sqrt { \frac { 3RT }{ M } } \) = \(\sqrt{\frac{3 \times 8.32 \times 300}{32 \times 10^{-3}}}\) = 483.73 ms^-1 ~ 484ms^-1

For Hydrogen molecule: Molar mass M = 2 × 10^-3 kg mol^-1
rms speed v_rms = \(\sqrt { \frac { 3RT }{ M } } \) = \(\sqrt{\frac{3 \times 8.32 \times 300}{2 \times 10^{-3}}}\) = 1934 ms^-1 = 1.93 K ms^-1

Note that the rms speed is inversely proportional to \(\sqrt{M}\) and the molar mass of oxygen is 16 times higher than molar mass of hydrogen. It implies that the rms speed of hydrogen is 4 times greater than rms speed of oxygen at the same temperature. \(\frac{1934}{484}\) ~ 4.

Question 28.
Explain about an angle of friction?
Answer:
The angle of friction is defined as the angle between the normal force (N) and the resultant force (R) of normal force and maximum friction force (\(f_{s}^{\max }\))

In the figure the resultant force is R = \(\sqrt{\left(f_{s}^{\max }\right)^{2}+\mathrm{N}^{2}}\)
tan θ = \(\frac{f_{s}^{\max }}{\mathrm{N}}\) ………………….. (1)

But from the frictional relation, the object begins to slide when \(f_{s}^{\max }=\mu_{\mathrm{s}} \mathrm{N}\)
or when \(\frac{f_{s}^{\max }}{\mathrm{N}}\) = µ_s ………………….. (2)

From equations (1) and (2) the coefficient of static friction is
µ_s = tan θ ……………………. (3)
The coefficient of static friction is equal to tangent of the angle of friction.

Question 29.
How does resolve a vector into its component? Explain?
Answer:
component of a resolve:
In the Cartesian coordinate system any vector \(\vec { A } \) can be resolved into three components along x, y and z directions. This is shown in figure. Consider a 3-dimensional coordinate system. With respect to this a vector can be written in component form as
\(\vec { A } \) = A_x \(\hat { i } \) + A_y\(\hat { j } \) + A_z\(\hat { k } \)
Components of a vector in 2 dimensions and 3 dimensions

Here Ax is the x-component of \(\vec { A } \), Ay is the y-component of \(\vec { A } \) and A_z is the z component of \(\vec { A } \).
In a 2-dimensional Cartesian coordinate system (which is shown in the figure) the vector \(\vec { A } \) is given by
\(\vec { A } \) = A_x \(\hat { i } \) + A_y \(\hat { j } \)

If \(\vec { A } \) makes an angle θ with x axis, and A_x and A_y are the components of A along x-axis and y-axis respectively, then as shown in figure,
A_x θ = A cos θ, A = A sin θ
where ‘A’ is the magnitude (length) of the vector \(\vec { A } \), A = \(\sqrt{\mathrm{A}_{x}^{2}+\mathrm{A}_{y}^{2}}\)

Question 30.
Derive an expression for energy of an orbiting satellite?
Answer:
The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is,
U = \(-\frac{\mathrm{GM}_{s} \mathrm{M}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\) ………………….. (1)

Here M_s -mass of the satellite, M_E -mass of the Earth, R_E – radius of the Earth.
The Kinetic energy of the satellite is
K.E = \(\frac{1}{2}\) M_sv² ………………….. (2)

Here v is the orbital speed of the satellite and is equal to
v = \(\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}}\)

Substituting the value of v in (2) the kinetic energy of the satellite becomes,
K.E = \(\frac{1}{2} \frac{\mathrm{GM}_{\mathrm{E}} \mathrm{M}_{s}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\)

Therefore the total energy of the satellite is
E = \(\frac{1}{2}\) \(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{M}_{s}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}-\frac{\mathrm{GM}_{s} \mathrm{M}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\)
E = \(-\frac{\mathrm{GM}_{s} \mathrm{M}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\)
The total energy implies that the satellite is bound to the Earth by means of the attractive gravitational force.

Note:
As h approaches ∞, the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth’s gravity and is not bound to Earth at large distance.

Question 31.
Explain in detail Newton’s law of cooling?
Answer:
Newton’s law of cooling:
Newton’s law of cooling body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{dQ}{dT}\) ∝(T – T_s) …………………. (1)
The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,

T = Temperature of the object
T_s = Temperature of the surrounding

From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.

Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount dT in time dt, then the amount of heat lost is,
dQ = msdT …………………….. (2)

Dividing both sides of equation (2) by dt
\(\frac{dQ}{dT}\) = \(\frac{msdT}{dt}\) ………………….. (3)

From Newton’s law of cooling
\(\frac{dQ}{dT}\) ∝(T – T_s)
\(\frac{dQ}{dT}\) = -a(T – T_s) ………………………… (4)

Where a is some positive constant.
From equation (3) and (4)
-a (T – T_s) = ms\(\frac{dT}{dt}\)
\(\frac{d \mathrm{T}}{\mathrm{T}-\mathrm{T}_{s}}\) = -a\(\frac{a}{ms}\) dt …………………… (5)

Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides we get,
T = T_s + \(b_{2} e^{\frac{-a}{m s} t}\) …………………….. (6)
Here b₂ = e^b1 Constant

Question 32.
Explain Laplace’s correction?
Answer:
Laplace’s correction: In 1816, Laplace satisfactorily corrected this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast.

Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat. Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson’s law (not Boyle’s law as Newton assumed), which is
PV^γ = Constant …………………. (1)
where, γ = \(\frac { C_{ p } }{ C_{ v } } \) which is the ratio between specific heat at constant pressure and specific heat at constant volume. Differentiating equation (1) on both the sides, we get
\(V^{ \gamma }dP+P(\gamma ^{ V\gamma -1 }dV)=0\)

or

\(\gamma \mathrm{P}=-\mathrm{V} \frac{d p}{d \mathrm{V}}=\mathrm{B}_{\mathrm{A}}\) ………………….. (2)
where, B_A is the adiabatic bulk modulus of air. Now, substituting equation (2) in equation
V = \(\sqrt{\frac{B}{\rho}}\), the speed of sound in air is
v_A = \(\sqrt{\frac{\mathrm{B}_{\mathrm{A}}}{\rho}}=\sqrt{\frac{\gamma \mathrm{P}}{\rho}}=\sqrt{\gamma v_{\mathrm{T}}}\)
Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take γ = 1.47. Hence, speed of sound in air is v_A = (\(\sqrt{1.4}\)) (280 m s^-1) = 331.30 ms^-1, which is very much closer to experimental data.

Question 33.
Explain types of equilibrium?
Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
What are the applications of dimensional analysis?
Verify s = ut + \(\frac{1}{2}\)at² by dimensional analysis?
Answer:

Given equation is dimensionally correct as the dimensions on the both side are same. Applications of dimensional analysis

1. Convert a physical quantity from one system of units to another.
2. Check the dimensional correctness of a given physical equation.
3. Establish relations among various physical quantities.

[OR]

(b) Explain the types of equilibrium with suitable examples?
Answer:
Translational motion – A book resting on a table.
Rotational equilibrium – A body moves in a circular path with constant velocity.
Static equilibrium – A wall-hanging, hanging on the wall.
Dynamic equilibrium – A ball decends down in a fluid with its terminal velocity.
Stable equilibrium – A table on the floor A pencil
Unstable equilibrium – standing on its tip.
Neutral equilibrium – A dice rolling on a game board.

Question 35 (a)
Explain the motion of block connected by a string in vertical motion?
Answer:
When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Vertical motion:
Consider two blocks of masses m₁ and m₂ (m₁ > m₂) connected by a light and inextensible string that passes over a pulley as shown in Figure 1.

Let the tension in the string be T and acceleration a.

When the system is released, both the blocks start, Two blocks connected by a string moving, m₂ vertically upward and m₁ downward with same acceleration a. The gravitational force m₁g on mass m₁ is used in lifting the mass m₂.

The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure 2.

Applying Newton’s second law for mass m₂
T\(\hat { j } \) – m₂\(\hat { j } \)g = m₂ a\(\hat { j } \)

The left hand side of the above equation is the total force that acts on m₂ and the right hand side is the product of mass and acceleration of m₂ in direction.

By comparing the components on both sides, we get
T -m₂g = m₂a ……………………. (1)

Similarly, applying Newton’s law second law of for mass m₁
T\(\hat { j } \) – m₁g\(\hat { j } \) = -m₁a\(\hat { j } \)

As mass m₁ moves downward (-\(\hat { j } \)), its accleration is along (-\(\hat { j } \))
By comparing the components on both sides, we get
T – m₁g = -m₁a
m₁g – T = m₁a …………………….. (2)

Adding equations (1) and (2), we get
m₁g – m₂g = m₁a + m₂a
(m₁g – m₂)g = (m₁ + m₂)a …………………….. (3)

From equation (3), the acceleration of both the masses is
a = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)g ……………………… (4)

If both the masses are equal (m₁ = m₂), from equation (4)
a = 0

This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest. To find the tension acting on the string, substitute the acceleration from the equation (4)
T – m₂g = m₂ \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)g
T = m₂g + m₂\(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)g ……………………… (5)
By taking m₂g common in the RHS of equation (5)

Equation (4) gives only magnitude of accleration.

For mass m₁ g the accleration vector is given by \(\vec{a}=-\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) g \hat{j}\)
For mass m₂ g the accleration vector is given by \(\vec{a}=-\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) g \hat{j}\)

[OR]

(b) Derive the kinematic equation of motion for constant acceleration?
Answer:
Consider an object moving in a straight line with uniform or constant acceleration ‘a’. Let u be the velocity of the object at time t = 0, and v be velocity of the body at a later time t.

Velocity – time relation:

(I) The acceleration of the body at any instant is given by the first derivative of the velocity with respect to time,
a = \(\frac{dv}{dt}\) or dv = a dt
Integrating both sides with the condition that as time changes from 0 to t, the velocity changes from u to v. For the constant acceleration,

Displacement – time relation:

(II) The velocity of the body is given by the first derivative of the displacement with respect to time.
v = \(\frac{ds}{dt}\) or ds = vdt
and since v = u + at,
We get ds = (u + at)dt
Assume that initially at time t = 0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time-independent, we have
\(\int_{0}^{s} d s=\int_{0}^{t} u d t+\int_{0}^{t} a t d t \text { or } s=u t+\frac{1}{2} a t^{2}\) ……………………. (2)
Velocity – displacement relation

(III) The acceleration is given by the first derivative of velocity with respect to time.
a = \(\frac{dv}{dt}\) = \(\frac{dv}{ds}\) \(\frac{ds}{dt}\) = \(\frac{dv}{ds}\) v [since ds/dt = v] where s is displacement traversed.
This is rewritten as a = \(\frac{1}{2}\) \(\frac { dv^{ 2 } }{ s } \) or ds = \(\frac{1}{2a}\) d(v²)
Integrating the above equation, using the fact when the velocity changes from u² to v², displacement changes from u² to v², we get

We can also derive the displacement s in terms of initial velocity u and final velocity v.
From equation we can write,
at = v – u
Substitute this in equation, we get
s = ut + \(\frac{1}{2}\) (v -u)t
s = \(\frac{(u+v)t}{2}\)

Question 36 (a).
State and prove perpendicular axis theorem?
Answer:
Perpendicular axis theorem: This perpendicular axis theorem holds good only for plane laminar objects.

The theorem states that the moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two perpendicular axes lying in the plane of the body such that all the three axes are mutually perpendicular and have a common point.

Let the X and Y-axes lie in the plane and Z-axis perpendicular to the plane of the laminar object. If the moments of inertia of the body about X and Y-axes are Ix and IY respectively and Iz is the moment of inertia about Z-axis, then the perpendicular axis theorem could be expressed as,
I_z = I_x + I_y

To prove this theorem, let us consider a plane laminar object of negligible thickness on which lies the origin (O). The X and Y-axes lie on the plane and Z-axis is perpendicular to it as shown in figure. The lamina is considered to be made up of a large number of particles of mass m. Let us choose one such particle at a point P which has coordinates (x, y) at a distance r from O.

The moment of inertia of the particle about Z-axis is, mr².
The summation of the above expression gives the moment of inertia of the entire lamina about Z-axis as,
I_z = Σ mr²
Here, r² = x² + y²
Then, I_z = Σm(x² + y²)
I_z = Σmx² + Σmy²
In the above expression, the term Σmx² is the moment of inertia of the body about the Y-axis and similarly the term Σmy² is the moment of inertia about X-axis. Thus,
I_X = Σmy² and I_Y = Σmx²
Substituting in the equation for I_Z gives, I_Z = I_X + I_Y
Thus, the perpendicular axis theorem is proved.

[OR]

(b) Explain in detail the triangle law of addition?
Answer:
Let us consider two vectors \(\vec { A } \) and \(\vec { B } \) as shown in figure.

To find the resultant of the two vectors we apply the triangular law of addition as follows:

Represent the vectors \(\vec { A } \) and \(\vec { B } \) by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.

To explain further, the head of the first vector \(\vec { A } \) is connected to the tail of the second vector \(\vec { B } \). Let θ be the angle between A and B. Then R is the resultant vector connecting the tail of the first vector \(\vec { A } \) to the head of the second vector \(\vec { B } \).

The magnitude of \(\vec { R } \) (resultant) is given geometrically by the length of \(\vec { R } \)(OQ) and the direction of the resultant vector is the angle between \(\vec { R } \) and \(\vec { A } \). Thus we write \(\vec { R } \) = \(\vec { A } \) + \(\vec { B } \).
\(\overline { OQ } \) = \(\overline { OP } \) + \(\overline { PQ } \)

1. Magnitude of resultant vector:
The magnitude and angle of the resultant vector are determined by using triangle law of vectors as follows.

From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure, let R is the magnitude of the resultant of \(\vec { A } \) and \(\vec { B } \).

For ∆OBN, we have OB² = ON² + BN²
⇒ R² = (A + B cos θ)² + (B sin θ)²
⇒ R² = A² + B² cos² θ + 2AB cos θ + B²sin² θ
⇒ R² = A² + B²(cos² θ + sin² θ) + 2AB cos θ
⇒R² = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

2. Direction of resultant vectors:
If θ is the angle between \(\vec { A } \) and \(\vec { B } \), then
|\(\vec { A } \) + \(\vec { B } \)| = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)
If \(\vec { R } \) makes an angle α with \(\vec { A } \), then in ∆OBN,
tan α = \(\frac{BN}{ON}\) = \(\frac{BN}{OA+AN}\)
tan α = \(\frac{B \sin \theta}{A+B \cos \theta}\) ⇒ α = tan^-1\(\left(\frac{B \sin \theta}{A+B \cos \theta}\right)\)

Question 37 (a).
Explain in detail the various types of errors?
Answer:
The uncertainty in a measurement is called an error. Random error, systematic error and gross error are the three possible errors.

(I) Systematic errors:
Systematic errors are reproducible inaccuracies that are consistently in the same direction. These occur often due to a problem that persists throughout the .experiment. Systematic errors can be classified as follows.

(1) Instrumental errors:
When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.

(2) Imperfections in experimental technique or procedure:
These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied

(3) Personal errors:
These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.

(4) Errors due to external causes:
The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity, or pressure during measurements may affect the result of the measurement.

(5) Least count error:
Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument’s resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.

(II) Random errors:
Random errors may arise due to random and unpredictable variations in experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer who performs the experiment. Random errors are sometimes called “chance error”.

When different readings are obtained by a person every time he repeats the experiment, personal error occurs. For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.
If n number of trial readings are taken in an experiment, and the readings are
a₁, a₂, a₃, ……………………….. a_n. The arithmetic mean is

[OR]

(b) To move an object, which one is easier, push or pull? Explain?
Answer:
When a body is pushed at an arbitrary angle θ (0 to \(\frac{π}{2}\)), the applied force F can be resolved into two components as F sin θ parallel to the surface and F cos θ perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is no acceleration along the vertical direction the normal force N is equal to
N_push = mg + F cos θ …………………. (1)
As a result the maximal static friction also increases and is equal to
\(f_{s}^{\max }=\mu_{r} N_{\text {push }}=\mu_{s}(m g+F \cos \theta)\) …………………. (2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.

When an object is pulled at an angle 0, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is
N_pull = mg – F cos θ …………………….. (3)

Question 38 (a).
Describe the method of measuring angle of repose?
Answer:
Angle of Repose Consider an inclined plane on which an object is placed, as shown in figure. Let the angle which this plane makes with the horizontal be θ. For small angles of θ, the object may not slide down. As θ is increased, for a particular value of θ, the object begins to slide down. This value is called angle of repose. Hence, the angle of repose is the angle of inclined plane with the horizontal such that an object placed on it begins to slide.

Let us consider the various forces in action here. The gravitational force mg is resolved into components parallel (mg sin θ) and perpendicular (mg cos θ) to the inclined plane. The component of force parallel to the inclined plane (mg sin θ) tries to move the object down. The component of force perpendicular to the inclined plane (mg cos θ) is balanced by the Normal force (N).
N = mg cos θ ……………… (1)

When the object just begins to move, the static friction attains its maximum value
f_s = \(f_{s}^{\max }\)

This friction also satisfies the relation
\(f_{s}^{\max }\) = µ_s = mg sin θ …………………… (2)

Equating the right hand side of equations (1) and (2),
\(\left(f_{s}^{\max }\right) / \mathrm{N}=\sin \theta / \cos \theta\)

From the definition of angle of friction, we also know that
tan θ = µ_s ……………………… (3) in which θ is the angle of friction.

Thus the angle of repose is the same as angle of friction. But the difference is that the angle of repose refers to inclined surfaces and the angle of friction is applicable to any type of surface.

[OR]

(b) A block of mass m slides down the plane inclined at an angle 60° with an acceleration g/2. Find the co-efficient of kinetic friction?
Answer:
Kinetic friction comes to play as the block is moving on the surface.
The forces acting on the mass are the normal force perpendicular to surface, downward gravitational force and kinetic friction along the surface.
Along the x-direction

There is no motion along the y-direction as normal force is exactly balanced by the mg cos θ.
mg cos θ = N = mg/2

[OR]

(c) Write a note on triangulation method and radar method to measure larger distances?
Answer:
Triangulation method for the height of an accessible object:
Let AB = h be the height of the tree or tower to 6e measured. Let C be the point of observation at distance x from B. Place a range finder at C and measure the angle of elevation, ACB = θ as shown in figure.

From right angled triangle ABC,
tan θ = \(\frac{AB}{BC}\) = \(\frac{h}{x}\)
(or) height h = x tan θ
Knowing the distance x. the height h can be detennined.

RADAR method:
The word RADAR stands for radio detection and ranging. A radar can be used to measure accurately the distance of a nearby planet such as Mars. In this method, radio waves are sent from transmitters which, after reflection from the planet, are detected by the receiver.

By measuring, the time interval (t) between the instants the radio waves are sent and received, the distance of the planet can be determined as where v is the speed of the radio wave. As the time taken (t) is for the distance covered during the forward and backward path of the radio waves, it is divided by 2 to get the actual distance of the object. This method can also be used to determine the height, at which an aeroplane flies from the ground.

[OR]

(d) Jupiter is at a distance of 824.7 million km from the earth. Its angular diameter is measured to be 35.75”. Calculate the diameter of jupiter?
Answer:
Given,
Distance of Jupiter = 824.7 × 10⁶ km = 8.247 × 10¹¹ m
angular diameter = 35.72 × 4.85 × 10^-6 rad = 173.242 × 10^-6 rad
= 1.73 × 10^-4rad
∴ Diameter of Jupiter D = θ × d = 1.73 × 10^-4 rad × 8.247 × 10¹¹ m
14.267 × 10⁷ m = 1.427 × 10⁸ m (or) 1.427 × 10⁵ km

Students can download 6th Social Science Term 3 History Chapter 3 The Age of Empires: Guptas and Vardhanas Questions and Answers, Notes, Samacheer Kalvi 6th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Social Science History Solutions Term 3 Chapter 3 The Age of Empires: Guptas and Vardhanas

Samacheer Kalvi 6th Social Science The Age of Empires: Guptas and Vardhanas Text Book Back Questions and Answers

I. Choose the correct Answer

Question 1.
………………. was the founder of Gupta dynasty.
(a) Chandragupta I
(b) Sri Gupta
(c) Vishnu Gopa
(d) Vishnugupta
Answer:
(b) Sri Gupta

Question 2.
Prayog Prashasti was composed by ________
(a) Kalidasa
(b) Amarasimha
(c) Harisena
(d) Dhanvantri
Answer:
(c) Harisena

Question 3.
The monolithic iron pillar of Chandragupta is at ………………
(a) Mehrauli
(b) Bhitari
(c) Gadhva
(d) Mathura
Answer:
(a) Mehrauli

Question 4.
________ was the first Indian to explain the process Of surgery.
(a) Charaka
(b) Sushruta
(c) Dhanvantri
(d) Agnivasa
Answer:
(b) Sushrutal

Question 5.
……………… was the Gauda ruler of Bengal.
(a) Sasanka
(b) Maitraka
(c) Rajavardhana
(d) Pulikesin II
Answer:
(a) Sasanka

II. Match the statement with the reason and tick the appropriate answer

Question 1.
Assertion (A) : Chandragupta I crowned himself as a monarch of a large kingdom after eliminating various small states in Northern India.
Reason (R) : Chandragupta I married Kumaradevi of Lichchavi family.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) A is correct but R is not correct.
(d) A is not correct but R is correct.
Answer:
(a) Both A and R are true and R is the correct explanation of A

Question 2.
Statement I : Chandragupta II did not have cordial relationship with the rules of South India.
Statement II : The divine theory of kingship was practised by the Gupta rulers.
(a) Statement I is wrong but statement II is correct.
(b) Statement II is wrong but statement I is correct.
(c) Both the statements are correct.
(d) Both the statements are wrong.
Answer:
(a) Statement I is wrong but statement II is correct.

Question 3.
Which of the following is arranged in chronological order?
(a) Srigupta – Chandragupta I – Samudragupta – Vikramaditya
(b) Chandragupta I – Vikramaditya – Srigupta – Samudragupta
(c) Srigupta – Samudragupta – Vikramaditya – Chandragupta I
(d) Vikramaditya – Srigupta – Samudragupta – Chandragupta I
Answer:
(a) Srigupta – Chandragupta I – Samudragupta -Vikramaditya

Question 4.
Consider the following statements and find out which of the following statements (s) is/are correct.
(1) Lending money at high rate of interest was practised.
(2) Pottery and mining were the most flourishing industries,
(a) 1. is correct
(b) 2. is correct
(c) Both 1 and 2 are correct
(d) Both 1 and 2 are wrong
Answer:
(a) 1. is correct

Question 5.
Circle the odd one

Answer:
Samudragupta.

Answer:
Harshacharita.

III. Fill in the blanks Answer

1. ……………., the king of Ceylon, was a contemporary of Samudragupta.
2. A Buddhist monk from China ……………., visited India during the reign of Chandragupta II.
3. ……………. invasion led to the downfall of the Gupta Empire.
4. ……………. was the main revenue to the Government.
5. The official language of the Guptas was …………….
6. ……………., the Pallava king was defeated by Samudragupta.
7. ……………. was the popular king of the Vardhana dynasty.
8. Harsha shifted his capital from ……………. to Kanauj.

Answer:

1. Reign of
2. Fahien
3. Huns
4. Land tax
5. Sanskrit
6. Vishnugopa
7. Harsha Vardhana
8. Thaneswar

IV. State whether True or False

1. Dhanvantri was a famous scholar in the field of medicine.
2. The structural temples built during the Gupta period resemble the Indo – Aryan style.
3. Sati was not in practice in the Gupta Empire.
4. Harsha belonged to the Hinayana school of thought.
5. Harsha was noted for his religious intolerance.

Answer:

1. True
2. False
3. False
4. False
5. False

V. Match the following

Question 1.

Answer:
b) 2, 4, 1, 3, 5

Question 2.

Answer:
c) 3, 5, 1, 2, 4

VI. Answer in one or two sentences

Question 1.
Who was given the title Kaviraja? Why?
Answer:

1. The title Kaviraja was given to Samudragupta.
2. He was a great lover of poetry and music.
3. In one of the gold coins, he is portrayed playing the harp (Veenai)

Question 2.
What were the subjects taught at Nalanda University?
Answer:

1. At Nalanda University Buddhism was the main subject of study.
2. Other subjects like Yoga, Vedic literature, and medicine were also taught.

Question 3.
Explain the Divine Theory of Kingship.
Answer:

1. The divine theory of Kingship meant that the king is the representative of God on earth.
2. He is answerable only to God and not to anyone else.

Question 4.
Highlight the achievements of Guptas in metallurgy.
Answer:

1. Mining and metallurgy were the most flourishing industries during the Gupta period.
2. The most important evidence of development in metallurgy was the Mehrauli Iron Pillar installed by King Chandragupta in Delhi.
3. This monolithic iron pillar has lasted through the centuries without rusting.

Question 5.
Who were the Huns?
Answer:

1. Huns were the nomadic tribe, who under their great Attila were terrorizing Rome and Constantinople.
2. They came to India through Central Asia, defeated Skandagupta, and spread across central India.
3. Their chief Toromana crowned himself asking.
4. After him, his son Mihirakula ruled and got finally defeated by Yasodharman, ruler of Malwa.

Question 6.
Name the three kinds of tax collected during Harsha’s reign.
Answer:
A Bhaga, Hiranya, and Bali were three kinds of tax collected during Harsha’s reign.

Question 7.
Name the books authored by Harsha.
Answer:
The books authored by Harsha were Ratnavali, Nagananda, and Priyadharshika.

VII. Answer the following briefly

Question 1.
Write a note on Prashasti.
Answer:

1. Prashasti is a Sanskrit word, meaning communication or in praise of.
2. Court poets flattered their kings listing out their achievements.
3. These accounts were later engraved on pillars so that the people could read them.

Question 2.
Give an account of Samudragupta’s military conquests.
Answer:

1. Samudragupta was a great general and he carried on a vigorous campaign all over the country.
2. He defeated the Pallava king Vishnugopa.
3. He conquered nine kingdoms in northern India.
4. He reduced 12 rulers of southern India to the status of feudatories and to pay tribute.
5. He received homage from the rulers of East Bengal, Assam, Nepal, the eastern part of Punjab, and various tribes of Rajasthan.

Question 3.
Describe the land classification during the Gupta period.
Answer:
Classification of land during the Gupta period.

1. Kshetra – Cultivable land
2. Khila – Wasteland
3. Aprahata – Jungle (or) Forest land
4. Vasti – Habitable land
5. Gapata saraha – Pastoral and

Question 4.
Write about Sresti and Sarthavaha traders.
Answer:

1. Sresti: Sresti traders were usually settled at a standard place.
2. Sarthavaha: Sarthavaha traders caravan traders who carried their goods to different places.

Question 5.
Highlight the contribution of Guptas to architecture.
Answer:

1. From the earlier tradition of rock – out shrines, the Guptas were the first to construct temples.
2. These temples, adorned with towers and elaborate carvings, were dedicated to all Hindu deities.
3. The most notable rock-cut caves are found at Ajanta and Ellora, Bagh, and Udaygiri.
4. The structural temples built during this period resemble the Dravidian style.

Question 6.
Name the works of Kalidasa.
Answer:

1. Kalidasa’s famous dramas were Sakunthala, Malavikagnimitra and Vikramaoorvashiyam.
2. Other significant works were Meghaduta, Raghuvamsa, Kumarasambava and Ritusamhara

Question 7.
Estimate Harshvardhana as a poet and a dramatist.
Answer:

1. Harsha himself was a poet and dramatist.
2. Around him gathered the best of poets and artists.
3. His popular works are Ratnavali, Nagananda and Priyadharshika
4. is royal court was adorned by Banabhatta, Mayura, Hardatta, and Jayasena.

VIII. HOTs

Question 1.
The gold coins issued by Gupta kings indicate ………………
Answer:
(a) the availability of gold mines in the kingdom
(b) the ability of the people to work with gold
(c) the prosperity of the kingdom
(d) the extravagant nature of kings.
Answer:
(c) the prosperity of the kingdom

Question 2.
The famous ancient paintings at Ajanta were painted on __________
a. walls of caves
b. ceilings of temples
c. Rocks
d. papyrus
Answer:
a. walls of caves

Question 3.
Gupta period is remembered for ……………….
(a) renaissance in literature and art
(b) expeditions to southern India
(c) invasion of Huns
(d) religious tolerance
Answer:
(a) renaissance in literature and art

Question 4.
What did Indian scientists achieve in astronomy and mathematics during the Gupta period?
Answer:

1. The invention of zero and the consequent evolution of the decimal system was the legacy of Guptas to the modem world.
2. Aryabhatta, Varahamihira, and Brahmagupta were the foremost astronomers and mathematicians of the time.
3. Aryabhatta, in his book ‘ Surya Siddhanta’, explained the true causes of solar and lunar eclipses.
4. He was the first Indian astronomer to declare that the earth revolves around its own axis.
5. Dhanvantri was a famous scholar in the field of medicine.
6. He was a specialist in Ayurveda.
7. Charaka was a medical scientist.
8. Susruta was the first Indian to explain the process of surgery.

IX. Student activity (For Students)

1. Stage any one of the dramas of Kalidasa in the classroom.
2. Compare and contrast the society of Guptas with that of Mauryas.

X. Life Skills (For Students)

1. Collect information about the contribution of Aryabhatta, Varahamihira and Brahmagupta to astronomy.
2. Visit a nearby ISRO centre to know more about satellite launching.

XI. Answer Grid

Question 1.
Who was Toromana?
Answer:
Toromana was the chief of White Huns.

Question 2.
Name the high-ranking officials of the Gupta Empire.
Answer:
Dandanayakas and Maha dandanayakas.

Question 3.
Name the Gupta kings who performed AsVamedha yagna.
Answer:
Samudragupta and Kumaragupta I

Question 4.
Name the book which explained the causes for the lunar and solar eclipses,
Answer:
Surya Siddhanta

Question 5.
Name the first Gupta king to find a place on coins.
Answer:
Samudragupta

Question 6.
Which was the main source of information to know about the Samudragupta’s reign?
Answer:
Allahabad Pillar

Question 7.
Harsha was the worshipper in the beginning.
Answer:
Shiva

Question 8.
University reached its fame during the Harsha period.
Answer:
The Nalanda

Samacheer Kalvi 6th Social Science The Post-Mauryan India Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
The successor of Sri Gupta ……………
(a) Kumaragupta I
(b) Skandagupta
(c) Vishnugupta
(d) Ghatotkacha
Answer:
(d) Ghatotkacha

Question 2.
Sri Gupta was succeeded by _______
(a) Chandra Gupta
(b) Samundra Gupta
(c) Ghatotkacha
(d) Skanda Gupta
Answer:
(c) Ghatotkacha

Question 3.
The Huhs chief crowned himself as king.
(a) Yasodharman
(b) Attila
(c) Mihirakula
(d) Toromana
Answer:
(d) Toromana

Question 4.
Srimeghavarman was the ruler of _______
(a) Singapore
(b) Ceylon
(c) Malaysia
(b) Hansena
Answer:
(b) Ceylon

Question 5.
The place Harsha went to participate in the great Kumbhamela held.
(a) Allahabad
(b) Kasi
(c) Ayodhya
(d) Prayag
Answer:
(d) Prayag

II. Match the statement with the reason and tick the appropriate answer

Question 1.
Assertion (A) : The last of the great Guptas Narasimha Gupta I was paying tribute to Mihirakula.
Reason (R) : He stopped paying tribute to Mihirakula’s hostility towards Buddhism.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is not correct
(d) A is not correct but R is correct
Answer:
(b) Both A and R are correct but R is not the correct explanation of A

Question 2.
Statement I : Criminal law was not more severe than that of the Gupta age.
Statement II : Death punishment was the punishment for violation of the laws and for plotting against the king.
(a) Statement I is wrong but statement II is correct
(b) Statement II is wrong but statement I is correct.
(c) Both the statements are correct.
(d) Both the statements are wrong.
Answer:
d) Both the statements are wrong

III. Fill in the blanks

1. In the assembly at ……………. Harsha distributed his wealth.
2. The capital of China ……………. was a great center of art and learning.
3. ……………. was the wife of Chandragupta I.
4. The military campaigns of kings were financed through revenue ……………..
5. The peasants were required to pay various taxes and were reduced to the position of ……………..

Answer:

1. Prayag
2. Xian
3. Kumaradevi
4. Surpluses
5. serfs

IV. Match the following

Answer:
b) 4, 5, 2, 1, 3

V. Answer in one or two sentences

Question 1.
Write a note on ‘Lichchhavi’.
Answer:

1. Lichchhavi was an old gana – Sanga and its territory lay between the Ganges and the Nepal Terai.
2. Chandragupta, I married Kumaradevi of the famous and powerful lichchhavi family.

Question 2.
How did Chandragupta I crown himself the monarch of a larger kingdom?
Answer:

1. Chandragupta, I married Kumaradevi of the famous and powerful Lichchhavi family.
2. With the support of this family, Chandragupta eliminated various small states and| crowned himself the monarch of a larger kingdom.

Question 3.
What did the travel accounts of Fahien provide information about the condi¬tions of the people of Magadha?
Answer:

1. According to Fahien the people of Magadha were happy and prosperous.
2. Gaya was desolated. Kapilvasthu had become a jungle, but at Pataliputra people were rich and prosperous.

VII. Answer the following briefly

Question 1.
Name the officials employed by the Gupta rulers.
Answer:

1. High – ranking officials were called dandanayakas and mahadandnayakas.
2. The provinces known as deshas or bhuktis were administered by the governors designated as Uparikas. The districts such as vaishyas, were controlled by vishyapatis. At the village level, gramika and gramadhyaksha were the functionaries.
3. The military designations.
4. Baladhikrita (Commander of infantry)
5. Mahabaladhikrita (Commander of the cavalry)
6. Dutakas (spies)

Question 2.
Mention the importance of Fahien’s travel accounts.
Answer:

1. During the reign of Chandragupta II, the Buddhist monk Fahien visited India.
2. His travel accounts provided us information about the socio-economic, religious and moral conditions of the people of the Gupta age.
3. According to Fahien, the people of Magadha were happy and prosperous.
4. Justice was mildly administered and there was no death penalty.
5. Gaya was desolated, Kapilavasthu had become a jungle, but at Pataliputra, people were rich and prosperous.

VIII. Mind map

Students can Download Tamil Nadu 11th Maths Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Maths Model Question Paper 4 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 90

PART – I

I. Choose the correct answer. Answer all the questions: [20 × 1 = 20]

Question 1.
If n(A) = 2 and n(B∪C) = 3 then n[(A × B) ∪ (A × C)] is ………………..
(a) 2³
(b) 3²
(c) 6
(d) 5
Answer:
(c) 6

Question 2.
For any two sets A and B, A∩(A∪B) = …………………….
(a) B
(b) ∅
(c) A
(d) none of these
Answer:
(c) A

Question 3.
cos 1° + cos 2° + cos 3° + cos 4° + cos 179° = …………………
(a) 0
(b) 1
(c) -1
(d) 89
Answer:
(a) 0

Question 4.
The value of log₉ 27 is ……………………
(a) \(\frac{2}{3}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{3}{4}\)
(d) \(\frac{4}{3}\)
Answer:
(b) \(\frac{3}{2}\)

Question 5.
The value of \(\frac{\sin 3 \theta+\sin 5 \theta+\sin 7 \theta+\sin 9 \theta}{\cos 3 \theta+\cos 5 \theta+\cos 7 \theta+\cos 9 \theta}\) = ………………..
(a) tan3θ
(b) tan6θ
(c) cot3θ
(d) cot6θ
Answer:
(b) tan6θ

Question 6.
In 3 fingers the number of ways 4 rings can be worn in ……………………. ways.
(a) 43 – 1
(b) 34
(c) 68
(d) 64
Answer:
(d) 64

Question 7.
Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is ……………….
(a) 11
(b) 12
(c) 10
(d) 6
Answer:
(b) 12

Question 8.
The H.M of two positive number whose AM and G.M. are 16, 8 respectively is ………………..
(a) 10
(b) 6
(c) 5
(d) 4
Answer:
(d) 4

Question 9.
The co-efficient of the term independent of x in the expansion of (2x+\(\frac{1}{3x}\))⁶ is …………………
(a) \(\frac{160}{27}\)
(b) \(\frac{160}{27}\)
(c) \(\frac{80}{3}\)
(d) \(\frac{80}{9}\)
Answer:
(a) \(\frac{160}{27}\)

Question 10.
The value of \(\left|\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right|^{2}\) is ………………..
(a) abc
(b) -abc
(c) 0
(d) a²b²c²
Answer:
(d) a²b²c²

Question 11.
The value of x for which the matrix A = \(\left[\begin{array}{cc}
e^{x-2} & e^{7+x} \\
e^{2+x} & e^{2 x+3}
\end{array}\right]\) is singular is …………………..
(a) 9
(b) 8
(c) 7
(d) 6
Answer:
(b) 8

Question 12.
If |\(\vec { a } \) + \(\vec { b } \)| = 60, |\(\vec { a } \) – \(\vec { b } \)| = 40 and |\(\vec { b } \)| = 46 then |\(\vec { a } \)| is …………………
(a) 42
(b) 12
(c) 22
(d) 32
Answer:
(c) 22

Question 13.
Given \(\vec { a } \) = 2\(\vec { i } \) + \(\vec { j } \) – 8\(\vec { k } \) and \(\vec { b } \) = \(\vec { i } \) + 3\(\vec { j } \) – 4\(\vec { k } \) then |\(\vec { a } \) + \(\vec { b } \)| = ………………….
(a) 13
(b) \(\frac{13}{3}\)
(c) \(\frac{4}{13}\)
(d) \(\frac{3}{13}\)
Answer:
(a) 13

Question 14.
If f(x) = \(\left\{\begin{array}{ccc}
k x & \text { for } & x \leq 2 \\
3 & \text { for } & 2
\end{array}\right.\) is continous at x = 2 then the value of k is ……………………
(a) \(\frac{3}{4}\)
(b) 0
(c) 1
(d) \(\frac{4}{3}\)
Answer:
(c) 1

Question 15.
If f: R→R is defined by f(x) = |x – 3| + |x – 4| for x∈R then \(\lim _{x \rightarrow 3^{-}}\) f(x) is equal to ………………..
(a) -2
(b) -1
(c) 0
(d) 1
Answer:
(c) 0

Question 16.
\(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+5 x+3}{x^{2}+x+3}\right)^{x}\) is ………………..
(a) e⁴
(b) e²
(c) e³
(d) 1
Answer:
(a) e⁴

Question 17.
\(\int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1}\) dx is ………………..
(a) e tan^-1(x + 1)
(b) tan^-1(e^x) + c
(c) e^x \(\frac{\left(\tan ^{-1} x\right)^{2}}{2}\) + c
(d) e^xtan^-1x + c
Answer:
(d) e^xtan^-1x + c

Question 18.
∫ \(\frac { secx }{ \sqrt { cos2x } } \) dx = …………………..
(a) tan^-1(sin x) + c
(b) 2 sin^-1(tan x) + c
(c) tan^-1(cos x) + c
(d) sin^-1 (tan x) + c
Answer:
(d) sin^-1 (tan x) + c

Question 19.
\(\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\) dx = ………………..
(a) x + c
(b) \(\frac { x^{ 3 } }{ 3 } \) + c
(c) \(\frac { 3 }{ x^{ 3 } } \) + c
(d) \(\frac { 1 }{ x^{ 2 } } \) + c
Answer:
(b) \(\frac { x^{ 3 } }{ 3 } \) + c

Question 20.
It is given that the events A and B are such that P(A) = \(\frac{1}{4}\), P(A/B) = \(\frac{1}{2}\), P(B/A) = \(\frac{2}{3}\) then P(B) = …………………
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{2}\)
Answer:
(b) \(\frac{1}{3}\)

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
In the set Z of integers, define mRn if m – n is a multiple of 12. Prove that R is an equivalence relation?
Answer:
As m – m = 0 and 0 = 0 × 12, we have 0 is a multiple of 12; hence mRm proving that R is reflexive.
Let mRn. Then m – n = 12k for some integer k; thus n – m = 12(-k) and hence nRm.
This shows that R is symmetric.
Let mRn and nRp: then m – n = 12k and n – p = 12l for some integers k and l.
So m – p = 12(k + l) and hence mRp. This shows that R is transitive.
Thus R is an equivalence relation.

Question 22.
Simplify:
\(\frac { 1 }{ 2+\sqrt { 3 } } \) + \(\frac { 3 }{ 4-\sqrt { 5 } } \) + \(\frac { 6 }{ 7-\sqrt { 8 } } \)
Answer:

Question 23.
Find the value of sin 22 \(\frac{1}{2}\)°?
Answer:
We know that cos θ = 1 – 2 sin² \(\frac{θ}{2}\) ⇒ sin \(\frac{θ}{2}\) = ±\(\sqrt { \frac { 1-cos2\theta }{ 2 } } \)
Take θ = 45°, we get sin \(\frac{45°}{2}\) = ±\(\sqrt { \frac { 1-cos45°}{ 2 } } \), (taking positive sign only, since 22\(\frac{1}{2}\)° lies in the first quadrant)
Thus, sin 22\(\frac{1}{2}\)° = \(\sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}\).

Question 24.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th
hour and wth hour?
Answer:
No. of bacteria at the beginning = 30
No. of bacteria after 1 hour = 30 × 2 = 60
No. of bactena after 2 hours = 30 × 2² = 30 × 4 = 120
No. of bacteria after 4 hours = 30 × 2⁴ = 30 × 16 = 480
No. of bacteria after n^th hour = 30 × 2ⁿ

Question 25.
If the sum of the distance of a moving point in a plane from the axis is 1, then find the locus of the point?
Answer:
Let coordinates of a moving point P be (x, y).
Given that the sum of the distances from the axis to the point is always 1.
∴|x| + |y| = 1 ⇒ x + y = 1
⇒- x- y = 1 ⇒ x + y = 1
⇒ x – y = 1
Hence, these equations give us the locus of the point P which is a square.

Question 26.
If (a, a + b, a + b + c) is one set of direction ratios of the line joining (1, 0, 0) and (0, 1, 0) then find a set of values of a, b, c?
Answer:
Let A be the point (1, 0, 0) and B be the point (0, 1, 0) (i.e.,) \(\overline { OA } \) = \(\hat { i } \) and \(\overline { OB } \) = \(\hat { j } \).
Then \(\overline { AB } \) = \(\overline { OB } \) – \(\overline { OA } \) = \(\hat { j } \) – \(\hat { i } \) = –\(\hat { i } \) + \(\hat { j } \)
= (-1, 1, 0)
= (a, a + b, a + b + c)
⇒ a = -1, a + b = 1 and a + b + c = 0
⇒ a = -1, ⇒ -1 + b = 1; a + b + c = 0
⇒ b = 2; -1 + 2 + c = 0 ⇒ c + 1 = 0
⇒ c = -1
∴ a = -1, b = 2, c = -1.
Note: if we taken \(\overline { BA } \) then we get a = 1, b = -2 and c = 1.

Question 27.
Find \(\frac{dy}{dx}\) for y = (x² + 4x + 6)⁵
Answer:
Let u = x² + 4x + 6
⇒ \(\frac{du}{dx}\) = 2x + 4
Now y = u⁵ = \(\frac{dy}{du}\) = 5u⁴
∴ \(\frac{dy}{dx}\) = \(\frac{dy}{du}\) × \(\frac{du}{dx}\) = 5u⁴ (2x + 4)
= 5(x² + 4x + 6)⁴ (2x + 4)
= 5(2x + 4) (x² + 4x + 6)⁴

Question 28.
Evaluate ∫\(\sqrt { 25x^{ 2 }-9 } \) dx?
Answer:

Question 29.
A bag contains 5 white and 7 black balls. 3 balls are drawn at random. Find the probability that

1. all are white
2. one white and 2 black.

Answer:
Number of white balls = 5
Number of black balls = 7
Total number of balls = 12
Selecting 3 from 12 balls can be done in
¹²C₃ = \(\frac{12 \times 11 \times 10}{3 \times 2 \times 1}\) = 220 ways
∴n(S) = 220

1. Let A be the selecting 3 white balls.
∴n(A) = ⁵C₃ = ⁵C₂ = \(\frac{5×3}{2×1}\) = 10
∴P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{10}{220}\) = \(\frac{1}{22}\)

2. Let B be the event of selecting one white and 2 black balls.
∴n(B) = ⁵C₁ × ⁷C₂ = (5) (\(\frac{7×6}{2×1}\)) = 5(21) = 105
∴P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{105}{220}\) = \(\frac{21}{44}\).

Question 30.
If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k?
Answer:
Area of ∆ with vertices (k, 2) (2, 4) and (3, 2) = \(\frac{1}{2}\) \(\left|\begin{array}{lll}
k & 2 & 1 \\
2 & 4 & 1 \\
3 & 2 & 1
\end{array}\right|\) = 4 given
⇒ \(\left|\begin{array}{lll}
k & 2 & 1 \\
2 & 4 & 1 \\
3 & 2 & 1
\end{array}\right|\) = 2(4) = 8
(i.e.,) k(4 – 2) – 2(2 – 3) + 1(4 – 12) ± 8
(i.e.,) 2k – 2(-1) + 1(-8) = ± 8
(i.e.,) 2k + 2 – 8 = 8
(i.e.,) 2k = 8 + 8 – 2 = 14
k = 14/2 = 7
∴k = 7
So k = 7 (or) k = -1.

2k + 2 – 8 = -8
⇒2k = – 8 + 8 – 2
2k = – 2
k = -1

PART – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
In the set Z of integers define mRn if m – n is a multiple of 12. Prove that R is an equivalence relation?

Question 32.
Prove that \(\frac{sin4x+sin2x}{cos4x+cos2x}\) = tan 3x?

Question 33.
A polygon has 90 diagonals. Find the number of its sides?

Question 34.
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)ⁿ are equal?

Question 35.
Find the equation of the line passing through the point (1, 5) and also divides the co-ordinate axes in the ratio 3 : 10?

Question 36.
Prove that \(\left|\begin{array}{ccc}
1 & x & x \\
x & 1 & x \\
x & x & 1
\end{array}\right|^{2}=\left|\begin{array}{ccc}
1-2 x^{2} & -x^{2} & -x^{2} \\
-x^{2} & -1 & x^{2}-2 x \\
-x^{2} & x^{2}-2 x & -1
\end{array}\right|\)

Question 37.
If G is the centroid of a traiangle ABC prove that \(\overline { GA } \) + \(\overline { GB } \) + \(\overline { GC } \) = 0

Question 38.
Find \(\frac{dy}{dx}\) for y = \(\sqrt { 1+tan2x } \)?

Question 39.
Evaluate \(\frac { \sqrt { x } }{ 1+\sqrt { x } } \) dx?

Question 40.
Find the relation between a and b if \(\underset { x\rightarrow 3 }{ lim } \) f(x) exists where f(x) = \(\left\{\begin{array}{cc}
a x+b & \text { if } x>3 \\
3 a x-4 b+1 \text { if } x<3
\end{array}\right.\)

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41 (a).
From the curve y = |x|, draw(i) y = |x-1| + 1

1. y = |x + 1| + 1
2. y = |x + 2| – 3

[OR]

(b) Resolve into partial fraction \(\frac { x+4 }{ (x^{ 2 }-4)(x+1) } \)

Question 42 (a).
Find the number of positive integers greater than 6000 and less than 7000 which arc divisible by 5, provided that no digit is to be repeated?

[OR]

(b) If ⁿP_r = ⁿP_r+1 and ⁿC_r = ⁿC_r-1, find the values of n and r?

Question 43 (a).
In a ∆ABC, prove that b² sin 2C + c² sin 2B = 2bc sin A?

[OR]

(b) Differentiate the following s(t) = \(\sqrt[4]{\frac{t^{3}+1}{t^{3}-1}}\)?

Question 44 (a).
Find the equation of the lines make an angle 60° with the positive x axis and at a distance 5\(\sqrt{2}\) units measured from the point (4, 7) along the line x – y + 3 = 0

[OR]

(b) If y = A cos4x + B sin 4x, A and B are constants then Show that y₂ + 16y = 0

Question 45 (a).
Find the sum up to the 17^th term of the series \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}\) + …………..

[OR]

(b) A shopkeeper in a Nuts and Spices shop makes gifi packs of cashew nuts, raisins and almonds?

1. Pack I contains 100 gm of cashew nuts, 100 gm of raisins and 50 gm of almonds.
2. Pack-II contains 200 gm of cashew nuts, 100 gm of raisins and 100 gm of almonds.
3. Pack-III contains 250 gm of cashew nuts, 250 gm of raisins and 150 gm of almonds.
4. The cost of 50 gm of cashew nuts is ₹50, 50 gm of raisins is ₹10. and 50gm of almonds is ₹60. What is the cost of each gift pack?

Question 46 (a).
Find matrix C if A = \(\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\), B = \(\begin{bmatrix} -3 & 2 \\ 4 & -1 \end{bmatrix}\) and 5C + 28= A?

[OR]

(b) The probability that a new railway bridge will get an award for its design is 0.48, the probability that it will get an award for the efficient use of materials is 0.36, and that it will get both awards is 0.2. What is the probability, that

1. it will get at least one of the two awards
2. it will get only one of the awards.

Question 47 (a).
\(\underset { \alpha \rightarrow 0 }{ lim } \) \(\frac { sin(\alpha ^{ n }) }{ (sin\alpha )^{ m } } \)

[OR]

(b) Evaluate I = sin^-1 (\(\frac { 2x }{ (1+x)^{ 2 } } \)) dx?

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 1 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is …………………….
(a) 0%
(b) 4.4%
(c) 16%
(d) 8.4%
MgCO₃ → MgO + CO₂↑
MgCO₃: (1 × 24) + (1 × 12) + (3 × 16) = 84g
CO₂: (1 × 12) + (2 × 16) = 44g
100% pure 84 g MgCO₃ on heating gives 44 g CO₂
Given that 1 g of MgCO₃ on heating gives 0.44 g CO₂
Therefore, 84 g MgCO₃ sample on heating gives 36.96 g CO₂
Percentage of purity of the sample = \(\frac { 100% }{ 44gCO_{ 2 } } \) × 36.96 g CO₂ = 84%
Percentage of impurity = 16%
Answer:
(c) 16%

Question 2.
The electronic configuration of Eu (atomic no. 63) Gd (atomic no. 64) and Tb (atomic no. 65) are …………………….. [NEET – Phase II]
(a) [Xe] 4f⁶ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁸ 5d¹ 6s²
(b) [Xe] 4f⁷ 6s² [Xe] 4f⁸ 6s² and [Xe] 4f⁹ 6s²
(c) [Xe] 4f⁷ 6s² [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
(d) [Xe] 4f ⁶ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
Solution:
Eu : [Xe] 4f⁷, 5d⁰, 6s²
Gd : [Xe] 4f⁷, 5d¹, 6s²
Tb : [Xe] 4f⁹, 5d⁰, 6s²
Answer:
(b) [Xe] 4f⁷ 6s² [Xe] 4f⁸ 6s² and [Xe] 4f⁹ 6s²

Question 3.
Match the following:

Answer:

Question 4.
Various successive ionization enthalpies (in kJ mol^-1) of an element are given below.

The element is ……………………….
(a) Phosphorus
(b) Sodium
(c) Aluminium
(d) Silicon
Answer:
(c) Aluminium

Question 5.
Non-stoichiometric hydrides are formed by ……………………..
(a) Palladium, vanadium
(b) Carbon, nickel
(c) Manganese, lithium
(d) Nitrogen, chlorine
Answer:
(a) Palladium, vanadium

Question 6.
Which is the function of sodium – potassium pump?
(a) Maintenance of ion balance
(b) Used in nerve impulse conduction
(c) Transmitting nerve signals
(d) Regulates the blood level
Answer:
(c) Transmitting nerve signals

Question 7.
∆S is expected to be maximum for the reaction ………………………
(a) Ca_(s) + 1/2O_2(g) → CaO_(s)
(b) C_(s) + O_2(g) → CO_2(g)
(c) N_2(g) + O_2(g) → 2NO_(g)
(d) CaCO_3(s) → CaO_(s) + CO_2(g)
Solution:
In CaCO_3(s) → CaO_(s) + CO_2(g) entropy change is positive. In (a) and (b) entropy change is negative; in (c) entropy change is zero.
Answer:
(d) CaCO_3(s) → CaO_(s) + CO_2(g)

Question 8.
Which one of the following is a reversible reaction?
(a) Ripening of a banana
(b) Rusting of iron
(c) Tarnishing of silver
(d) Transport of oxygen by Hemoglobin in our body
Solution:
All the other three reactions are irreversible reactions. But the hemoglobin combines with O₂ in lungs to form oxyhemoglobin. The oxyhemoglobin has a tendency to form hemoglobin by releasing O₂. So it is a reversible reaction.
Answer:
(d) Transport of oxygen by Hemoglobin in our body

Question 9.
P₁ and P₂ are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x₁ represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be ………………………….
Solution:
(a) P₁ + x₁ (P₂ – P₁)
(b) P₂ – x₁ (P₂ + P₁)
(c) P₁ – x₂ (P₁ – P₂)
(d) P₁ + x₂ (P₁ – P₂)
P_total = P₁ + P₂
= P₁x₁ + P₂ x₂
= P₁(1 – x₂) + P₂x₂ [∵x₁ + x₂ = 1, x₁ = 1 – x₂]
= P₁ – P₁x₂ + P₂x₂ = P₁ – x₂ = P₁ – x₂ (P₁ – P₂)
Answer:
(c) P₁ – x₂ (P₁ – P₂)

Question 10.
Which of the following molecule contain no n bond?
(a) SO₂

(b) NO₂

(c) CO₂

(d) H₂O

Solution:
Water (H₂O) contains only σ bonds and no π bonds.
Answer:
(d) H₂O

Question 11.
IUPAC name of  is …………………….
(a) Trimethylheptane
(b) 2 -Ethyl -3, 3- dimethyl heptane
(c) 3, 4, 4 – Trimethyloctane
(d) 2 – Butyl -2 -methyl – 3 – ethyl-butane.
Answer:
(c) 3, 4, 4 – Trimethyloctane

Question 12.
Which one of the following is an example for free radical initiators?
(a) Benzoyl peroxide
(b) Benzyl alcohol
(c) Benzyl acetate
(d) Benzaldehyde.
Answer:
(a) Benzoyl peroxide

Question 13.
Some meta-directing substituents in aromatic substitution are given. Which one is most – deactivating?
(a) – COOH
(b) – NO₂
(C) – C ≡ N
(d) – SO₃H
Answer:
(b) – NO₂

Question 14.
Consider the following statements.
(I) S_N² reaction is a bimolecular nucleophilic first order reaction.
(II) S_N² reaction take place in one step.
(III) S_N² reaction involves the formation of a carbocation. Which of the above statements is/are not correct?
(a) (II)
(b) (I) only
(c) (I) & (III)
(d) (III)
Answer:
(c) (I) & (III)

Question 15.
The questions given below consists of an assertion and the reason. Choose the correct option out of the choices given below each question.
Assertion (A): If BOD level of water in a reservoir is more than 5 ppm it is highly polluted.
Reason(R): High biological oxygen demand means high activity of bacteria in water.
(a) Both (A) and R are correct and (R) is the correct explanation of (A)
(b) Both (A) and R are correct and (R) is not the correct explanation of (A)
(c) Both (A) and R are not correct
(d) (A) is correct but( R) is not correct
Answer:
(d) (A) is correct but( R) is not correct

PART – II

Answer any six questions in which question No. 24 is compulsory. [6 × 2 = 12]

Question 16.
What is the actual configuration of copper (Z = 29)? Explain about its stability?
Answer:
Copper (Z = 29)
Expected configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²
Actual configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
The reason is that fully filled orbitals have been found to have extra stability.

Copper has the electronic configuration [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁹ 4s² due the symmetrical distribution and exchange energies of d electrons.

Symmetry leads to stability. The full filled configuration have symmetrical distribution of electrons and hence they are more stable than unsymmetrical configuration.

Question 17.
What is screening effect?
Answer:
Screening effect:
The repulsive force between inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus the inner shell elections act as a shield between the nucleus and the valence electrons. This effect is called shielding effect (or) screening effect.

Question 18.
Ice is less dense than water at 0°C. Justify this statement?
Answer:
In ice, each oxygen atom is surrounded tetrahedrally by through hydrogen bonds to four water molecules. That is, the presence of two hydrogen atoms and two lone electron pairs (on oxygen) in each water molecule results in a three-dimensional structure. The arrangements creates an open structure, which in turn accounts for the fact that ice is less dense than water at 0°C.

Question 19.
Aerosol cans carry clear warning of heating of the can. Why?
Answer:
Aerosol cans carry clear warning of heating of the can. As the temperature rises, pressure in the can will increase and ambient temperatures about 120°F may lead to explosions.

So aerosol cans should always be stored in dry areas where they will not be exposed to excessive temperatures. You should never throw an aerosol can onto a fire, or leave it in the direct sunlight, even it is empty. This is because the pressure will build up so much that the can will burst. It is due to 2 reasons.

1. The gas pressure increases.
2. More of the liquefied propellant turns into a gas.

Question 20.
One mole of PCl₅ is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant?
Answer:
Given that [PCl₅]_initial = \(\frac { 1mol }{ dm^{ 3 } } \)
[Cl₂]_eq = 0.6 mol dm^-3
PCl₅⇄ PCl₃ + Cl₂
[PCl₅]_eq = 0.6 mole dm^-3
[PCl5]_eq = 0.4 mole dm^-3
∴ K_C = \(\frac{0.6×0.6}{0.4}\)
K_C = 0.9

Question 21.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Mostly dissolution of gases in liquid is an exothermic process. It is because the fact that this process involves decrease of entropy.

Thus, increase of temperature tends to push the equilibrium towards backward direction as a result of which solubility of the gas decrease with rise in temperature.
(Gas + Solvent ⇄ Solution + Heat)

Question 22.
Give the general formula for the following classes of organic compounds?
(a) Aliphatic monohydric alcohol
(b) Aliphatic ketones
(c) Aliphatic amines.
Answer:
(a) Aliphatic monohydric alcohol

(b) Aliphatic ketones

(c) Aliphatic amines

Question 23.
Identify which of the following shows +1 and -I effect?

(I) -NO₂
(II) -SO₃H
(III) -I
(IV) -OH
(V) CH₃O^–
(VI) CH_3-

Answer:

Question 24.
Write the products A & B for the following reaction?

Answer:

PART – III

Answer any six questions in which question No. 33 is compulsory. [6 × 3 = 18]

Question 25.
Balance by oxidation number method: Mg + HNO₃ → Mg(N0₃)₂ + NO₂ + H₂O.
Answer:
Step 1:

Step 2:
Mg + 2HNO₃ → Mg(NO₃)₂ + NG₂ + H₂O

Step 3:
To balance the number of oxygen atoms and hydrogen atoms 2HNO₃ is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + H₂O

Step 4:
To balance the number of hydrogen atoms, the H₂O molecule is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + 2H₂O

Question 26.
For each of the following, give the sub level designation, the allowable m values and the number of orbitals?
Answer:

1. n = 4, l = 2
2. n = 5, l = 3
3. n = 7, l = 0

1. n = 4, l = 2
If l = 2, ‘m’values are -2,-1, 0, +1, +2. So, 5 orbitals such as d_xy, d_yz ,d_xz, \(d_{ x^{ 2 }-y^{ 2 } }\) and \(d_{ z^{ 2 } }\)

2. n = 5, l = 3
If l = 3, ‘m’ values are -3, -2, -1, 0, +1, +2, +3
So, 7 orbitals such as \(\mathrm{f}_{\mathrm{z}}^{3}, \mathrm{f}_{\mathrm{xz}}^{2}, \mathrm{f}_{\mathrm{yz}}^{2}, \mathrm{f}_{\mathrm{xyz}}, \mathrm{f}_{\mathrm{z}\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{x}\left(\mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{y}\left(3 \mathrm{x}^{2}-\mathrm{z}^{2}\right)}\)

3. n = 7, l = 0
If l = 0, ‘m’ values are 0. Only one value. So, 1 orbital such as 7s orbital.

Question 27.
Prove that ionization energy is a periodic property?
Answer:
Variation in a period:
On moving across a period from left to right, the ionization enthalpy value increases.
This is due to the following reasons:

1. Increase of nuclear charge in a period
2. Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus, thus ionization enthalpy increases. Hence, ionization energy is a periodic property.

Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons:

1. A gradual increase in atomic size
2. Increase of screening effect on the outermost electrons due. to the increase of number of inner electrons.

Hence, ionization enthalpy is a periodic property.

Question 28.
Explain how heat absorbed at constant pressure is measured using coffee cup calorimeter with neat diagram?
Answer:

1. Measurement of heat change at constant pressure can be done in a coffee cup calorimeter.
2. We know that ∆H = qp (at constant P) and therefore, heat absorbed or evolved, q_p at constant pressure is also called the heat of reaction or enthalpy of reaction, ∆H_r
3. In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and ∆H_r will Reaction also be negative. Mixture
4. Similarly in an endothermic reaction, heat is absorbed, q_p is positive and ∆H_r will also be positive.

Question 29.
Consider the following reactions,
(a) H₂(g) + I₂(g) ⇄ 2 HI(g)
(b) CaCO₂(s) ⇄ CaO(s) + CO₂(g)
(c) S(s) + 3F₂(g) ⇄ SF₆(g)
In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product?
Answer:
(a) H₂(g) + I₂2(g) ⇄ 2HI(g)
In the above equilibrium reaction, volume of gaseous mdlecules is equal on both sides. So increase or decrease the volume will not affect the equilibrium and there will be no change in the yield of product.

(b)
Volume is greater in product side. By decreasing the pressure, volume will increase thus, to get more of product CO₂, the pressure should be decreased or volume should be increased.

(c)
Volume is lesser in product side. So by increasing the pressure, equilibrium shifts to the product side.

Question 30.
You are provided with a solid ‘A’ and three solutions of A dissolved in water one saturated, one unsaturated, and one super saturated. How would you determine each solution?
Answer:
(I) Saturated solution:
When maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called a saturated solution.

(II) Unsaturated solution:
When minimum amount of solute is dissolved in a solvent at a given temperature and pressure is called an unsaturated solution.

(III) Super saturated solution:
It is a solution that holds more solute than it normally could in its saturated form.

Example:

1. A saturated solution where the addition of more compound would not dissolve in the solution. 359 g of NaCl in 1 litre of water at 25°C.
2. An unsaturated solution has the capacity to dissolve more of the compound. 36 g of NaCl in 1 litre of water at 25°C.
3. A super saturated solution is the solution in which crystals can start growing. 500 g of NaCl in 1 litre of water at 25°C.

Question 31.
Explain about the salient features of molecular orbital theory?
Answer:

• When atoms combine to fonn molecules, their individual atomic orbitals lose their identity and form new orbitals called molecular orbitals.
• The shape of molecular orbitals depend upon the shapes of combining atomic orbitals.
• The number of molecular orbitals formed is the same as the number of combining atomic orbitals. Half the number of molecular orbitals formed will have lower energy and are called bonding orbitals, while the remaining half molecular orbitals will have higher energy and are called anti-bonding molecular orbitals.
• The bonding molecular orbitals are represented as σ (sigma), π (pi), δ (delta) and the corresponding anti-bonding orbitals are called C*, 7t* and 5*.
• The electrons in the molecule are accommodated in the newly formed molecular orbitals. The filling of electrons in these orbitals follow Aufbau’s Principle, Pauli’s exclusion principle and Hund’s rule as in the case of filling of electrons in the atomic orbitals.
• Bond order gives the number of covalent bonds between the two combining atoms.

The bond order of a molecule can be calculated using the following equation:
Bond order = \(\frac { N_{ b }-N_{ a } }{ 2 } \)
N_b = Number of electrons in bonding molecular orbitals.
N_a = Number of electrons in anti-bonding molecular orbitals.
(vii) A bond order of zero value indicates that the molecule does not exist.

Question 32.
Explain the types of addition reactions?
Answer:
Addition reactions are classified into three types they are,

1. Electrophilic addition reaction
2. Nucleophilic addition reaction
3. Free radical addition reaction

1. Electrophilic addition reaction:
An electrophilic addition reaction can be described as an addition reaction in whifth a reactant with multiple bonds as in a double or triple bond undergoes has its n bond broken and two new a bond are formed.

(ethane) Br Bf

2. Nucleophilic addition reaction:
A nucleophilic addition reaction is an addition reaction where a chemical compound with an electron deficient or electrophilic double or triple bond, a n bond, reacts with a nucleophilic which is an electron rich reactant with the disappearance of the double bond and creation of two new single or a bonds.

3. Free radical addition reaction:
It is an addition reaction in organic chemistry involving free radicals. The addition may occur between a radical and a non radical or between two radicals.

Question 33.
Complete the following reaction and identify A, B and C?

Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
(I) How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How many angular nodes?
(II) How many unpaired electrons are present in the ground state of

(a) Cr³⁺ (Z = 24)
(b) Ne (Z = 10)

[OR]

(b) (I) The electronic configuration of an atom is one of the important factor which affects the value of ionization potential and electron gain enthalpy. Explain?
(II) Explain why cation are smaller and anions are larger in radii than their parent atoms?
Answer:
(a) (I) Formula for total number of nodes = n – 1
For 2s orbital: Number of radial nodes = 1.
For 4p orbital: Number of radial nodes = n – l – 1.
= 4 – 1 – 1 = 2
Number of angular nodes = l
∴Number of angular nodes = l.
So, 4p orbital has 2 radial nodes and 1 angular node.

For 5d orbital:
Total number of nodes = n – 1
= 5 – 1 =4 nodes
Number of radial nodes = n – l – 1
= 5 – 2 – 1 = 2 radial nodes.
Number of angular nodes = l = 2
∴5d orbital have 2 radial nodes and 2 angular nodes.

For 4f orbital:
Total number of nodes = n – 1
= 4 – 1 = 3 nodes
Number of radial nodes = n – l – 1
= 4 – 3 – 1 = 0 node.
Number of angular nodes = l
= 3 nodes
∴ 5d orbital have 0 radial node and 3 angular nodes.

(II) (a) Cr (Z = 24) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
Cr³⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴.
It contains 4 unpaired electrons.
(b) Ne(Z=10) 1s² 2s² 2p⁶. No unpaired electrons in it.

[OR]

(b) (I)

• Electronic configuration of an atom affects the value of ionization potential and electron gain enthalpy.
• Half filled valence shell electronic configuration and completely filled valence shell electronic configuration are more stable than partially filled electronic configuration.
• For e.g. Beryllium (Z = 4) 1s² 2s² (completely filled electronic configuration)
  Nitrogen (Z = 7) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹ (half filled electronic configuration)
  Both beryllium and nitrogen have high ionization energy due to more stable nature.
• In the case of beryllium (1s² 2s²), nitrogen (1s² 2s² 2p³) the addition of extra electron will disturb their stable electronic configuration and they have almost zero electron affinity.
• Noble gases have stable ns² np⁶ configuration and the addition of further electron is unfavorable and they have zero electron affinity.

(II) A cation is smaller than the parent atom because it has fewer electrons while its nuclear charge remains the same. The size of anion will be larger than that of parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Question 35 (a).
(I) Arrange NH₃, H₂O and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement?
(II) Can we use concentrated sulphuric acid and pure zinc in the preparation of dihydrogen?

[OR]

(b) (I) Beryllium halides are covalent whereas magnesium halides are ionic. Why?
(II) What happens when
Sodium metal is dropped in water?

1. Sodium metal is heated in free supply of air?
2. Sodium peroxide dissolves in water?

Answer:
(a) 1. Increasing magnitude of hydrogen bonding among NH₃, H₂O and HF is
HF > H₂O > NH₃

2. The extent of hydrogen bonding depends upon electronegativity and the number of hydrogne atoms available for bonding.

3. Among N, F and O the increasing order of their electronegativities are
N < O < F 4. Hence the expected order of the extent of hydrogen bonding is HF > H₂O > NH₃

(II) Conc. H₂SO₄ cannot be used because it acts as an oxidizing agent also and gets reduced to SO₂.
Zn + 2H₂SO₄ (Conc) → ZnSO₄ + 2H₂O + SO₂ Pure Zn is not used because it is non-porous and reaction will be slow. The impurities in Zn help in constitute of electrochemical couple and speed up reaction.

[OR]

(b)
(I) Beryllium ion (Be²⁺) is smaller in size and it is involved in equal sharing of electrons with halogens to form covalent bond, whereas magnesium ion (Mg²⁺) is bigger and it is involved in transfer of electrons to form ionic bond.

(II)

1. 2Na + 2H₂O → 2NaOH + H₂
2. 2Na + O₂ → Na₂O₂
3. Na₂O₂ + 2H₂O → 2NaOH + H₂O₂

Question 36 (a).
(I) Define Gibb’s free energy?
(II) You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below?
Answer:

[OR]

(b) (I) Define mole fraction.
(II) Differentiate between ideal solution and non-ideal solution.
Answer:
(a) (I) Gibbs free energy is defined as the part of total energy of a system that can be converted (or) available for conversion into work.
G = H – TS, where G = Gibb’s free energy
H – enthalpy;
T = temperature;
S = entropy

(II) For ethanol:
Given: T_b = 78.4°C = (78.4 + 273) = 351.4 K

[OR]

(b) (I) Mole fraction of a component is the ratio of number of moles of the component to the total number of moles of all components present in the solution.

(II)
Ideal Solution:

1. An ideal solution is a solution in which each component obeys the Raoult’s law over entire range of concentration.
2. For an ideal solution,
   ΔH_missing = 0, ΔV_mixing = 0
3. Example: Benzene and toulene n – Hexane and n – Heptane.

Non – Ideal Solution:

1. The solution which do not obey Raults’s law over entire range of concentrations are called non-ideal solution.
2. For an ideal solution,
   ΔH_mixing ≠ 0, ΔV_mixing ≠ 0
3. Example: Ethyl alcohol and Cyclo hexane, Benzene and aceton.

Question 37 (a).
(I) What is dipole moment?
(II) Describe Fajan’s rule?

[OR]

(b) (I) How does Huckel rule help to decide the aromatic character of a compound?
(II) Draw cis-trans isomers for the following compounds
(a) 2- chloro-2-butene
(b) CH₃-CCl=CH-CH₂CH₃
Answer:
(a)
(I)

1. The polarity of a covalent bond can be measured in terms of dipole moment which is defined as: m = q × 2d, where m is the dipole moment, q is the charge, 2d is the distance between the two charges.
2. The dipole moment is a vector quantity and the direction of the dipole moment points from the negative charge to positive charge.
3. The unit of dipole moment is Coulomb metre (C m). It is usually expressed in Debye unit (D).
4. 1 Debye = 3.336 × 10^-3o C m

(II)

1. The ability of a cation to polarise an anion is called its polarising ability and the tendency of ah anion to get polarised is called its polarisibility. The extent of polarisation in an ionic compound is given by the Fajans rule.

2. To show greater covalent character, both the cation and anion should have high charge on them. Higher the positive charge on the cation greater will be the attraction on the electron cloud of the anion. Similarly higher the magnitude of negative charge on anion, greater is its polarisability. For example, Na⁺ < Mg²⁺ < Al³⁺, the covalent character also follows the order:
NaCl < MgCl₂ < AlCl₃.

3. The smaller cation and larger anion show greater covalent character due to the greater extent of polarisation, e.g., LiCl is more covalent than NaCl.

4. Cation having ns² np⁶ nd⁰ configuration shows greater polarising power than the cations with ns² np⁶ configuration, e.g., CuCl is more covalent than NaCl.

[OR]

(b) (I) A compound is said to be aromatic, if it obeys the following rules:

1. The molecule must be cyclic.
2. The molecule must be co-planar.
3. Complete delocalisation of rc-electrons in the ring.
4. Presence of (4n + 2)π electrons in the ring where n is an integer (n = 0, 1, 2 …)

This is known as Huckel’s rule.
Example –  – Benzene

1. It is cyclic one.
2. It is a co-planar molecule.
3. It has six delocalised n electrons.
4. 4n + 2 = 6

4n = 6 – 2
4n = 4
⇒ n = 1
It obey Huckel’s rule, with n = 1, hence benzene is aromatic in nature.

(II) (a) 2-Chloro-2-butene:

(b)

Question 38 (a).
(I) Reagents and the conditions used in the reactions are given below. Complete the table by writing down the product and the name of the reaction?
Answer:

(II) What is the IUPAC name of the insecticide DDT? Why is their use banned in most of the countries?

[OR]

(b) (I) Explain about green chemistry in day-to-day life?
(II) How acetaldehyde is commercially prepared by green chemistry?
Answer:
(a)
(I)

(II)

•  The IUPAC name of the insecticide DDT is p, p’-dichloro-diphenyl trichloroethane.
• Even DDT is an effective insecticide. Now-adays it is banned because of its long term toxic effects.
• DDT is very persistent in the environment and it has a high affinity for fatty tissues. As a result, DDT gets accumulated in animal tissue fat, in particular that of birds of prey with subsequent thinning of their eggs shells and impacting their rate of reproduction. That is why DDT is banned in most of the countries.

[OR]

(b)
(I)
1. Dry cleaning of clothes:
Solvents like tetrachloroethylene used in dry cleaning of clothes, pollute the ground water and are carcinogenic. In place of tetrachloro ethylene, liquefied CO₂ with suitable detergent is an alternate solvent used. Liquefied CO₂ is not harmful to the ground water. Nowadays H₂O₂ is used for bleaching clothes in laundry, gives better result and utilises less water.

2. Bleaching of paper:
Conventional method of bleaching was done with chlorine. Nowadays H₂O₂ can be used for bleaching paper in the presence of catalyst.

1. Instead of petrol, methaftol is used as a fuel in automobiles.
2. Neem based pesticides have been synthesised, which are more safer than the chlorinated hydrocarbons.

(II) Acetaldehyde is commericially prepared by one step oxidation of ethene in the presence of ionic catalyst in aqueous medium with 90% yield.

Students can download 6th Social Science Term 3 History Chapter 2 The Post-Mauryan India Questions and Answers, Notes, Samacheer Kalvi 6th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Social Science History Solutions Term 3 Chapter 2 The Post-Mauryan India

Samacheer Kalvi 6th Social Science The Post-Mauryan India Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
The last Mauryan emperor was killed by ……………
(a) Pushyamitra
(b) Agnimitra
(c) Vasudeva
(d) Narayana
Answer:
(a) Pushyamitra

Question 2.
___________ was the founder Of Satavahana dynasty.
(a) Simuka
(b) Satakarani
(c) Kanha
(d) Sivasvati
Answer:
(a) Simuka

Question 3.
…………… was the greatest of all the Kushana emperors.
(a) Kanishka
b) Kadphises I
(c) Kadphises II
(d) Pan – Chiang
Answer:
(a) Kanishka

Question 4.
The Kantara School of Sanskrit flourished in the
(a) Deccan
(b) North-west India
(c) Punjab
(d) Gangetic valley
Answer:
(a) Deccan

Question 5.
Sakas ruled over Gandhara region …………… as their capital.
(a) Sirkap
(b) Taxila
(c) Mathura
(d) Purushpura
Answer:
(a) Sirkap

II. Match the statement with the reason and tick the appropriate answer

Question 1.
Assertion (A) : Colonies of Indo – Greeks and Indo – Parthians were established along the north-western part of India.
Reason (R) : The Bactrian and Parthian settlers gradually intermarried and intermixed with the indigenous population.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is not correct.
(d) A is not correct but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 2.
Statement I : Indo – Greek rulers introduced die system and produced coins with inscription and symbols, engraving figures on them.
Statement II : Indo – Greek rule was ended by the Kushanas.
(a) Statement I is wrong, but statement II is correct.
(b) Statement II is wrong, but statement I is correct
(c) Both the statements are correct.
(d) Both the statements are wrong.
Answer:
(b) Statement II is wrong, but statement I is correct

Question 3.
Circle the odd one

Answer:
Kanishka

Question 4.
Answer the following in a word:

Question 1.
Who was the last Sunga ruler?
Answer:
Devabhuti

Question 2.
Who was the most important and famous king of Sakas?
Answer:
Rudradaman

Question 3.
Who established Kanva dynasty in Magadha?
Answer:
Vasudeva

Question 4.
Who converted Gondophemes into Christianity?
Answer:
St.Thomas

III. Fill in the blanks

1. …………… was the founder of Indo – Parthian Kingdom.
2. In the South, Satavahanas became independent after …………… death.
3. Hala is famous as the author of ……………
4. …………… was the last ruler of Kanva dynasty.
5. Kushana’s later capital was ……………

Answer:

1. Arsaces
2. Susarman
3. Sattasai (Saptasati)
4. Susarman
5. Peshawar(Purushpura)

IV. State whether True or False Answer

1. Magadha continued to be a great centre of Buddhist culture True even after the fall of the Mauryan Empire.
2. We get much information about Kharavela from Hathigumba True inscription.
3. Simuka waged a successful war against Magadha.
4. Buddhacharita was written by Asvaghosha.

Answer:

1. True
2. True
3. False
4. True

V. Match the following

Answer:
b) 3451 2

VI. Find out the wrong statement from the following

(1) The Kushanas formed a section of the yueh-chi tribes who inhabited north-western China.
(2) Kanishka made Jainism the state religion and built many monasteries.
(3) The Great Stupa of Sanchi and the railings which enclose it belog to the Sunga period.
(4) Pan – Chiang was the Chinese general defeated by Kanishka.
Answer:
2) Kanishka made Jainism the state religion and built many monasteries.

VII. Answer in one or two sentences

Question 1.
What happened to the last Mauryan emperor?
Answer:

1. The last Mauryan emperor was Brihadratha.
2. He was assassinated by his own general, Pushyamitra Sunga.

Question 2.
Write a note on Kalidasa’s Malavikagnimitra.
Answer:

1. Pushyamitra’s son Agnimitra is said to be the hero of Kalidasa’s Malavikagnimitra.
2. This drama also refers to the victory of Vasumitra, Agnimitra’s son, over the
   Greeks on the banks of the Sindhu river.

Question 3.
Name the ruler of Kanva dynasty.
Answer:

1. Vasudeva
2. Bhumi Mitra
3. Narayana
4. Susarman.

Question 4.
Highlight the literary achievements of Satavahanas.
Answer:

1. The Satavahana king Hala was himself a great scholar of Sanskrit.
2. The Kantara school of Sanskrit flourished in the Deccan in Second Century B.C.
3. Hala is famous as the author of Sattasai (Saptasati), 700 stanzas in Prakrit.

Question 5.
Name the places where Satavahana’smounments are situated.
Answer:

1. Gandhara
2. Madhura
3. Amaravati
4. BodhGaya
5. Sanchi
6. Bharhut

Question 6.
Give an account of the achievements of Kadphises
Answer:

1. Kadphises I was the first famous military and political leader of the Kushanas.
2. He overthrew the Indo-Greek and Indo-Parthian rulers.
3. He established himself as a sovereign ruler of Bactria.
4. He extended his power in Kabul, Gandhara and up to the Indus.

Question 7.
Name the Buddhist saints and scholars who adorned the court of Kanishka
Answer:

1. Asvaghosha
2. Vasumitra
3. Nagarjuna

VIII. Answer the following

Question 1.
Who invaded India after the decline of the Mauryan empire.
Answer:

1. The break-up of the Mauryan empire resulted in the invasions of Sakas, Scythians, Parthians, Indo-Greeks or Bactria Greeks, and Kushanas from the north-west.
2. In the South, Satavahanas became independent after Asoka’s death.
3. There were Sunga and Kanvas in the north before the emergence of the Gupta dynasty.
4. Chedis (Kalinga) declared their independence.
5. Though Magadha ceased to be the premier state of India, it continued to be a great center of Buddhist culture.

Question 2.
Give an account of the conquests of Pushyamitra Sunga.
Answer:

1. The last Mauryan emperor, Brihadratha, was assassinated by his own general, Pushyamitra Sunga.
2. He established his Sunga dynasty in Magadha. His capital was Pataliputra.
3. Pushyamitra successfully repulsed the invasion of Bactria king Menander. He also conquered Vidarbha.
4. He was a staunch follower of Vedic religion. He performed two Asvamedbayagnas (horse sacrifices) to assert his imperial authority.

Question 3.
Write a note on GautamiputraSatakarni
Answer:

1. GautamiputraSatakarni was the greatest ruler of the family.
2. In the Nasik prashasti, published by his mother GautamiBalasri, Gautamiputra Satakami is described as the destroyer of Sakas, Yavanas (Greeks), and Pahlavis (Parthians).
3. The extent of the empire is also mentioned in the record.
4. Their domain included Maharashtra, north Konkan, Berar, Gujarat, Kathiawar, and Malwa.
5. His ship coins are suggestive of Andhras’ skill in seafaring and their naval power.
6. The Bogor inscriptions suggest that South India played an important role in the process of early state formation in Southeast Asia.

Question 4.
What do you know of the Gondopharid dynasty?
Answer:

1. Indo-Parthians came after the Indo-Greeks and the Indo-Scythians who were, in turn, defeated by the Kushanas in the second half of the first century A.D.
2. Indo-Parthian kingdom or Gondopharid dynasty was founded by Gondophemes.
3. The domain of Indo-Parthians comprised Kabul and Gandhara.
4. The name of Gonodophemes is associated with the Christian apostle St. Thomas.
5. According to Christian tradition, St.Thomas visited the court of Gondophemes and converted him to Christianity.

Question 5.
Who was considered the best known Indo-Greek king? Why?
Answer:

1. Menander was one of the best known Indo-Greek kings.
2. He is said to have ruled a large kingdom in the north-west of the country.
3. His coins were found over an extensive area ranging from Kabul valley and Indus river to western Uttar Pradesh.
4. MilindaPanha, a Buddhist text, is a discourse between Bactrian king Milinda and the learned Buddhist scholar Nagasena.
5. This Milinda is identified with Menander.
6. Menander is believed to have become a Buddhist and promoted Buddhism.

Question 6.
Who was Sakas?
Answer:

1. The Indo-Greek rule in India was ended by the Sakas. Sakas as nomads came in huge number and spread all over northern and western India.
2. The Sakas were against the tribe of Turki nomads.
3. Sakas were Scythians, nomadic ancient Iranians, and known as Sakas in Sanskrit.
4. Saka rule was founded by Maos or Mogain in the Gandhara region and his capital was ‘Sirkap’. His name is mentioned in Mora inscription. His coins bear images of Buddha and Siva.
5. Rudradaman was the most important and famous king of Sakas. His Junagadh/ Gimar inscription was the first inscription in chaste Sanskrit.
6. In India, the Sakas were assimilated into Indian society. They began to adopt Indian names and practice Indian religious beliefs.
7. The Sakas appointed kshetras or satraps as provincial governors to administer their territories.

Question 7.
Give an account of the religious policy of Kanishka.
Answer:

1. Kanishka was an ardent Buddhist.
2. His empire was a Buddhist empire.
3. He adopted Buddhism under the influence of Asvaghosha, a celebrated monk from Pataliputra.
4. He was as equal as the exponent and champion of Mahayanism.
5. He made Buddhism the state religion.
6. He built many stupas and monasteries in Mathura, Taxila, and many other parts of his kingdom.
7. He sent Buddhist missionaries to Tibet, China, and many countries of Central Asia for the propagation of Buddha’s gospel.
8. He organised the fourth Buddhist Council at Kundalavana near Srinagar to sort out the differences between the various schools of Buddhism. It was only in this council that Buddhism was split into Hinayanism and Mahayanism.

IX. HOTS

Question 1.
The importance of the Gandhara School of Art.
Answer:

1. The Gandhara School of Indian Art is heavily indebted to Greek influence.
2. The Greeks were good cave builders.
3. The Mahay ana Buddhists learnt the art of carving out caves from them.
4. They became skilled in rock-cut architecture.
5. This Gandhara art flourished during Kanishka time. The most favourite subject was the carving of Sculptures of Buddha.

Question 2.
Provide an account of trade and commerce during the Post-Mauryan period in South India.
Answer:

1. Kadphises II maintained a friendly relationship with the emperors of China and Rome.
2. He encouraged trade and commerce with foreign countries.
3. His coins contained the inscribed figures of Lord Siva and his imperial titles.
4. The inscriptions in the coins were in the Kharosthi language.

X. Activity (For Students)

1. Prepare an album with centres of archaeological monuments of Satavahanas and Kushanas.
2. Arrange a debate in the classroom on the cultural contribution of Indo-Greeks, Sakas, and Kushanas.

XI. Answer Grid

Question 1.
Who wrote Brihastkatha?
Answer:
Gunadhya

Question 2.
Name the Satavahana ruler who performed two Asvamedha sacrifices.
Answer:
Satakarni

Question 3.
How many years did the Satavahanas rule the Deccan?
Answer:
450 years

Question 4.
Who laid the foundation of the Saka era?
Answer:
Mao’s (or) Mogain

Question 5.
What was the favourite subject of Gandhara artists?
Answer:
Carving of Sculptures of Buddha

Question 6.
Where did Kanishka organise the fourth Buddhist Council?
Answer:
Kundalavana (near Srinagar)

Samacheer Kalvi 6th Social Science The Post-Mauryan India Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
The Chinese Buddhist monk and traveller who wrote si-yu-ki …………….
(a) Fahien
(b) Hiuen Tsang
(c) Yuch – Chi
(d) Pan – Chiang
Answer:
(b) Hiuen Tsang

Question 2.
Asvahosha wrote
(a) Brihastkatha
(b) Mahabhasya
(c) Buddhacharita
(d) Harshacharita
Answer:
(c) Buddhacharita

Question 3.
During the Sunga period stone was replaced by railings.
(a) wood
(b) iron
(c) copper
(d) brick
Answer:
(a) wood

Question 4.
MilindaPanha is a Buddhist
(a) Statue
(b) Cave
(c) Text
(d) Monastry
Answer:
(c) Text

Question 5.
……………. gradually gained ascendancy and became the court language.
(a) Sanskrit
(b) Kharasthi
(c) Kannada
(d) Prakrit
Answer:
(a) Sanskrit

II. Match the statement with the reason and tick the appropriate answer

Question 1.
Assertion (A): The Greek rulers of Bactria and Parthia started encroaching into the northwestern borderlands of India.
Reason (R): There was a decline in the Mauryan empire.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct and R is not the correct explanation of A.
(c) A is correct but R is not correct.
(d) Both the statements are wrong.
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 2.
Statement I: The Mahayana Buddhists learned the art of carving out caves from the Greeks.
Statement II: The Greeks were good cave builders.
(a) Statement I is wrong, but statement II is correct.
(b) Statement II is wrong, but the statement I is correct.
(c) Both the statements are correct.
(d) Both the statements are wrong.
Answer:
(c) Both the statements are correct.

III. Fill in the blanks

1. Simuka’s successor was his brother …………….
2. A bronze statue of the standing Buddha was discovered in …………….
3. The Indo – Greek rule in India was ended by the ……………..
4. The Saka ruler Mogain’s capital was ……………..
5. Rudradaman’s ……………. inscription was the first inscription in chaste Sanskrit.

Answer:

1. Krishna
2. OC – EO
3. Sakas
4. Sirkap
5. Junagadh / Girnar

IV. True or False

1. Satakarni was Simuka’s nephew.
2. A stone seal discovered in Nakhon Pathom was in Thailand.
3. Mao’s name is mentioned in the Mora inscription.
4. The Kushanas appointed satraps as provincial governors.
5. Kushana rulers were Buddhists.

Answer:

1. True
2. True
3. True
4. False
5. True

V. Answer in one or two sentences

Question 1.
Write a short note on king Kharavela of Kalinga
Answer:

1. King Kharavela was a contemporary of the Sungas.
2. Hathigumba Inscription gave information about Kharavela.

Question 2.
Who laid the foundation of the Satavahana dynasty?
Answer:

1. The last Kanva ruler Susarman was assassinated by his powerful feudatory chief of Andhra named Simuka.
2. Simuka laid the foundation of the Satavahana dynasty.

Question 3.
How did the Sakas assimilate into Indian Society?
Answer:

1. The Sakas began to adopt Indian names.
2. They practiced Indian religious beliefs.

Question 4.
Under whom did the satrapies Bactria and Parthia become independent,
Answer:
Satrapies Bactria became independent under the leadership of Diodotus I and Parthia under Arsaces.

VII. Answer the following

Question 1.
Write a note on the Conquests of Kanishka.
Answer:

1. Kanishka conquered and annexed Kashmir.
2. He waged a successful war against Magadha.
3. He also waged a war against a ruler of Parthia to maintain safety and integrity in his vast empire on the western and south-western border.
4. After the conquest of Kashmir and Gandhara, he turned his attention towards China.
5. He defeated the Chinese general Pan-Chiang and safeguarded the northern borders of India from Chinese intrusion.
6. His empire extended from Kashmir down to Benaras, and the Vindhya mountain in the south. It included Kashgar, Yarkand touching the borders of Persia and Parthia.

VIII. Mind map

Students can download 6th Social Science Term 3 History Chapter 1 Society and Culture in Ancient Tamizhagam: The Sangam Age Questions and Answers, Notes, Samacheer Kalvi 6th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Social Science History Solutions Term 3 Chapter 1 Society and Culture in Ancient Tamizhagam: The Sangam Age

Samacheer Kalvi 6th Social Science Society and Culture in Ancient Tamizhagam: The Sangam Age Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
Pattini cult in Tamil Nadu was introduced by …………….
(a) PandyanNeduncheliyan
(b) CheranSenguttuvan
(c) IlangoAdigal
(d) Mudathirumaran
Answer:
(b) CheranSenguttuvan

Question 2.
Which dynasty was not in power during the Sangam Age?
(a) Pandyas
(b) Cholas
(c) Pallavas
(d) Cheras
Answer:
(c) Paliavas

Question 3.
The rule of Pandyas was followed by ……………..
(a) Satavahanas
(b) Cholas
(c) Kalabhras
(d) Pallavas
Answer:
(c) Kalabhras

Question 4.
The lowest unit of administration during the Sangam Age was
(a) Mandalam
(b) Nadu
(c) Ur
(d) Pattinam
Answer:
(c) Ur

Question 5.
What was the occupation of the inhabitants of the Kurinji region?
(a) Plundering
(b) Cattle rearing
(c) Hunting and gathering
(d) Agriculture
Answer:
(c) Hunting and gathering

II. Read the statement and tick the appropriate answer

Question 1.
Assertion (A) : The assembly of the poets was known as Sangam.
Reason (R) : Tamil was the language of Sangam literature.
(a) Both A and R are true. R is the correct explanation of A.
(b) Both A and R are true. R is not the correct explanation of A.
(c) A is true but R is false.
(d) Both A and R is not true.
Answer:
(a) Both A and R are true. R is the correct explanation of A

Question 2.
Which of the following statements are not true?
a. Karikala won the battle of Talayalanganam.
b. The Pathitrupathu provides information about Chera Kings.
c. The earliest literature of the Sangam age was written mostly in the form of prose
a. 1 only
b. 1 and 3 only
c. 2 only
Answer:
(b) 1 and 3 only

Question 3.
The ascending order of the administrative division in the ancient Tamizhagam was
(a) Ur < Nadu < Kurram < Mandalam
(b) Ur < Kurram < Nadu < Mandalam
(c) Ur < Mandalam < Kurram < Nadu
(d) Nadu < Kurram < Mandalam < Ur
Answer:
(b) Ur < Kurram < Nadu < Mandalam

Question 4.
Match the following dynasties with the Royal Insignia

Answer:
A) 3 2 1

III. Fill in the blanks

1. The battle of Venni was won by ………………
2. The earliest Tamil grammar work of the Sangam period was ………………
3. ……………… built Kallanai across the river Kaveri.
4. The chief of the army was known as ……………….
5. Land revenue was called ………………

Answer:

1. Karikal Valavan
2. Tholkappiyam
3. Karikalan
4. Thanai thalaivan
5. Irai

IV. True or False

1. Caste system developed during the Sangam period.
2. Kizhar was the village chief.
3. Puhar was the general term for city.
4. Coastal region was called Marudham.

Answer:

1. False
2. True
3. False
4. False

V. Match

Answer:
1. – d
2. – a
3. – b
4. – c

VI. Answer in one or two sentences

Question 1.
Name any two literary sources to reconstruct the history of ancient Tamizhagam?
Answer:
Tholkappiyam, Ettuthogai, and Patthupattu are some of the literary sources to reconstruct the history of ancient Tamizhagam.

Question 2.
What was Natukkal or Virakkal?
Answer:

1. The ancient Tamils had great respect for the heroes who died on the battlefield.
2. The hero stones were created to commemorate heroes who sacrificed their lives in war. These hero stones were known as Natukkal or Virakkal.

Question 3.
Name five things mentioned in the Sangam literature.
Answer:

1. Kurinji.
2. Mullai.
3. Marutham.
4. Neithal.
5. Palai.

Question 4.
Name any two archaeological sites related to Sangam period.
Answer:
The excavated materials from Adichanallur, Arikamedu, Kodumanal, Puhar, Korkai, Alankulam, Urariyur etc.

Question 5.
Name the seven patrons (KadaiyeluVallalgal).
Answer:
The seven patrons were

1. Pari
2. Kari
3. Ori
4. Pegan
5. Ay
6. Adiyaman
7. Nalli

Question 6.
Name any three Tamil poetic works of Kalahhra period.
Answer:
Periapuranam, Seevakachinthamani, and Kundalakesi were written during the Kalabhra

VII. Answer the following

Question 1.
Discuss the status of women in the Sangam Society.
Answer:

1. In the Sangam Society, there were learned and wise women.
2. Forty women poets had lived
3. Marriage was a matter of choice.
4. Chastity (Karpu) was considered the highest virtue.
5. In their Parents’ property sons and daughters had equal shares.

VIII. HOTS

Question 1.
KarikalValavan is regarded as the greatest Chola king. Justify.
Answer:

1. KarikalValavan or Karikalan was the most famous of the Chola kings.
2. He defeated the combined army of the Cheras, Pandyas and the eleven Velir Chieftains who supported them at Venni, a small village in the Thanjavur region.
3. He converted forests into cultivable lands.
4. He built Kallanai across the river Kaveri to develop agriculture.
5. Their port Puhar attracted merchants from various regions of the Indian Ocean.
6. The Pattinapaalai a poetic work in the pathinenkeezhkanakku gives elaborate information of the trading activity during the rule of Karikalan.

Question 2.
The period ilabhra is not a dark age. Give reasons.
Answer:

1. Following the Sangam period, the Kalabhras had occupied Tamil Country for about two and half centuries.
2. There is evidence of their rule in literary texts.
3. The literary sources for this period include Tamil Navalar Charithai, Yapemkalam and Periapuranam.
4. Seevakachinthamani and Kundalakesi were also written during this period.
5. In Tamizhagam, Jainism and Buddhism became prominent during this period
6. Introduction of Sanskrit and Prakrit languages had resulted in the development of a new script called Vattezhuththu.
7. Many works under Pathinenkeezhkanakku were composed.
8. Trade and commerce continued to flourish during this period.
9. So the Kalabhra period is not a dark age, as it is portrayed.

X. Life skill (For Students)

Collect and paste the pictures of landscape and find out the eco – region to which belongs. Write the important crops grown and occupation of the people there.

XI. Answer Grid

Question 1.
Mention two epics of the Sangam period.
Answer:

1. Silapathikaram
2. Manimegalai

Question 2.
Name the two groups of officials who assisted the king.
Answer:

1. Aimperunguzhu
2. Enberayam

Question 3.
Name any two women poets of the Sangam period.
Answer:

1. Awaiyar, Velli
2. Veethiyar

Question 4.
Name any three major ports of Sangam age.
Answer:

1. Musiri
2. Tondi
3. Korkai

Question 5.
What constituted Muthamizh?
Answer:

1. Iyal
2. Isai
3. Naatakam

Question 6.
Silapathikaram was written by ……………
Answer:
Ilango Adigal

Question 7.
Talayalanganam is related to whish Pandya king?
Answer:
Nedunchezhiyan

Question 8.
Which ecoregion was called menpulam?
Answer:
Marutham

Question 9.
The lighthouses in the ports are called ……………
Answer:

1. Kalangari
2. flangu Sudar

Samacheer Kalvi 6th Social Science Society and Culture in Ancient Tamizhagam: The Sangam Age Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
According to Prof.George L.Hart Tanial is as old as ……………
(a) Chinese
(b) Greek
(c) Latin
(d) English
Answer:
(c) Latin

Question 2.
The epic character from Silappathikaram
(a) Pallavas
(b) Cheras
(c) Pandyas
(d) Cholas
Answer:
(d) Cholas

Question 3.
…………… is the port of Pandvas.
(a) Puhar
(b) Korgai
(c) Muziri
(d) Tondi
Answer:
(b) Korgai

Question 4.
Bow and arrow as the symbol of
(a) Kalabhras
(b) Cholas
(c) Cheras
(d) Pandyas
Answer:
(c) Cheras

II. Read the statement and tick the appropriate answer

Question 1.
Assertion (A) : The Kalabhra period is not a dark age.
Reason (R) : It is known about the literary sources, new script and flourishing of trade and commerce.
(a) Both A and R are true. R is not the correct explanation of A.
(b) Both A and R are true. R is the correct explanation of A
(c) A is true but R is false.
(d) Both A and R is not true.
Answer:
(b) Both A and R are true. R is the correct explanation of A.

Question 2.
Which of the following statements are not true?
(1) Pandyas garlanded Fig (Athi) flower.
(2) The deity of the kurinji people is Indiran.
(3) The author of ‘Natural History’ is Pliny the younger
(a) 1, 2 and 3
(b) 2 and 3
Answer:
(b) 2 and 3

III. Answer the following

Question 1.
What were the ornaments made of?
Answer:

1. Gold
2. Silver
3. Pearls
4. Precious stones
5. Conch shells
6. Beads

Question 2.
What were the main imports?
Answer:

1. Topaz
2. Tin
3. Wine
4. Glass
5. Horses

Question 3.
What do you know about Indian silk?
Answer:

1. The silk supplied by Indians, merchants to the Roman Empire was very important. .
2. The Roman emperor Aurelian declared it to be worth its weight in gold.

Question 4.
What is the Royal Insignia?
Answer:

1. Sceptre
2. Drum
3. White umbrella

VII. Answer the following

Question 1.
Explain the Religious Beliefs and Social Divisions in the Sangam Society. The primary deity of the Tamils was Seyon or Murugan.
Answer:

1. The other worshipped gods were Sivan, Mayon (Vishnu), Indiran, Varunan and Kotravai.
2. The Hero stone (natukkal) worship was in practice
3. Buddhism and Jainism also co-existed.
4. As it did in northern India caste did not develop in Tamizhagam.
5. Varuna system came to the Dravidian south comparatively late.

Question 2.
What was said by George L Hart about the Tamil language?
Answer:

1. George L. Hart, Professor of Tamil language at the University of California, has said that Tamil is as old as Latin.
2. The language arose as an entirely independent tradition with no influence of other languages.

VII. Mind map