Prasanna

Students can download Maths Chapter 8 Statistics and Probability Ex 8.2 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.2

Question 1.
The standard deviation and mean of a data are 6.5 and 12.5 respectively. Find the coefficient of variation.
Answer:
Standard deviation of a data (σ) = 6.5
Mean of the data (\(\bar{x}\)) = 12.5
Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100 \%\)
= \(\frac{6.5}{12.5} \times 100 \%=52 \%\)
Coefficient of variation = 52%

Question 2.
The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Answer:
Standard deviation (σ) = 1.2
Coefficient of variation = 25.6

Question 3.
If the mean and coefficient of variation of a data are 15 and 48 respectively, then find the value of standard deviation.
Answer:
Mean (\(\bar{x}\)) = 15
Co efficient of variation = 48

Question 4.
If n = 5, \(\bar{x}\) = 6, Σx² = 765, then calculate the coefficient of variation.
Answer:

Question 5.
Find the coefficient of variation of 24, 26, 33, 37, 29, 31.
Answer:
Arrange in ascending order we get 24, 26, 29, 31, 33, 37
Assumed mean = 29

Question 6.
The time taken (in minutes) to complete homework by 8 students in a day are given by 38, 40, 47, 44, 46, 43, 49, 53. Find the coefficient of variation.
Answer:
Arrange in ascending order we get, 38, 40, 43, 44, 46, 47, 49, 53.
Assumed mean = 46


= \(\frac{453}{45}\)%
= 10.066
Coefficient of variation = 10.07%

Question 7.
The total marks scored by two students Sathya and Vidhya in 5 subjects are 460 and 480 with standard deviation of 4.6 and 2.4 respectively. Who is more consistent in performance?
Answer:
Total marks scored by sathya = 460
Total marks scored by vidhya = 480
Number of subjects = 5
Mean marks of sathya = \(\frac{460}{5}\)
\(\bar{x}\) = 92%
Given standard deviation, (σ) = 4.6

Vidhya coefficient of variation is less than Sathya.
Vidhya is more consistent.

Question 8.
The mean and standard deviation of marks obtained by 40 students of a class in three subjects Mathematics, Science and Social Science are given below.

Which of the three subjects shows the highest variation and which shows the lowest variation in marks?
Answer:
(i) Mathematics:
Mean (\(\bar{x}\)) = 56
Standard deviation (σ) = 12
Coefficient variation (CV₁) = \(\frac{\sigma}{\bar{x}} \times 100=\frac{12}{56} \times 100\)

Science shows the highest variation
Social science shows the lowest variation

Question 9.
The temperature of two cities A and B in the winter season are given below.

Find which city is more consistent in temperature changes?
Answer:
(i) city A:
Assumed mean = 22




C.V of city A < C.V of city B
City A is more consistent in temperature change.

Students can download Maths Chapter 8 Statistics and Probability Ex 8.1 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.1

Question 1.
Find the range and coefficient of range of the following data.
(i) 63, 89, 98, 125, 79, 108, 117, 68
(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8
Answer:
(i) Here the largest value (L) = 125
The smallest value (S) = 63
Range = L – S = 125 – 63 = 62

Question 2.
If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.
Solution:
If the range = 36.8 and
the smallest value = 13.4 then
the largest value = L = R + S
= 36.8 + 13.4 = 50.2

Question 3.
Calculate the range of the following data.

Answer:
Smallest value (S) = 400
Largest value (L) = 650
Range = L – S = 650 – 400 = 250

Question 4.
A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages yet to be completed by them.
Answer:
The remaining number of pages to be completed is 60 – 32; 60 – 35; 60 – 37; 60 – 30; 60 – 33; 60 – 36; 60 – 35 and 60 – 37
The pages to be completed are, 28, 25, 23, 30, 27, 24, 25, and 23
Arrange in ascending order we get, 23, 23, 24, 25, 25, 27, 28 and 30

Question 5.
Find the variance and standard deviation of the wages of 9 workers given below:
₹ 310, ₹ 290, ₹ 320, ₹ 280, ₹ 300, ₹ 290, ₹ 320, ₹ 310, ₹ 280.
Answer:
Arrange in ascending order we get,
280, 280, 290, 290, 300, 310, 310, 320 and 320
Assumed mean = 300


Variance = 222.222
Standard deviation = √Variance = √222.222 = 14.907 = 14.91
Variance = 222.22
Standard deviation = 14.91

Question 6.
A wall clock strikes the bell once at 1 o’clock, 2 times at 2 o’clock, 3 times at 3 o’clock and so on. How many times will it strike in a particular day? Find the standard deviation of the number of strikes the bell make a day.
Answer:
Wall clock strikes the bell in 12 hours
1, 2, 3, 4, 5,… ,12
Wall clock strikes in a day (24 hours)
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
Assumed mean = 14



The standard deviation of bell strike in a day is 6.9

Question 7.
Find the standard deviation of the first 21 natural numbers.
Answer:
Here n = 21
The standard deviation of the first ‘n’ natural numbers,

The standard deviation of the first 21 natural numbers = 6.06

Question 8.
If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.
Solution:
If the standard deviation of a data is 4.5 and each value of the data decreased by 5, the new standard deviation does not change and it is also 4.5.

Question 9.
If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
Answer:
The standard deviation of the data = 3.6
Each value of the data is divided by 3
New standard deviation = \(\frac{3.6}{3}\) = 1.2
New Variance = (1.2)² = 1.44 [∴ Variance = (S.D)²]
New standard Deviation = 1.2
New variance = 1.44

Question 10.
The rainfall recorded in various places of five districts in a week are given below.

Find its standard deviation.
Answer:
Assumed mean = 60

Question 11.
In a study about viral fever, the number of people affected in a town were noted as

Find its standard deviation.
Answer:
Assumed mean = 35

Question 12.
The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find the standard deviation.

Answer:
Assumed mean = 34.5

Question 13.
The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.

Answer:
Assumed mean = 11


Standard deviation (σ) = 1.24

Question 14.
For a group of 100 candidates, the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on, it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.
Answer:
Number of candidates = 100
n = 100
Mean (\(\bar{x}\)) = 60
standard deviation (σ) = 15
Mean (\(\bar{x}\)) = \(\frac{\Sigma x}{n} \Rightarrow 60=\frac{\Sigma x}{100}\)
Σx = 6000
Correct total = 6000 + (45 – 40) + ( 72 – 27) = 6000 + 5 + 45 = 6050
Correct mean (\(\bar{x}\)) = \(\frac{6050}{100}\) = 60.5
Given standard deviation = 15


Correct mean = 60.5
Correct standard deviation (σ) = 14. 61

Question 15.
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.
Answer:
Let the missing two observations be ‘a’ and ‘b’
Arithmetic mean = 8

Students can download Maths Chapter 7 Mensuration Additional Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 7 Mensuration Additional Questions

I. Multiple Choice Questions:

Question 1.
The curved surface area of a right circular cylinder of radius 1 cm and height 1 cm is equal to ______
(1) π cm²
(2) 2π cm²
(3) 3π cm²
(4) 2 cm²
Answer:
(2) 2π cm²
Hint:
C.S.A of a cylinder = 2πrh sq. units = 2 × π × 1 × 1 cm² = 2π cm²

Question 2.
The total surface area of a solid right circular cylinder whose radius is half of its height h is equal to ______ sq. units.
(1) \(\frac{3}{2} \pi h\)
(2) \(\frac{2}{3} \pi h^{2}\)
(3) \(\frac{3}{2} \pi h^{2}\)
(4) \(\frac{2}{3} \pi h\)
Answer:
(3) \(\frac{3}{2} \pi h^{2}\)
Hint:
T.S.A = 2πr(h + r)
[radius is half of the height]
= \(2 \pi \times \frac{h}{2}\left(\frac{h}{2}+h\right)\)
= \(=\pi h\left(\frac{3 h}{2}\right)=\frac{3}{2} \pi h^{2}\) sq. units

Question 3.
Base area of a right circular cylinder is 80 cm². If its height is 5 cm, then the volume is equal to _______
(1) 400 cm³
(2) 16 cm³
(3) 200 cm³
(4) \(\frac{400}{3}\) cm³
Answer:
(1) 400 cm³
Hint:
Volume of a cylinder = πr²h cu. units
[Base area (πr²) = 80 cm² = 80 × 5 cm³ = 400 cm³

Question 4.
If the total surface area of a solid right circular cylinder is 200π cm² and its radius is 5 cm, then the sum of its height and radius is ______
(1) 20 cm
(2) 25 cm
(3) 30 cm
(4) 15 cm
Answer:
(1) 20 cm
Hint:
T.S.A of a cylinder = 200π cm²
2πr (h + r) = 200π
2 × 5 (h + r) = 200
(h + r) = 20 cm

Question 5.
The curved surface area of a right circular cylinder whose radius is a units and height is b units, is equal to ______
(1) πa²b sq.cm
(2) 2πab sq.cm
(3) 2π sq.cm
(4) 2 sq.cm
Answer:
(2) 2πab sq.cm .
Hint:
C.S.A. of a cylinder = 2πrh sq. units = 2 × π × a × b sq. cm = 2πab sq. cm

Question 6.
Radius and height of a right circular cone and that of a right circular cylinder are respectively, equal. If the volume of the cylinder is 120 cm³, then the volume of the cone is equal to _______
(1) 1200 cm³
(2) 360 cm³
(3) 40 cm³
(4) 90 cm³
Answer:
(3) 40 cm³
Hint:
Volume of the cone = \(\frac{1}{3}\) × volume of the cylinder
= \(\frac{1}{3}\) × 120 cm³
= 40 cm³

Question 7.
If the diameter and height of a right circular cone are 12 cm and 8 cm respectively, then the slant height is
(1) 10 cm
(2) 20 cm
(3) 30 cm
(4) 96 cm
Answer:
(1) 10 cm
Hint:
Slant height of a cone

Question 8.
If the circumference at the base of a right circular cone and the slant height are 120π cm and 10 cm respectively, then the curved surface area of the cone is equal to ______
(1) 1200π cm²
(2) 600π cm²
(3) 300π cm²
(4) 600 cm²
Answer:
(2) 600π cm²
Hint:
Circumference (2πr) = 120π cm
Slant height (l) = 10 cm;
Curved surface area of a cone = πrl sq. units
= \(\frac{120 \pi}{2}\) × 10 cm² = 600π cm²

Question 9.
If the volume and the base area of a right circular cone are 48π cm and 12π cm respectively, then the height of the cone is equal to ______
(1) 6 cm
(2) 8 cm
(3) 10 cm
(4) 12 cm
Answer:
(4) 12 cm
Hint:
Volume of a cone = 48π cm³
[Base area (πr²) = 12π]
\(\frac{1}{3}\) πr²h = 48π
\(\frac{1}{3}\) × 12π × h = 48π
[Substitute πr² = 12π]
h = \(\frac{48}{4}\) = 12 cm

Question 10.
If the height and the base area of a right circular cone are 5 cm and 48 sq.cm respectively, then the volume of the cone is equal to _______
(1) 240 cm³
(2) 120 cm³
(3) 80 cm³
(4) 480 cm³
Answer:
(3) 80 cm³
Hint:
Volume of a cone (V) = \(\frac{1}{3}\) πr²h sq. units
Base area (πr²) = 48 sq. cm
V = \(\frac{1}{3}\) × 48 × 5 = 80 cm³

Question 11.
The ratios of the respective heights and the respective radii of two cylinders are 1 : 2 and 2 : 1 respectively. Then their respective volumes are in the ratio _______
(1) 4 : 1
(2) 1 : 4
(3) 2 : 1
(4) 1 : 2
Answer:
(3) 2 : 1
Hint:
h₁ : h₂ = 1 : 2
r₁ : r₂ = 2 : 1
Ratio of their volumes
= \(\frac{1}{3} \pi r_{1}^{2} h_{1}: \frac{1}{3} \pi r_{2}^{2} h_{2}\)
= 22 × 1 : 12 × 2 = 4 : 2 = 2 : 1

Question 12.
If the radius of a sphere is 2 cm, then the curved surface area of the sphere is equal to ________
(1) 8π cm²
(2) 16 cm²
(3) 12π cm²
(4) 16π cm²
Answer:
(4) 16π cm²
Hint:
C.S.A of a sphere = 4πr² sq. units
[radius = 2 cm]
= 4 × π × 22 cm²
= 16π cm²

Question 13.
The total surface area of a solid hemisphere of diameter 2 cm is equal to _______
(1) 12 cm²
(2) 12π cm²
(3) 4π cm²
(4) 3π cm²
Answer:
(4) 3π cm²
Hint:
Radius of a hemisphere = \(\frac{2}{2}\) = 1 cm
Total surface area of a hemisphere = 3πr² sq. units = 3 × π × 12 cm² = 3π cm²

Question 14.
If the volume of a sphere is \(\frac{9}{16} \pi\) cu.cm, then its radius is ________
(1) \(\frac{4}{3}\) cm
(2) \(\frac{3}{4}\) cm
(3) \(\frac{3}{2}\) cm
(4) \(\frac{2}{3}\) cm
Answer:
(2) \(\frac{3}{4}\) cm
Hint:
Volume of the sphere = \(\frac{9}{16} \pi\)

Question 15.
The surface areas of two spheres are in the ratio of 9 : 25. Then their volumes are in the ratio _______
(1) 81 : 625
(2) 729 : 15625
(3) 27 : 75
(4) 27 : 125
Answer:
(4) 27 : 125
Hint:
Ratio of their surface area = 9 : 25

Question 16.
The total surface area of a solid hemisphere whose radius is a units, is equal to ________
(1) 2πa² sq. units
(2) 3πa² sq. units
(3) 3πa sq. units
(4) 3a² sq. units
Answer:
(2) 3πa² sq. units
Hint:
T.S.A. of a solid hemisphere = 3πr² sq. units
= 3 × π × a × a sq.units
= 3πa² sq. units

Question 17.
If the surface area of a sphere is 100π cm², then its radius is equal to ______
(1) 25 cm
(2) 100 cm
(3) 5 cm
(4) 10 cm
Answer:
(3) 5 cm
Hint:
Surface area of a sphere = 100π cm²
4πr² = 100π
r² = 25
r = √25 = 5 cm

Question 18.
If the surface area of a sphere is 36π cm², then the volume of the sphere is equal to _______
(1) 12π cm³
(2) 36π cm³
(3) 72π cm³
(4) 108π cm³
Answer:
(2) 36π cm³
Hint:
Surface area of a sphere = 36π cm²
4πr² = 36π
r² = 9
r = 3 cm
Volume of a sphere = \(\frac{4}{3} \pi r^{3}\) cu. units
= \(\frac{4}{3} \pi\) × 3 × 3 × 3 cm³ = 36π cm³

Question 19.
If the total surface area of a solid hemisphere is 12π cm² then its curved surface area is equal to ______
(1) 6π cm²
(2) 24π cm²
(3) 36π cm²
(4) 8π cm²
Answer:
(4) 8π cm²
Hint:
T.S.A of a hemisphere = 12π cm²
3πr² = 12π
r² = 4
r = 2
Curved surface area of a hemisphere = 2πr² = 2 × π × 4 = 8π cm²

Question 20.
If the radius of a sphere is half of the radius of another sphere, then their respective volumes are in the ratio _____
(1) 1 : 8
(2) 2 : 1
(3) 1 : 2
(4) 8 : 1
Answer:
(1) 1 : 8
Hint:
\(r_{1}=\frac{r_{2}}{2} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{1}{2} \Rightarrow r_{1}: r_{2}=1: 2\)

II. Answer the following questions:

Question 1.
Curved surface area and circumference at the base of a solid right circular cylinder are 4400 sq.cm and 110 cm respectively. Find its height and diameter.
Answer:
Given, Circumference of the base of a cylinder = 110 cm
2πr = 110 ……. (1)
Curved surface area = 4400 cm²
2πrh = 4400 cm² ……. (2)
From (1) & (2), \(\frac{(2)}{(1)} \Rightarrow \frac{2 \pi r h}{2 \pi r}=\frac{4400}{110}=40 \mathrm{cm}\)
Height of the cylinder (h) = 40 cm
From (1), 2πr = 110
2 × \(\frac{22}{7}\) × r = 110
r = \(\frac{35}{2}\)
We know that, diameter (d) = 2 × radius
d = 2 × \(\frac{35}{2}\) = 35 cm
Diameter of the Circular cylinder = 35 cm

Question 2.
A mansion has 12 right cylindrical pillars each having radius 50 cm and height 3.5 m. Find the cost of painting the lateral surface of the pillars at ₹ 20 per square metre.
Answer:
Given, Radius of a cylinder (r) = 50 cm = 0.5 m
Height of a cylinder (h) = 3.5 m
Curved surface area of a pillar = 2πrh sq. units
Curved surface area of 12 pillars = 12 × 2πrh
= 12 × 2 × \(\frac{22}{7}\) × 0.5 × 3.5 m²
= 132 sq. m.
Cost for painting the lateral surface of pillars per metre = ₹ 20
Cost of painting = 132 × ₹ 20 = ₹ 2640

Question 3.
The total surface area of a solid right circular cylinder is 231 cm². Its curved surface area is two thirds of the total surface area. Find the radius and height of the cylinder.
Answer:
Given, Total surface area of a cylinder (T.S. A) = 231 sq.cm
Curved surface area = \(\frac{2}{3}\) × T.S.A = \(\frac{2}{3}\) × 231 = 154 cm²
2πrh = 154 cm² …… (1)
Total surface area = 231 cm²
2πr (h + r) = 231
2πrh + 2πr²= 231
154 + 2πr² = 231 [from (1)]
2πr² = 231 – 154 = 77

Radius of the cylinder = 3.5 cm
Height of the cylinder = 7 cm

Question 4.
The total surface area of a solid right circular cylinder is 1540 cm². If the height is four times the radius of the base, then find the height of the cylinder.
Answer:
Given, Let the radius of the cylinder be ‘r’
Height of a cylinder (h) = 4r (by given condition)
Total surface area = 1540 cm²
2πr(h + r) = 1540 cm²

Height of the cylinder = 4r = 4 × 7 = 28 cm

Question 5.
If the vertical angle and the radius of a right circular cone are 60° and 15 cm respectively, then find its height and slant height.
Answer:
Given, In the figure, OAB is a cone and OC ⊥ AB
∠AOC = \(\frac{60^{\circ}}{2}\) = 30°
In the right ∆OAC, tan 30° = \(\frac{\mathrm{AC}}{\mathrm{OC}}\)

Slant Height of the cone (l) = 15 × 2 = 30 cm

Question 6.
The central angle and radius of a sector of a circular disc are 180° and 21 cm respectively. If the edges of the sector are joined together to make a hollow cone, then find the radius of the cone.
Answer:

Given, Radius of a sector (r) = 21 cm
The angle of the sector (θ) = 180°
Let “R” be the radius of the cone.
Circumference of the base of a cone = Arc length of the sector

Radius of the cone (R) = 10.5 cm

Question 7.
If the curved surface area of a solid hemisphere is 2772 sq.cm, then find its total surface area.
Answer:
Given, the Curved surface area of a solid hemisphere = 2772 cm²
2πr² = 2772

Total surface area = 3πr² sq. units
= 3 × \(\frac{22}{7}\) × 21 × 21
= 4158 sq.cm
Aliter:
C.S.A of a hemisphere = 2772 cm²
2πr² = 2772 cm²
πr² = \(\frac{2772}{2}\) = 1386 cm
T.S.A of a hemisphere = 3πr² sq.units = 3 × 1386 cm² = 4158 cm²

Question 8.
An inner curved surface area of a hemispherical dome of a building needs to be painted. If the circumference of the base is 17.6 m, find the cost of painting it at the rate of ₹ 5 per sq. m.
Answer:
Given, Circumference of the dome = 17.6 m
2πr = 17.6
\(r=\frac{17.6 \times 7}{2 \times 22}=\frac{8.8 \times 7}{22}=2.8 \mathrm{m}\)
Curved surface area of the dome = 2πr² sq. units
= 2 × \(\frac{22}{7}\) × 2.8 × 2.8 m²
= 49.28 m²
Cost of painting for one sq.metre = ₹ 5
Cost of painting the curved surface = 49.28 × ₹ 5 = ₹ 246.40

Question 9.
Volume of a solid cylinder is 62.37 cu.cm. Find the radius if its height is 4.5 cm.
Answer:
Given, Height of a cylinder (h) = 4.5 cm
Volume of a solid cylinder = 62.37 cu. cm

Radius of a cylinder (r) = 2.1 cm

Question 10.
A rectangular sheet of metal foil with dimension 66 cm × 12 cm is rolled to form a cylinder of height 12 cm. Find the volume of the cylinder.
Answer:
After rolling the rectangular sheet into a cylinder

Volume of the cylinder = 4158 cm³

Question 11.
The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.
Answer:
Given, Height of the wooden solid cone (h) = 12 m
Circumference of the base = 44 m
2πr = 44
r = \(\frac{44 \times 7}{2 \times 22}\) = 7 m
Volume of the wooden solid = \(\frac{1}{3} \pi r^{2} h\) cu. units
= \(\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 12 \mathrm{m}^{3}\)
= 88 × 7
= 616 m³
Volume of the solid = 616 m³

Question 12.
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 14 cm.
Answer:
Given, Edge of the cube = 14 cm
The largest circular cone is cut out from the cube.
Radius of the cone (r) = \(\frac{14}{2}\) = 7 cm
Height of the cone (h) = 14 cm
Volume of a cone

Volume of a cone = 718.67 cm³

Question 13.
The thickness of a hemispherical bowl is 0.25 cm. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. (Take π = \(\frac{22}{7}\))
Answer:
Let r, R and w be the inner and outer radii and thickness of the hemispherical bowl respectively.

Given that r = 5 cm, w = 0.25 cm
R = r + w = 5 + 0.25 = 5.25 cm
Now, outer surface area of the bowl = 2πR²
= 2 × \(\frac{22}{7}\) × 5.25 × 5.25
= 173.25 sq. cm
Thus, the outer surface area of the bowl = 173.25 sq. cm

Question 14.
Volume of a hollow sphere is \(\frac{11352}{7}\) cm3. If the outer radius is 8 cm, find the inner radius of the sphere. (Take π = \(\frac{22}{7}\))
Answer:
Let R and r be the outer and inner radii of the hollow sphere respectively.
Let V be the volume of the hollow sphere.

Hence, the inner radius r = 5 cm

Question 15.
How many litres of water will a hemispherical tank whose diameter is 4.2 m?
Answer:
Radius of the tank = \(\frac{4.2}{2}\) = 2.1 m
Volume of the hemisphere
= \(\frac{2}{3} \pi r^{3}\) cu.units
= \(\frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 \mathrm{m}^{3}\)
= 19.404 m³
= 19.404 x 1000 lit
= 19,404 litres

III. Answer the following questions.

Question 1.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Answer:

For cylindrical part:
Radius (r) = 7 cm
Height (h) = 6 cm
Curved surface area = 2πrh = 2 × \(\frac{22}{7}\) × 7 × 6 cm² = 264 cm²
For hemispherical part:
Radius (r) = 7 cm
Surface area (h) = 2πr²
= 2 × \(\frac{22}{7}\) × 7 × 7 cm²
= 308 cm²
Total surface area = (264 + 308) = 572 cm²

Question 2.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Answer:

Question 3.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Answer:

For cylinderical part:
Height (h) = 2.4 cm
Diameter (d) = 1.4 cm
Radius (r) = 0.7 cm
Total surface area of the cylindrical part

For conical part:
Base area (r) = 0.7 cm
Height (h) = 2.4 cm

Question 4.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Answer:
Diameter of the cylindrical well = 7 m
Radius of the cylinder (r) = \(\frac{7}{2}\) m
Depth of the well (h) = 20 m
Volume = πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 \mathrm{m}^{3}\)
= 22 × 7 × 5 m³
Volume of the earth taken out = 22 × 7 × 5 m³
Now this earth is spread out to form a cuboidal platform having
Length (l) = 22 m
Breadth (b) = 14 m
Let ‘h’ be the height of the platform.
Volume of the platform = 22 × 14 × h m³
Volume of the platform = Volume of the earth taken out
22 × 14 × h = 22 × 7 × 5
\(h=\frac{22 \times 7 \times 5}{22 \times 14}=\frac{5}{2} \mathrm{m}=2.5 \mathrm{m}\)
Thus, the required height of the platform is 2.5 m.

Question 5.
The perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum is 8 cm, find the whole surface area of the frustum. [Use π = 3.14]
Answer:

Let the radii of circular ends are R and r [R > r]
Perimeter of circular ends are 207.24 cm and 169.56 cm
2πR = 207.24 cm

The whole surface area of the frustum = π [(R² + r²) + (R + r) l]
Required whole surface area of the frustum
= 3.14 [33² + 27² + (33 + 27) × 10] cm²
= 3.14 [1089 + 729 + 600] cm²
= 3.14 [2418] cm²
= 7592.52 cm²

Question 6.
A cuboid-shaped slab of iron whose dimensions are 55 cm × 40 cm × 15 cm is melted and recast into a pipe. The outer diameter and thickness of the pipe are 8 cm and 1 cm respectively. Find the length of the pipe. (Take π = \(\frac {22}{7}\))
Answer:

Let h₁ be the length of the pipe
Let R and r be the outer and inner radii of the pipe respectively.
Iron slab:
Volume = lbh = 55 × 40 × 15 cm³
Iron pipe:
Outer diameter, 2R = 8 cm
Outer radius, R = 4 cm
Thickness, w = 1 cm
Inner radius, r = R – w = 4 – 1 = 3 cm
Now, the volume of the iron pipe = Volume of the iron slab

Time is taken by the pipe to empty half of the tank = 3 hours 12 minutes.

Question 7.
The perimeter of the ends of a frustum of a cone are 44 cm and 8.4π cm. If the depth is 14 cm., then find its volume.
Answer:
Given let the radius of the top of the frustum be “R” and the radius of the bottom of the frustum be “r”

Question 8.
A tent is in the shape of a right circular cylinder surmounted by a cone. The total height and the diameter of the base are 13.5 m and 28 m. If the height of the cylindrical portion is 3 m, find the total surface area of the tent.
Answer:

Given, Total height of solid = 13.5 cm
Diameter of the cylinder (d) = 28 m
Height of a cylinder (h) = 3 m
Height of a conical portion = 13.5 – 3 = 10.5 m
From the diagram, Radius of a cone = Radius of a cylinder
Radius (r) = 14 m

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 3 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 3 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.
40 ml of methane is completely burnt using 80ml of oxygen at room temperature. The volume of gas left after cooling to room temperature. The volume of gas left after cooling to room temperature is …………………….
(a) 40 ml of CO₂ gas
(b) 40 ml of CO₂ gas and 80 ml of H₂O gas
(c) 60 ml of CO₂ gas and 60 ml H₂O gas
(d) 120 ml of CO₂ gas
Solution:
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(l)

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct.
Answer:
(a) 40 ml of CO₂ gas

Question 2.
What are the values of n, l, m and s for 3p_x electron?
(a) 3, 2, 1,0
(b) 3, 1,-1, +\(\frac{1}{2}\)
(c) 3, 2, +1, –\(\frac{1}{2}\)
(d) 3, 0, 0, +\(\frac{1}{2}\)
Solution:
3p_x electron; n = 3 (main shell)
for p_x orbital, l = 1, m = -1, s = \(\frac{1}{2}\)
Answer:
(b) 3, 1,-1, +\(\frac{1}{2}\)

Question 3.
Which of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having high n value.
Answer:
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having high n value.

Question 4.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 5.
Lithium shows diagonal relationship with ………………………
(a) Sodium
(b) Magnesium
(c) Calcium
(d) Ahuninium
Answer:
(b) Magnesium

Question 6.
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion ie α = \(\frac{1}{V}\) \((\frac { \partial V }{ \partial T } )\)
(a) T
(b) 1/T
(c) P
(d) None of these
Solution:

Answer:
(b) 1/T

Question 7.
Heat of combustion is always …………………….
(a) Positive
(b) Negative
(c) Zero
(d) Either positive or negative
Answer:
(b) Negative

Question 8.
If in a mixture where Q = K, then what happens?
(a) The reaction shift towards products
(b) The reaction shift towards reactants
(c) Nothing appears to happen, but forward and reverse reactions are continuing at the same rate
(d) Nothing happens
Answer:
(c) Nothing appears to happen, but forward and reverse reactions are continuing at the same rate

Question 9.
Which one of the following gases has the lowest value of Henry’s law constant?
(a) N₂
(b) He
(c) CO₂
(d) H₂
Solution:
Carbon dioxide; most stable gas and has lowest value of Henry’s Law constant.
Answer:
(c) CO₂

Question 10.
In the molecule O_A = C = O_B, the formal charge on O_A, C and O_B are respectively.
(a) – 1, 0, + 1
(b) + 1, 0, -1
(c) – 2, 0, + 2
(d) 0, 0, 0
Solution:

Formal charge of O_A/O_B = N_V – (N_e + \(\frac { N_{ b } }{ 2 } \)) = 6 – (4 + \(\frac{4}{2}\)) = 6 – 6 = 0
Formal charge of C = 4 – (0 + \(\frac{8}{2}\)) = 4 – 4 = 0
Answer:
(d) 0, 0, 0

Question 11.
In the hydrocarbon CH₃ – CH₃ – CH = CH – CH₂ – C = CH the state of hybridisation of carbon 1, 2, 3, 4 and 7 are in the following sequence.
(a) sp, sp, sp³, sp², sp³
(b) sp², sp, sp³, sp², sp³
(c) sp, sp, sp², sp, sp³
(d) None of these
Answer:
(a) sp, sp, sp³, sp², sp³

Question 12.
Enzyme present in apple is ………………………..
(a) Polyphenol oxidase
(b) Polyphenol reductase
(c) Polyphenol
(d) Polyphenol hydrolase
Answer:
(a) Polyphenol oxidase

Question 13.
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is ………………………….
(a) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
(b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
(c) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
(d) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
Answer:
(b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

Question 14.
In Finkelstein reaction, the mechanism followed is ……………………….
(a) S_N1
(b) E₁
(c) E₂
(d) S_N2
Answer:
(d) S_N2

Question 15.
Which sequence for greenhouse gases is based on GWP?
(a) CFC > N₂O > CO₂ > CH₄
(b) CFC > CO₂ > N₂O > CH₄
(c) CFC > N₂O > CH₄ > CO₂
(d) CFC > CH₄ > N₂O > CO₂
Answer:
(c) CFC > N₂O > CH₄ > CO₂

PART – II

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

Question 16.
Define Avogadro Number?
Answer:
Avogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = 6.023 × 10²³.

Question 17.
Explain the meaning of the symbol 4P. Write all the four quantum numbers for these electrons?
Answer:
4f²: It means that the element has 2 electrons in outermost 4f shell.
Quantum number values are,
n = principal quantum number = 4
l = azimuthal quantum number = 3
m = magnetic quantum number = – 3, -2
s = spin quantum number = +\(\frac{1}{2}\), –\(\frac{1}{2}\)

Question 18.
Is the definition given below for ionization enthalpy is correct?
“Ionization enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”?
Answer:
No, It is not correct. Thq accurate and absolute definition is as follows:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 19.
What are the uses of calcium hydroxide?
Calcium hydroxide is used
Answer:

1. In the preparation of mortar, a building material.
2. In white wash due to its disinfectant nature.
3. In glass making and tanning industry.
4. For the preparation of bleaching powder and for the purification of sugar.

Question 20.
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^-1 mol^-1. Calculate the melting point of sodium chloride?
Answer:
Given:
∆H_f = 30.4 kJ = 30400 J mol^-1
∆S_f (NaCl) = 28.4 kJ^-1 mol^-1
T_f = ?
∆S_f = \(\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{f}}}\); T_f = \(\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{S}_{\mathrm{f}}}\)
T_f = \(\frac{30400 \mathrm{J} \mathrm{mol}^{-1}}{28.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}}\) = 1070.4 K

Question 21.
How is a gas-solution equilibrium exist?
Answer:
When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecules in the gaseous state and those dissolved in the liquid.
Example: In carbonate beverages the following equilibrium exists.
CO₂(g) ⇄ CO₂ (solution).

Question 22.
What type of hybridisations are possible in the following geometeries?

1. Octahedral
2. Tetrahedral
3. Square planar

Answer:

1. Octahedral geometry is possible by sp³d² (or) d²sp³ hybridisation.
2. Tetrahedral geometry is possible by sp³ hybridisation.
3. Square planar geometry is possible by dsp² hybridisation.

Question 23.
How will you prepare Lassaigne’s extract?
Lassagine’s extract preparation:
Answer:

1. A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is gently heated. When it melts to a shining globule, a pinch of organic compound is added.
2. The tube is heated till reaction ceases and become red hot. Then it is plunged in 50 ml of distilled water taken in a china dish and the bottom of the tube is broken by striking it against the dish.
3. The contents of the dist is boiled for 10 minutes and then it is filtered. The filtrate is known as Lassaigne’s extract.

Question 24.
Complete the reactions and identify the products?

Answer:

PART – III

Answer any six questions in which question No. 32 is compulsory. [6 × 3 = 18]

Question 25.
The balanced equation for a reaction is given below 2x + 3y → 41 + m
When 8 moles of x react with 15 moles of y, then

1. Which is the limiting reagent?
2. Calculate the amount of products formed.
3. Calculate the amount of excess reactant left at the end of the reaction.

Answer:
2x + 3y → 41 + m
1. 2x reacts with 3y to give products.
8x reacts with 15y means, y is the excess because 8 moles of x should react with 4 × 3y = 12y moles of y to give products.
In this reaction 15y moles are used.
Therefore, 3 moles of y is excess and x is the limiting agent.

2. When 8 moles of x react with 12 moles of y, the product formed will be 4 × 41 i.e. 161 and 4m as product.
8x+ 12y → 161 + 4m

3. At the end of the reaction, the excess reactant left is 3 moles of y.

Question 26.
Which would you expect to have a higher melting point, magnesium oxide or magnesium fluoride? Explain your reasoning.
Answer:

1. Magnesium oxide has very strong ionic bonds as compared to magnesium fluoride.
2. Mg²⁺ and O^2- have charges of +2 and -2, respectively.
3. Oxygen ion is smaller than fluoride ion.
4. The smaller the ionic radii, the smaller the bond length in MgO and the bond is stronger than MgF₂.
5. Due to more strong bond nature in MgO, it has high melting point than MgF₂.

Question 27.
A sample of solid KClO₃ (potassium chlorate) was heated in a test tube to obtain O₂ according to the reaction
2KClO_3(s) → 2KCl + 3O₂
The oxygen gas was collected by downward displacement of water at 295 K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mm of Hg at 300K. What is the partial pressure of the oxygen gas?
Answer:

Question 28.
List the characteristics of entropy?
Characteristics of entropy:
Answer:

• Entropy is a thermodynamic state function that is a measure of the randomness or disorderliness of the system.
• In general, the entropy of gaseous system is greater than liquids and greater than solids. The symbol of entropy is S.
• Entropy is defined as “for a reversible change taking place at a constant temperature (T), the change in entropy (∆S) of the system is equal to heat energy absorbed or evolved (q) by the system divided by the constant temperature (T).
  \(\Delta S_{\mathrm{sys}}=\frac{q_{\mathrm{rev}}}{T}\)
• If heat is absorbed, then ∆S is positive and there will be increase in entropy. If heat is evolved, ∆S is negative and there is a decrease in entropy,
• The change in entropy of a process represented by ∆S and is given by the equation,
  ∆S_sys = S_f – S_i
• If S_f > S_i, ∆S is positive, the reaction is spontaneous and reversible.
  If S_f < S_i, ∆S is negative, the reaction is non-spontaneous and irreversible.
• Unit of entropy: SI unit of entropy is J K^-1.

Question 29.
Derive the values of K_C and K_P for the synthesis of HI.
Answer:
H₂(g) + I₂(g) ⇄ 2HI(g)
Let us consider the formation of HI in which V moles of hydrogen, ‘b’ moles of iodine gas are allowed to react in an container of volume ‘V’.
Let ‘x’ moles of each of H₂ and I₂ react together to form 2x moles of HI.

Applying mass of action

Caluculation of K_P: K_P = K_C.RT^∆ng
Here ∆n_g = n_p – n_r = 2 -2 = 0
Hence, K_P = K_C
K_P = \(\frac{4 x^{2}}{(a-x)(b-x)}\)

Question 30.
Describe the classification of organic compounds based on their structure?
Answer:
Classification of organic compounds based on the structure

Question 31.
For the following bond cleavages use curved-arrows to show the electron flow and classify each as homolytic or heterolytic fission. Identify reactive intermediate produced as free radical, carbocation and carbanion?

Answer:

Question 32.
An alkyl halide with molecular formula C₆H₁₂Br on dehydrohalogenation gave two isomeric alkenes X and,Y with molecular formula C₆H₁₂. On reductive ozonolysis, X and Y gave four compounds CH₃COCH₃, CH₃CHO, CH₃CH₂CHO and (CH₃)₂ CHCHO. Find the alkyl halide?
Answer:

1. C₆ H₁₃ Br is 3 – Bromo – 4 – methylpentane.
2. 3 – Bromo – 4 – methylpentane on dehydration give two isomers X and Y as follows:

Therefore C₆H₁₃ Br is 3 – Bromo – 4 – methylpentane

Question 33.
What do you mean by ozone hole? What are its consequences?
Answer:
Depletion of ozone layer creates some sort of holes in the blanket of ozone which surrounds us in the atmosphere and this is known as ozone hole.

1. With the depletion of the ozone layer, UV radiations filters into the troposphere which leads to ageing of skin, cataract, sunburn, skin cancer etc.
2. By killing many of the phytoplanktons, it can damage the fish productivity.
3. Evaporation rate increases through the surface and stomata of leaves which can decrease the moisture content of the soil.

PART – IV

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

Question 34 (a).
(I) Balance the following equations by ion electron method?
KMnO₄ + SnCl₂ + HCl → MnCl₂ + SnCl₄ + H₂O + KCl
(II) Boric acid, H₃BO₃ is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H₃BO₃. What is the mass of boric acid in the sample?

[OR]

(b)
(I) How many unpaired electrons are present in the ground state of Fe³⁺ (z = 26), Mn²⁺ (z = 25) and argon (z = 18)?
(II) Explain about the significance of de Broglie equation?
Answer:
KMnO₄ + SnCl₂ + HCl → MnCl₂ + SnCl₄ + H₂O + KCl
Oxidation half reaction: (loss of electron)

Reduction of halfa reaction: (gain of electron)

Add H₂O to balance oxygen atoms

Add Hcl to balance hydrogen atoms
KMnO₄ + 5e^– + 8 HCl → MnCl₂ + 4H₂O …………………. (4)

To equalize the number of electrons equation (1) × 5 and equation (2) × 2

(II) Molecular mass of H₃BO₃ = (1 × 3) + (11 × 1) + (16 × 3) = 62
Boric acid sample contains 0.543 mole.
Mass of 0.543 mole of Boric acid = Molecular mass x mole
= 62 × 0.543
= 33.66 g

[OR]

(b) (I) Fe → Fe³⁺ + 3e^–
Fe (Z = 26)
Fe³⁺ = number of electrons = 23
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion.
So, it contain 5 unpaired electrons.
Mn (Z = 25) Electronic configuration is
1s² 2s² 2p⁵ 3s² 3p⁶ 4s² 3d⁵
Mn → Mn²⁺ + 2e^–
Number of unpaired electrons in Mn²⁺ = 5
Ar (Z = 18). Electronic configuration is
1s² 2s² 2p⁶ 3s² 3p⁶.
All orbitals are completely filled. So, no unpaired electrons in it.

(II) Significance of de Broglie equation:

1. λ = \(\frac{h}{mv}\) This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle.
2. For a particle with high linear momentum (mv) the wavelength will be so small and cannot be observed.
3. For a microscopic particle such as an electron, the mass is of the order of 10^-31 kg, hence the wavelength is much larger than the size of atom and it becomes significant.
4. For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

Question 35 (a).
(I) Mention any two anomalous properties of second period elements?
(II) Arrange Na⁺, Mg²⁺ and Al³⁺ in the increasing order of ionic radii. Give reason?

[OR]

(b)
(I) How do you expect the metallic hydrides to be useful for hydrogen storage?
(II) Write a note about ortho water and para water?
Answer:
(a) (I) Anomalous properties of second period elements:

1. In the 1^st group, lithium forms compounds with more covalent character while the other elements of this group form only ionic compounds.
2. In the 2^nd group, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.

(II) Na⁺, Mg²⁺ and Al³⁺ are isoelectronic cations.

The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is, \(\mathbf{r}_{\mathrm{Na}^{+}}\), > \(\mathbf{r}_{\mathrm{Mg}^{2+}}\), > \(\mathbf{r}_{\mathrm{Al}^{3+}}\).

[OR]

(b)
(I) In metallic hydrides, hydrogen is adsorbed as H-atoms. Due to the adsorption of H atoms the metal lattice expands and become unstable. Thus, when metallic hydride is heated, it decomposes to form hydrogen and finely divided metal. The hydrogen evolved can be used as fuel.

(II)
1. Water exists in space in the interstellar clouds, in proto-planetary disks, in the comets and icy satellites on the solar system, and on the Earth.
2. In particular, the ortho-to-para ratio (OPR) of water in space has recently received attention. Like hydrogen, water can also be classified into ortho – H₂O and para – H₂O, in which the directions are antiparallel.

3. At the temperature conditions of the Earth (300 K), the OPR of H₂O is 3.
4. At low temperatures below (< 50 K) the amount of para – H₂O increases. It is known that the OPR of water in interstellar clouds and comets has more para – H₂O (OPR = 2.5) than on Earth.

Question 36 (a).
Derive the values of critical constants from the Van der Waals constants?

[OR]

(b) Derive the values of K_p and K_C for dissociation of PCl₅?
Answer:
(a) Derivation of critical constants from the Van der Waals constants:
Van der Waals equation is,
\(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT for 1 mole.
From this equation, the values of critical constant P_C, V_C and T_C are derived in terms of a and b the Vander Waals constants.

The above equation (4) is an cubic equation of V, which can have three roots. At the critical point, all the three values of V are equal to the critical volume Vc. i.e. V = V_C„
i.e; V = V_C
V – V_C = 0 ………………… (5)
(V – V_C)³ ………………….. (6)
(V³ – 3V_CV²) + 3V_C²V – V_C³ …………………. (7)
As the equation (4) is identical with equation (7), comparing the ‘V’ terms in (4) and (7),

Divide equation (11) by (10)

When equation (12) is substituted in (10)

Substituting the values of V_C and P_C in equation (9)

Critical constants a and b can be calculated rising Vander Waals constants as follows:

(b) Consider that V moles of PCl₅is taken in container of volume‘V’
Let x moles of PCl₅ be dissociated into x moles of PCl₃ and x moles of Cl₂.

Applying law of mass action

K_p caluculation: K_p = K_C. \(\mathrm{RT}^{\mathrm{An}}\); ∆n_g = 2 – 1 = 1
We know that PV = nRT
RT = \(\frac{PV}{n}\)
Where ‘n’ is the total number of moles at equilibrium
n = a – x + x + x = a + x

Question 37 (a).
(I) Solubility of a solid solute in a liquid solvent increases with increase in temperature. Justify this statement?
(II) Explain how non-ideal solutions shows positive deviation from Raoult’s law?

[OR]

(b)
(I) How will you distinguish between electrophiles and nucleophiles?
(II) Complete the following reactions and identify the products?
(a)
(b)
Answer:
(a) (I) When the temperature is increased,the average kinetic energy of the molecules of the solute and the solvent increases. The increase in the kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.
(II)

1. Let us consider the positive deviation shown by a solution of ethyl alcohol and water
2. In this solution, the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alco¬hol and water-water interaction).
3. This results in the increased evaporation of both components from the aqueous solution of ethanol.
4. Consequently, the vapour pressure of the solution is greater than the vapour pressure
   predicted by Raoult’s law.
5. Here, the mixing process is endothermic i.e., DHmixing > 0 and there will be a slight increase in volume (DVmixing > 0)
6. 

(b)
(I)
Electrophiles:

1. They are electron deficient.
2. They are cations.
3. They are lewis acids.
4. Accept an electron pair.
5. Attack on electron rich sites.

Nucleophiles:

1. They are electron rich.
2. They are anions.
3. They are lewis bases.
4. Donate an electron pair.
5. Attack on electron deficient sites.

(II)
(a)
(b)

Question 38 (a).
(I) Is it possible to prepare methane by Kolbe’s electrolytic method?
(II) Explain how 2-butyne reacts with
(a) Lindlar’s catalyst
(b) Sodium in liquid ammonia.

[OR]

(b) (I) Discuss the aromatic nucleophilic substitution reactions of chlorobenzene?
(II) CCl₄ > CHCl₃ > CH₂C₁₂ > CH₃Cl is the decreasing order of boiling point in haloalkanes. Give reason?
Answer:
(a) (I) Kolbe’s electrolytic method is suitable for the preparation of symmetrical alkanes, that is alkanes containing even number of carbon atoms. Methane has only one carbon, hence it cannot be prepared by Kolbe’s electrolytic method.
(II) (a) 2-butyne reacts with Lindlar’s catalyst: 2-butyne can be reduced to cis – 2 butene using CaCO₃ supported in Pd -metal partially deactivated with sulphur.
This reaction is stereo specific giving only the cis – 2 – butene.

(b) 2-butyne reacts with sodium in liquid ammonia:
2-butyne can also be reduced to trans – 2 – butene using sodium in liquid ammonia. This reaction is stereospecific giving only the trans – 2 – butene.

[OR]

(b) (I) Aromatic nucleophilic substitution reactions:
Dow’s process:

1.

2.

3.

(II) The boiling point of chloro, bromo and iodoalkanes increases with increase in the number of halogen atoms. So the correct decreasing order of boiling point of haloalkanes is:
CCl₄ > CHCl₃ > CH₂Cl₂ > CH₃Cl.

Students can download Maths Chapter 8 Statistics and Probability Unit Exercise 8 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 8 Statistics and Probability Unit Exercise 8

Question 1.
The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f₁ and f₂.

Answer:
Arithmetic Mean (\(\bar{x}\)) = 62.8
Sum of all the frequencies (Σf_i) = 50
Let the missing frequencies be f₁ and f₂


Substitute of the value of f₂ in (1)
f₁ + 12 = 20
⇒ f₁ = 20 – 12 = 8
The Missing frequency is 8 and 12.

Question 2.
The diameter of circles (in mm) drawn in the design are given below.

Calculate the standard deviation.
Answer:
Assumed mean = 42. 5

Question 3.
The frequency distribution is given below

In table k is a positive integer, has a variance of 160. Determine the value of k.
Answer:
Assumed mean = 3k



The value of k = 7

Question 4.
The standard deviation of some temperature data in degree Celsius (°C) is 5. If the data were converted into degree Fahrenheit (°F) then what is the variance?
Solution:
F° = (C° × 1.8) + 32
\(\begin{array}{l}{\sigma_{c}=5^{\circ} \mathrm{C}} \\ {\sigma_{\mathrm{F}}=\left(1.8 \times 5^{\circ} \mathrm{C}\right) \cdot 9^{\circ} \mathrm{F}}\end{array}\)
Adding value to data doesn’t affect standard deviation.
New variance = \(\sigma_{\mathrm{F}}^{2}=81^{\circ} \mathrm{F}\)

Question 5.
If for a distribution, Σ (x – 5) = 3, Σ (x – 5)² = 43, and total number of observations is 18, find the mean and standard deviation.
Answer:


(i) Arithmetic mean (\(\bar{x}\)) = 5.17
(ii) Standard deviation (σ) = 1.53

Question 6.
Prices of peanut packets in various places of two cities are given below. In which city, prices were more stable?

Answer:
Coefficients of variation of prices in city A.
Arrange in ascending order we get, 16, 19, 20, 22, 23
Assumed mean = 20

Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100\)
= \(\frac{2.45}{20} \times 100\)
= 12. 25%
Coefficient of variation = 12.25%
Coefficients of variation of prices in city B.
Arrange in ascending order we get, 10, 12, 15, 18, 20
Assumed mean = 15


Prices in city A is more stable (since 12.25 < 24.6 %)

Question 7.
If the range and coefficient of range of the data are 20 and 0.2 respectively, then find the largest and smallest values of the data.
Answer:
Range of the data (R) = 20
L – S = 20 ……(1)
Coefficient of range = 0.2
Coefficient of range = \(\frac{L-S}{L+S}\)
0.2 = \(\frac{20}{L+S}\)

substituting the value of L = 60 in (2)
60 + S = 100
S = 100 – 60 = 40
Largest value = 60
Smallest value = 40

Question 8.
If two dice are rolled, then find the probability of getting the product of face value 6 or the difference of face values 5.
Answer:
Sample space = {(1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),(2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6), (3, 1), (3, 2), (3, 3),(3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1),(6, 2), (6, 3), (6, 4),(6, 5), (6, 6)}
n(S) = 36
(i) Let A be the event of getting product of face value 6.
A = {(1, 6), (2, 3), (3, 2) (6, 1)}
n(A) = 4
P (A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{4}{36}\)
(ii) Let B be the event of getting difference of face value is 5.
B = {(6, 1)}
n(B) = 1

The probability is \(\frac{1}{9}\)

Question 9.
In a two children family, find the probability that there is at least one girl in a family.
Answer:
Sample space (S) = {(Boy, Boy) (Boy, Girl) (Girl, Boy) (Girl, Girl)}
n(S) = 4
Let A be the event of getting atleast one Girl
A = {(Boy, Girl) (Girl, Boy) (Girl, Girl)}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{4}\)
Probability of atleast one girl in a family is \(\frac{3}{4}\)

Question 10.
A bag contains 5 white and some black balls. If the probability of drawing a black ball from the bag is twice the probability of drawing a white ball then find the number of black balls.
Answer:
Let the number of black balls be “x”
Sample space (S) = x + 5
n(S) = x + 5
Let A be the event of drawing a black ball
n (A) = x
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{x}{x+5}\)
Let B be the event of getting a white ball
n(B) = 5
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{5}{x+5}\)
By the given condition,
\(\frac{x}{x+5}=2 \times\left(\frac{5}{x+5}\right)\)
⇒ \(\frac{x}{x+5}=\left(\frac{10}{x+5}\right)\)
⇒ 10x + 50 = x² + 5x
⇒ x² + 5x – 10x – 50 = 0
⇒ x² – 5x – 50 = 0
⇒ (x – 10)(x + 5) = 0
⇒ x = 10 or x = -5
Number of balls will not be negative.
Number of black balls = 10

Question 11.
The probability that a student will pass the final examination in both English and Tamil is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Tamil examination?
Answer:
Let A be the event of getting student pass in English
Let B be the event of getting student pass in Tamil


Probability of passing the tamil examination is \(\frac{13}{20}\)

Question 12.
The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting
(i) a diamond
(ii) a queen
(iii) a spade
(iv) a heart card bearing the number 5.
Answer:
Total number of cards = 52
3 cards are removed
Remaining number of cards = 52 – 3 = 49
n(S) = 49
(i) Let A be the event of getting a diamond card.
n(A) = 13
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{13}{49}\)
(ii) Let B be the event of getting a queen card.
n(B) = (4 – 1) = 3
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{3}{49}\)
(iii) Let C be the event of getting a spade card.
n(C) = (13 – 3) = 10
P(C) = \(\frac{n(C)}{n(S)}=\frac{10}{49}\)
(iv) Let D be the event of getting a 5 of heart card.
n(D) = 1
P(D) = \(\frac{n(\mathrm{D})}{n(\mathrm{S})}=\frac{1}{49}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.5 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.5

Multiple Choice Questions.

Question 1.
Which of the following is not a measure of dispersion?
(1) Range
(2) Standard deviation
(3) Arithmetic mean
(4) Variance
Answer:
(3) Arithmetic mean
Hint:
Measures of dispersion are,
(i) Range
(ii) Mean deviation
(iii) Quartile deviation
(iv) Standard deviation
(v) Variance
(vi) coefficient of variation

Question 2.
The range of the data 8, 8, 8, 8, 8.. . 8 is
(1) 0
(2) 1
(3) 8
(4) 3
Solution:
(1) 0
Hint
R = L – S = 8 – 8 = 0

Question 3.
The sum of all deviations of the data from its mean is _______
(1) always positive
(2) always negative
(3) zero
(4) non-zero integer
Answer:
(3) zero

Question 4.
The mean of 100 observations is 40 and their standard deviation is 3. The sum of squares of all deviations is
(1) 40000
(2) 160900
(3) 160000
(4) 30000
Solution:
(2) 160900
Hint:
σ = 3

Question 5.
Variance of first 20 natural numbers is ______
(1) 32.25
(2) 44.25
(3) 33.25
(4) 30
Answer:
(3) 33.25
Hint:
Variance of 20 natural numbers is

Question 6.
The standard deviation of a data Is 3. If each value is multiplied by 5 then the new variance is
(1) 3
(2) 15
(3) 5
(4) 225
Solution:
σ = 3. 1f each is multiplied by 5. The new standard variation is also multiplied by 3.
∴ The new S.D = 5 × 3 = 15
Variance = 15² = 225

Question 7.
If the standard deviation of x, y, z is p then the standard deviation of 3x + 5, 3y + 5, 3z + 5 is ________
(1) 3p + 5
(2) 3p
(3) p + 5
(4) 9p + 15
Answer:
(2) 3p
Hint:
(i) Each value is added by any constant there is no change in standard deviation.
(ii) Each value is multiplied by 3 standard deviations also multiplied by 3.
The standard deviation is 3p.

Question 8.
If the mean and coefficient of variation of a data are 4 and 87.5% then the standard
deviation is
(1) 3.5
(2) 3
(3) 4.5
(4) 2.5
Solution:
(1) 3.5
Hint:
\(\bar{x}\) = 4, coefficient of variation is = 87. 5%

Question 9.
Which of the following is incorrect?
(1) P(A) > 1
(2) 0 ≤ P(A) ≤ 1
(3) P(Φ) = 0
(4) P(A) + P(\(\bar{A}\)) = 1
Answer:
(1) P(A) > 1
Hint:
Probability is always less than one or equal to one.

Question 10.
The probability a red marble selected at random from a jar containing p red, q blue and r green marbles is _________
(1) \(\frac{q}{p+q+r}\)
(2) \(\frac{p}{p+q+r}\)
(3) \(\frac{p+q}{p+q+r}\)
(4) \(\frac{p+r}{p+q+r}\)
Answer:
(1) \(\frac{q}{p+q+r}\)
Hint:
Sample spaces = p + q + r
Let A be the event of getting red
n(A) = p
P(A) = \(\frac{q}{p+q+r}\)

Question 11.
A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 7 is _______
(1) \(\frac{3}{10}\)
(2) \(\frac{7}{10}\)
(3) \(\frac{3}{9}\)
(4) \(\frac{7}{9}\)
Answer:
(2) \(\frac{7}{10}\)
Hint:
Here n(S)= 10 (given digit at imit place. It has two digit)
n(A) = 7
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{7}{10}\)

Question 12.
The probability of getting a job for a person is \(\frac{x}{3}\). If the probability of not getting the job is \(\frac{2}{3}\) then the value of x is.
(1) 2
(2) 1
(3) 3
(4) 1.5
Solution:
(2) 1
Hint:
Probability of getting a job = \(\frac{x}{3}\)
Probability of not getting a job = 1 – \(\frac{x}{3}\)

Question 13.
Kamalam went to play a lucky draw contest. 135 tickets of the lucky draw were sold. If the probability of Kamalam winning is \(\frac{1}{9}\), then the number of tickets bought by Kamalam is _______
(1) 5
(2) 10
(3) 15
(4) 20
Answer:
(3) 15
Hint:
n(S) = 135
P(A) = \(\frac{1}{9}\)

Question 14.
If a letter is chosen at random from the English alphabets {a, b, …, z}, then the probability that the letter chosen precedes x.
(1) \(\frac{12}{13}\)
(2) \(\frac{1}{13}\)
(3) \(\frac{23}{26}\)
(4) \(\frac{3}{26}\)
Solution:
(3) \(\frac{23}{26}\)
Hint:
n(S) = 26
Let A denote the letter chosen precedes x
A= {a, b, c, d, …, x}
n(A) = 23
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{23}{26}\)

Question 15.
A purse contains 10 notes of ₹ 2000, 15 notes of ₹ 500, and 25 notes of ₹ 200. One note is drawn at random. What is the probability that the note is either a ₹ 500 note or ₹ 200 note?
(1) \(\frac{1}{5}\)
(2) \(\frac{3}{10}\)
(3) \(\frac{2}{3}\)
(4) \(\frac{4}{5}\)
Answer:
(4) \(\frac{4}{5}\)
Hint:
Sample space (S) = 10 + 15 + 25 = 50
n(S) = 50
Let A be the event of getting ₹ 500 note
n (A) = 15
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{15}{50}\)
Let B be the event of getting ₹ 200 note
n (B) = 25
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{25}{50}\)
Probability of the note is either a ₹ 500 note or ₹ 200 note
P(A) + P(B) = \(\frac{15}{50}+\frac{25}{50}\) = \(\frac{40}{50}\) = \(\frac{4}{5}\)

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 4 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 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: 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.
When 6.3 g of sodium bicarbonate is added to 30 g of acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is ……………………
(a) 3
(b) 0.75
(c) 0.075
(d) 0.3
Solution:

The amount of CO₂ released, x = 3.3 g
No. of moles of CO₂ released = 3.3/4.4 = 0.075 mol
Answer:
(c) 0.075

Question 2.
Two electrons occupying the same orbital are distinguished by …………………
(a) Azimuthal quantum number
(b) Spin quantum number
(c) Magnetic quantum number
(d) Orbital quantum number
Solution:
Spin quantum number For the first electron ms = + \(\frac{1}{2}\)
For the second electron ms = – \(\frac{1}{2}\)
Answer:
(b) Spin quantum number

Question 3.
Statement – 1: Ionization enthalpy of N is greater than that of O.
Statement – II: N has exactly half filled electronic configuration which is more stable than electronic configuration of O.
(a) Statement – I is wrong but statement – II is correct
(b) Statement – I is correct but statement – II is wrong.
(c) Statement – I and II are correct and statement – II is the correct explanation of statement – I.
(d) Statement – I and II are correct but statement – II is not the correct explanation of statement – I.
Answer:
(c) Statement – I and II are correct and statement – II is the correct explanation of statement – I.

Question 4.
Water gas is …………………….
(a) H₂O_(g)
(b) CO + H₂O
(c) CO + H₂
(d) CO + N₂
Answer:
(c) CO + H₂

Question 5.
Among the following the least thermally stable is ……………………
(a) K₂CO₃
(b) Na₂CO₃
(c) BaCO₃
(d) Li₂CO₃
Li₂CO₃ is least stable.
Answer:
(d) Li₂CO₃

Question 6.
Match the following.

Answer:

Question 7.
C(diamond) → C(graphite), ∆H = -ve, this indicates that …………………
(a) Graphite is more stable than diamond
(b) Graphite has more energy than diamond
(c) Both are equally stable
(d) Stability cannot be predicted
Answer:
(a) Graphite is more stable than diamond

Question 8.
In the equilibrium, 2A(g) ⇄ 2B(g) + C₂(g)
the equilibrium concentrations of A, B and C₂ at 400 K are 1 × 10^-4M, 2.0 × 10^-3M, 1.5 × 10⁴M respectively. The value of K_C for the equilibrium at 400 K is ……………………..
(a) 0.06
(b) 0.09
(c) 0.62
(d) 3 × 10^-2
Solution:
[A] = 1 × 10^-4M; [B] = 2 × 10^-3M; [C] = 1.5 × 10^-4M
2A(g) ⇄ 2B(g) + C₂(g)

= 6.0 × 10^-2 = 0.06
Answer:
(a) 0.06

Question 9.
Which of the following is a non-aqueous solution?
(a) Salt solution
(b) Sugar solution
(c) Br₂ in CCl₄
(d) Ethanol dissolved in water
Answer:
(c) Br₂ in CCl₄

Question 10.
Which of the following molecule does not exist due to its zero bond order?
(a) \(\mathrm{H}_{2}^{-}\)
(b) \(\mathrm{He}_{2}^{+}\)
(c) He₂
(d) \(\mathrm{H}_{2}^{+}\)
Answer:
(c) He₂

Question 11.
Which of the following is optically active?
(a) 3 – Chloropentane
(b) 2 – Chloropropane
(c) Meso – tartaric acid
(d) Glucose
Answer:
(d) Glucose

Question 12.
Which of the following represent a set of nucleophiles?
(a) BF₃, H₂O, NH^2-
(b) AlCl₃, BF₃, NH₃
(c) CN^–, RCH₂^–, ROH
(d) H⁺, RNH₃⁺, CCl₂
Answer:
(c) CN^–, RCH₂^–, ROH

Question 13.
Propyne on passing through red hot iron tube gives ……………………
(a)
(b)
(c)
(d) one of these
Answer:
(a)

Question 14.
Consider the following statements:

(I) E₂ reaction is a bimolecular elimination reaction of second order
(II) E₂ reaction takes place in two steps.
(III) E₂ reaction generally Jakes place in primary alkyl halides.

Which of the above statements is/are not correct?
(a) (I) only
(b) (II) only
(c) (III) only
(d) (I) & (III)
Answer:
(II) E₂ reaction takes place in two steps.

Question 15.
Photo chemical smog formed in congested metropolitan cities mainly consists of ………………….
(a) Ozone, SO₂ and hydrocarbons
(b) Ozone, PAN and NO₂
(c) PAN, smoke and SO₂
(d) Hydrocarbons, SO₂ and CO₂
Answer:
(b) Ozone, PAN and NO₂

PART – II

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

Question 16.
Why interstitial hydrides have a lower density than the parent metal?
Answer:

1. d block elements form metallic or interstitial hydrides, on heating with dihydrogen under pressure.
2. Hydrogen atom being small in size occupy some in the metallic lattice producing distortion without any change in its type.
3. The densities of these hydrides are lower than those of metals from which they are formed since the crystal lattice expands due to the inclusion of dihydrogen.

Question 17.
Prove that calcium oxide is a basic oxide?
Answer:
Calcium oxide is a basic oxide. It combines with acidic oxides at high temperature.

1. 
2. 

Question 18.
Give the mathematical expression that relates gas volume and moles. Describe in words what the mathematical expression means?
Answer:

1. The mathematical relationship between the volume of a gas and the number of moles is V ∝ n
2. \(\frac { V_{ 1 } }{ n_{ 1 } } \) = \(\frac { V_{ 2 } }{ n_{ 2 } } \) Constant, where V₁ and n₁ are the volume and number of moles of a gas and V₂ and n₂
   and n₂ are the values of volume and number of moles of same gas at a different set of conditions.
3. If the volume of the gas increase then the number of moles of the gas also increases.
4. At a certain temperature and pressure, the volume of a gas is directly proportional to the number of the moles of the gas.

Question 19.
Why pressure has no effect on the synthesis of HI?
Answer:
When the total number of moles of gaseous reactants and gaseous products are equal, the change in pressure has no effect on system at equilibrium.
H₂(g) + I₂(g) ⇄ 2HI(g)
Here the number of moles of reactants and products are equal. So the pressure has no effect on such equilibrium with ∆n_g = 0.

Question 20.
Draw the lewis structure of PCl₅ and SF₆
Answer:

Question 21.
How are naphthalene and camphor purified?
Answer:

1. Naphthalene, camphor and benzoic acid when heated, pass directly from solid to vapour without melting. On cooling the vapours will give back solid. This phenomenon is known as sublimation. This technique is used to purify naphthalene, camphor from non volatile impurities.

2. Substances to be purified is taken in a beaker. It is covered with a watch glass. The beaker is heated for a while and the resulting vapours condense on the bottom of the watch glass. Then the watch glass is removed and the crystals are collected.

Question 22.
How will you convert ethyl chloride into
(I) Ethane
(II) n – butane
Answer:
(I) Conversion of ethyl chloride into ethane:

(II) Conversion of ethyl chloride into n – butane:
Wurtz reaction:

Question 23.
Chloroform is kept with a little ethyl alcohol in a dark coloured bottle, why?
Answer:
(I) Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carboxyl chloride (phosgene), it is therefore stored in closed dark coloured bottles completely filled so that air is kept out.

(II) With the use of 1 % ethanol we can stabilise chloroform, because ethanol can convert the poisonous COCl₂ gas into non poisonous diethyl carbonate.
COCl₂ + 2C₂H₅OH → CO(OC₂H₅)₂ + 2HCl.

Question 24.
How does classical smog differ from photochemical smog?
Answer:
Classical smog:

1. Classical smog is caused by coal-smoke and fog.
2. It occurs in cold humid climate.
3. The chemical composition is the mixture of SO₂, SO₃ gases and humidity.
4. Chemically it is reducing in nature because of high concentration of SO₂ and so it is also called reducing smog.
5. It is primarily responsible for acid rain.
6. It also causes bronchial irritation.

Photochemical smog:

1. Photochemical smog is cause by photochemical oxidants.
2. It occurs in warm, dry and sunny climate.
3. The chemical composition is the mixture of NO₂ and O₃ gases.
4. Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₃ and so it is also called oxidising smog.
5. It causes irritation to eyes, skin and lungs and increase the chances of asthma.
6. It causes corrosion of metals, stones and painted surfaces.

PART – III

Answer any six questions in which question No. 28 is compulsory. [6 × 3 = 18]

Question 25.
An ice cube at 0°C is placed in some liquid water at 0°C, the ice cube sinks – Why?
Answer:

1. In an ice cube, each atom is surrounded tetrahedrally by four water molecules through hydrogen bond and its density is low.
2. Liquid water at 0°C has the density as 999.82 kg/cm³. Maximum density is attained by water only at 4°C as 1000 kg/cm³.
3. When the temperature changed from 4°C to 0°C, the density of water decreases rather than increases. This is called anomalous expansion of water.
4. The reason for this phenomenon lies in the structure of ice lattice and hydrogen bonding in water,
5. At 0°C, ice cube sinks in liquid water at 0°C because of the lesser density and greater volume of water.

Question 26.
Write the chemical equations for the reactions involved in Solvay process of preparation of sodium carbonate?
Answer:
Solvay process:
The Solvay process is represented by the below chemical equations:

Question 27.
Explain whether a gas approaches ideal behaviour or deviates from ideal behaviour if
(a) It Is compressed to a smaller volume at constant temperature
(b) The temperature is raised while keeping the volume constant
(c) More gas is introduced into the same volume and at the same temperature
Answer:
(a) If a gas is compressed to a smaller volume at constant temperature, pressure is increased. At high pressure with a smaller volume, the gas deviates from ideal behaviour.

(b) If a gas temperature is raised keeping the volume constant, the pressure of the gas will increase. At high pressure, the gas deviates from ideal behaviour.

(c) If more gas is introduced into the same volume and at the same temperature, the number of moles are increasing. If the volume remains same, the increased number of moles collide with each other and kinetic energy increases and pressure decreases. At increased pressure, the gas deviates from ideal behaviour.

Question 28.
Calculate \(\Delta \mathrm{H}_{\mathrm{r}}^{0}\) for the reaction
CO₂(g) + H₂(g) → CO(g) + H₂O (g)
given that \(\Delta \mathrm{H}_{\mathrm{f}}^{0}\) for CO₂(g), CO(g) and H₂O(g) are – 393.5, – 111.31 and – 242 kJ mol^-1 respectively.
Answer:
Given:
\(\Delta \mathrm{H}_{\mathrm{f}}^{0}\) CO₂ = -393.5 KJ mol^-1

Question 29.
Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O₂ is paramagnetic?
Answer:
(I) Electronic configuration of O atom is 1s² 2s² 2p⁴

(II) Electronic configuration of O₂ molecule is

(III) Bond order = \(\frac { N_{ b }-N_{ a } }{ 2 } \) = \(\frac{10-6}{2}\) = 2

(IV) Molecule has two unpaired electrons, hence it is paramagnetic.

Question 30.
Give the principle involved in the estimation of halogen in an organic compound by Carius method?
Answer:
Estimation of halogens: Carius method
(I) A known mass of the organic compound is heated with fuming HNO₃ and AgNO₃.

(II) C, H and S gets oxidised to CO₂, H₂O and SO₂ and halogen combines with AgNO₃ to form a precipitate of silver halide.

(III) The precipate AgX is filtered, washed, dried and weighted.

(IV) From the mass of AgX and the mass of organic compound taken, the percentage of halogen are caluculated.
(V)

Question 31.
What polymerisation? Explain with suitable example?
Answer:
A polymer is a larga molecule formed by the combination of large number of small molecules (monomers). This process is known as polymerisation, a few examples are:

Question 32.
Compare \(\mathbf{S}_{\mathrm{N}^{1}}\) and \(\mathbf{S}_{\mathrm{N}^{2}}\) reaction mechanisms?
Answer:

Question 33.
From where does ozone come in the photochemical smog?
Answer:
(I) Photochemical smog is formed by the combination of smoke, dust and fog with air pollutants in the presence of sunlight.

(II) Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₃. So it is also called oxidising smog.

(III) Photochemical smog is formed by following reactions:
N₂ + O₂ 2NO
2NO + O₂ 2NO₂

(O) + O₂ O₃
O₃ + NO NO₂ + (O)

(IV) NO and O₃ are strong oxidising agents and they can react with unbumt hydrocarbons in polluted air to form formaldehyde, acrolein and PAN.

PART – IV

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

Question 34 (a).
(I) An atom of an element contains 35 electrons and 45 neutrons. Deduce

1. The number of protons
2. The electronic configuration for the element
3. All the four quantum numbers for the last electron

(II) How many unpaired electrons are present in the ground state of Fe²⁺ (z = 26), Mn²⁺ (z = 25) and argon (z=18)?

[OR]

(b)
(I) Explain why hydrogen is not placed with the halogen in the periodic table.
(II) Complete the following reactions.
Al₄C₃ + D₂O → ?
CaC₂ + D₂O → ?
Mg₃N₂, + D₂O → ?
Ca₃P₂ + D₂O → ?
Answer:
(a) (I) An element X contains 35 electrons and 45 neutrons

1. The number of protons must be equal to the number of electrons. So the number of protons = 35.
2. Number of electrons = 35. So the electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵.
3. The last electron i.e. 5^th electron in 4p orbital has the following quantum numbers. n = 4, l = 1, m = +1, s = –\(\frac{1}{2}\)

(II) Fe → Fe²⁺ + 3e^–
Fe (Z = 26) Fe³⁺ = number of electrons = 23
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion.
So, it contain 5 unpaired electrons.
Mn (Z = 25). Electronic configuration is
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵
Mn → Mn²⁺ + 2e^–
Number of unpaired electrons in Mn²⁺ = 5
Ar (Z = 18). Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶.
All orbitals are completely filled. So, no unpaired electrons in it.

[OR]

(b) (I)

1. Hydrogen resembles alkali metals as well as halogens.
2. Hydrogen resembles more alkali metals than halogens.
3. Electron affinity of hydrogen is much less than that of halogen atom. Hence the tendency to form hydride ion is low compared to that of halogens.
4. In most of its compounds hydrogen exists in +1 oxidation state. Therefore it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

(II)

Question 35 (a).
(I) Why alkafr metals have high chemical reactivity? How this changes along the group?
(II) Distinguish between alkali metals and alkaline earth metals?

[OR]

(b)
(I) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude?
(II) Explain the graphical representation of Charles’ law?
Answer:
(a) (I) Alkali metals exhibit high chemical reactivity because of their low ionization enthalpy and their larger size.
The reactivity of alkali metals increases from Li to Cs, since the value of ionization energy decreases down the group (Li to Cs). All the alkali metals are highly reactive towards the more electronegative elements such as oxygen and halogens.

(II)
Alkali Metals:

1. Alkali metals are soft.
2. They have a single electron in the valence shell and their electronic configuration is [noble gas] ns¹.
3. They have low melting points.
4. Hydroxides are strongly basic.
5. Carbonates do not decompose.
6. Nitrates give corresponding nitrites and oxygen as products.
7. They show +1 oxidation states.
8. Their carbonates are soluble in water except Li₂CO₃.
9. Except Li, alkali metals do not form complex compounds.

Alkaline earth metals:

1. Alkaline earth metals are hard.
2. They have two electrons in the valence shell and their electronic configuration is [noble gas] ns².
3. They have relatively high melting points.
4. Hydroxides are less basic.
5. Carbonates decompose to form oxide, when heated to high temperatures.
6. Nitrates give corresponding oxides, nitrogen dioxide and oxygen as products.
7. They show +2 oxidation states.
8. Their carbonates are insoluble in water.
9. They can form complex compounds.

[OR]

(b)
(I) The volume of the gas is inversely proportional to pressure at a given temperature according to Boyle’s law. As the weather balloon ascends, the pressure tends to decrease.
As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

(II)

1. Variation of volume of the gas sample with temperature at constant pressure.
2. Each line (iso bar) represents the variation of volume with temperature at certain pressure. The pressure increases from P₁ to P₅.
3. i.e. P₁ < P₂ < P₃ < P₄ < P₅. When these lines are extrapolated to zero volume, they intersect at a temperature of -273.15°C.
4. All gases are becoming liquids, if they are cooled to sufficiently low temperatures.
5. In other words, all gases occupy zero volume at absolute zero. So the volume of a gas can be measured over only a limited temperature range.

Question 36 (a).
(I) Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero?
(II) Derive the relationship between standard free energy (∆G°) and equilibrium constant (K_eq)

[OR]

(b)
(I) 2.56g of Sulphur is dissolved in 100 g of carbon disulphide. The solution boils at 319. 692 K. What is the molecular formula of Sulphur in solution. The boiling point of CS₂ is 319. 450K. Given that K_b for CS₂ = 2.42 K kg mol^-1
(II) Show that the sum of mole fraction of a solution is equal to one?
Answer:
(a) (I) A substance has a perfectly ordered arrangement only at absolute zero. Hence, entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from its constituents means no heat change.

(II)

1. In a reversible process, system is at all times in perfect equilibrium with its surroundings.

2. A reversible chemical reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up.

3. This means that the reactions in both the directions should proceed with decrease in free energy, which is impossible.

4. It is possible only if at equilibrium, the free energy of a systepi is minimum.

5. Lets consider a general equilibrium reaction,
A + B ⇄ C + D
The free energy change of the above reaction in any state (∆G) is related to the standard free energy change of the reaction (∆G°) according to the following equation.
∆G = ∆G° +RTIn Q ………………. (1)
where Q is reaction quotient and is defined as the ratio of concentrajion of the products to the concentration of the reactants under non-equilibrium condition.

6. When equilibrium is attained, there is no further free energy change i.e. ∆G = 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes, ∆G° =-RTln K_eq ……………… (2)
This equation is known as Van’t Hoff equation.
∆G° = -2.303 RTlogKeq ………………. (3)
We also know that,
∆G° = ∆H° -T∆S° = – RT In K …………….. (4)

[OR]

W₂ = 2.56 g; W₁ = 100 g
T = 319.692; K_b = 2.42 K kg mol^-1
∆T_b = (319.692 – 319.450) K = 0.242 K
M₂ = image 28
M₂ = 256 g mol^-1
Molecular mass of sulphur in solution = 256 g mol^-1
Atomic mass of one mole of sulphur atom = 32
No. of atoms in a molecule of sulphur = \(\frac{256}{2}\) = 8
Hence, molecular formula of sulphur is S₈.

(II) Consider a solution containing two components A and B whose mole fractions are x_A and x_B respectively. Let the number of moles of two components A and B are n_A
and n_B respectively.

Question 37 (a).
(I) Explain about the procedure and calculation behind the carius method of estimation of sulphur?
(II) What is the difference between distillation, distillation under reduced pressure and steam distillation?

[OR]

(b)
(I) An organic compound (A) of a molecular formula C₂H₄ which is a simple alkene. A reacts with dil H₂SO₄ to give B. A again reacts with Cl₂ to give C. Identify A, B and C and write the equations.
(II) Why chloro acetic acid is stronger acid than acetic acid?
Answer:
(a) (I) Carius method

• Procedure: A known mass of the organic compound is taken in a clean carius tube and few mL of fuming HNO₃ is added and then the tube is sealed. It is then placed in an iron tube and heated for 5 hours. The tube is allowed to cool and a hole is made to allow gases to escape.
• The carius tube is broken and the content collected in a beaker. Excess of BaCl₂ is added to the beaker. H₂SO₄ formed is converted to BaSO₄ (white ppt.) The precipitate is filtered, washed, dried and weight. From the mass of BaSO₄, percentage of S is calculated.

(II) Calculation:
Mass of organic compound = Wg
233 g of BaSO₄ contains 32 g of sulphur
Percentage of sulphur = (\(\frac{32}{233}\) × \(\frac{x}{w}\) × 100)%

(II) Distillation is used in case of volatile liquid mixed with a non-volatile impurities.
Distillation under reduced pressure:
This method is used to purify such liquids which have very high boiling points and which decompose at or below their boiling points.
Steam distillation is used to purify steam volatile liquids associated with water immiscible impurities.

[OR]

(b) (I)

1. C₂H₄ is CH₂ = CH₂ is a simple alkene. A is ethylene.
2. Ethylene (A) reacts with dil H₂SO₄ to give ethanol (B)

3. Ethylene (A) reacts with Cl₂ to give 1, 2 dichloro ethane (C)

(II) Chloro acetic acid: image 31
Chloro acetic acid has Cl – group and it has high electronegativity and shows -I effect. Therefore Cl – atom to facilitate the dissociation of O – H bond very fastly. Whereas in the case of acetic acid, has CH₃ group and it shows +1 effect, therefore dissociation of O – H bond will be more difficult. Thus chloro acetic acid is stronger acid than acetic acid.

Question 38 (a).
(I) Write a chemical reaction useful to prepare the following:

1. Freon – 12 from carbon tetrachloride.
2. Carbon tetrachloride from carbon disulphide.

(II) What are ambident nucleophiles? Explain with an example.

[OR]

(I) Write about hydrosphere (or) Why Earth is called as Blue planet?
(II) Even though the use of pesticides increases the crop production, they adversely affect the living organisms. Explain the function and the adverse effects of the pesticides.
Answer:
1. Freon-12 from carbon tetrachloride:
Freon-12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of catalytic amount of antimony pentachloride. This reaction is called “Swarts reaction.”

2. Carbon tetrachloride from carbon disulphide:
Carbon disulphide reacts with chlorine gas in the presence of anhydrous AlCl₃ as catalyst to give carbon tetrachloride.

(II) Nucleophiles which can attack through two different sites are called ambident nucleophiles. For example, cyanide group is a resonance hybrid of two contributing structures and therefore it can act as a nucleophile in two different ways:

It can attack through carbon to form cyanides and through nitrogen to form isocyanides or carbylamines.

[OR]

(b)

1. Hydrosphere include all types of water sources like oceans, seas, rivers, lakes, streams, underground water, polar ice – caps, clouds etc.
2. It covers about 75% of the earth’s surface. Hence earth is called as Blue planet:

(II) Pesticides are the chemicals that are used to kill or stop the growth of unwanted organims. But these pesticides can affect the health of human beings. Pesticides are classified as

1. Insecticides
2. Fungicides and
3. Herbicides.

1. Insecticides:
Insecticides like DDT, BHC, Aldrin can stay in soil for a long period of time and are absorbed by soil. They contaminate root crops like carrot, radish..

2. Fungicides:
Organomercury compounds dissociate in soil to produce mercury which is highly toxic.

3. Herbicides:
They are used to control unwanted plants and are also known as weed killers. Eg, Sodium chlorate, sodium nitrate. They are toxic to mammals.

Students can Download Tamil Nadu 11th Maths Model Question Paper 5 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 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 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 × B) ∩ (A × C)) = 8 and n(B ∩ C) = 2 then n(A) = …………………
(a) 6
(b) 4
(c) 8
(d) 16
Answer:
(b) 4

Question 2.
The value of log₃ \(\frac{1}{81}\) is ……………….
(a) -2
(b) -8
(c) -4
(d) -9
Answer:
(c) -4

Question 3.
The value of log₃ 11 log₁₁ 13 log₁₃ 15 log₁₅ 27 log₂₇ 81 is ……………………
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 4.
The value of sin(45° + θ) – cos (45° – θ) is …………………..
(a) 2 cos θ
(b) 1
(c) 0
(d) 2 sin θ
Answer:
(c) 0

Question 5.
If tanα and tan β are the roots of x² + ax + b = 0 then \(\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}\) is equal to ……………………..
(a) \(\frac{b}{a}\)
(b) \(\frac{a}{b}\)
(c) –\(\frac{a}{b}\)
(d) –\(\frac{b}{a}\)
Answer:
(c) –\(\frac{a}{b}\)

Question 6.
If a² – aC₂ = a² – aC₄ then the value of a is …………………….
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 7.
If ⁿP_r = 840, ⁿC_r= 35 then n = …………………..
(a) 1
(b) 6
(c) 5
(d) 4
Answer:
(a) 1

Question 8.
If 2x² + 3xy – cy² = 0 represents a pair of perpendicular lines then c = …………………….
(a) -2
(b) \(\frac{1}{2}\)
(c) – \(\frac{1}{2}\)
(d) 2
Answer:
(d) 2

Question 9.
The number of terms in the expansion of [(a + b)²]¹⁸ = …………………..
(a) 19
(b) 18
(c) 36
(d) 37
Answer:
(d) 37

Question 10.
The point on the line 2x – 3y = 5 is equidistance from (1, 2) and (3, 4) is …………………..
(a) (7, 3)
(b) (4, 1)
(c) (1,-1)
(d) (3, 4)
Answer:
(b) (4, 1)

Question 11.
Let A and B be two symmetric matrices of same order. Then which one of the following statement is not true?
(a) A + B is a symmetric matrix
(b) AB is a symmetric matrix
(c) (AB) = (BA)^T
(d) A^TB = AB^T
Answer:
(d) A^TB = AB^T

Question 12.
If [3 -1 2] B = [5, 6] then the order of B is ………………….
(a) 3 × 2
(b) 2 × 3
(c) 3 × 1
(d) 1 × 1
Answer:
(a) 3 × 2

Question 13.
1f \(\underset { x\rightarrow 0 }{ lim } \) \(\frac{sin px}{tan 3x}\) = 4 then the value of p is …………………….
(a) 6
(b) 9
(c) 12
(d) 4
Answer:
(c) 12

Question 14.
For \(\vec { a } \) = \(\vec { i } \) + \(\vec { j } \) – 2\(\vec { k } \), \(\vec { b } \) = \(\vec { i } \) + 2\(\vec { j } \) + \(\vec { k } \) and \(\vec { c } \) = \(\vec { i } \) – 2\(\vec { j } \) + 2\(\vec { k } \) the unit vector parallal to is \(\vec { a } \)
+ \(\vec { b } \) + \(\vec { c } \) is ……………………….
(a) \(\frac{\vec{i}+\vec{j}-\vec{k}}{\sqrt{3}}\)
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)
(c) \(\frac{\vec{i}+\vec{j}+\vec{k}}{3}\)
(d) \(\frac{\vec{i}-\vec{j}+\vec{k}}{\sqrt{6}}\)
Answer:
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)

Question 15.
The differential co-efficient of log₁₀ x with respect to log_x 10 is …………………….
(a) 1
(b) -(log₁₀x)²
(c) (log_x10)²
(d) \(\frac { x^{ 2 } }{ 100 } \)
Answer:
(b) -(log₁₀x)²

Question 16.
\(\frac{d}{dx}\)(e^x+5logx) is …………………..
(a) e^xx⁴(x + 5)
(b) e^xx(x + 5)
(c) e^x + \(\frac{5}{x}\)
(d) e^x – \(\frac{5}{x}\)
Answer:
(a) e^xx⁴(x + 5)

Question 17.
If f(x) = x tan^-1x thenf'(1) = ……………………..
(a) 1 + \(\frac { \pi }{ 4 } \)
(b) \(\frac{1}{2}\) + \(\frac { \pi }{ 4 } \)
(c) \(\frac{1}{2}\) – \(\frac { \pi }{ 4 } \)
(d) 2
Answer:
(b) \(\frac{1}{2}\) + \(\frac { \pi }{ 4 } \)

Question 18.
∫cosec xdx = …………………….
(a) log tan \(\frac{x}{2}\) + c
(b) -log (cosec x + cot x) + c
(c) log (cosecx – cot x) + c
(d) all of them
Answer:
(d) all of them

Question 19.
Ten coins are tossed; The probability of getting atleast 8 heads is …………………..
(a) \(\frac{7}{64}\)
(b) \(\frac{7}{32}\)
(c) \(\frac{7}{128}\)
(d) \(\frac{7}{16}\)
Answer:
(c) \(\frac{7}{128}\)

Question 20.
Two items are chosen from a lot containing twelve items of which four are defective. Then the probability that atleast one of the item is defective is …………………..
(a) \(\frac{19}{33}\)
(b) \(\frac{17}{33}\)
(c) \(\frac{23}{33}\)
(d) \(\frac{13}{34}\)
Answer:
(a) \(\frac{19}{33}\)

PART – II

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

Question 21.
For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A?
Answer:
A × A = 16 elements = 4 × 4
⇒ A has 4 elements
∴ A = {0, 1, 2, 3}

Question 22.
Prove that \(\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}\) = \(\frac{1+sinθ}{cosθ}\)
Answer:

Question 23.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Answer:
No. of non-collincar points = 15
To draw a Triangle we need 3 points
∴ Selecting 3 from 15 points can be done in ¹⁵C₃ ways.
∴ No. of Tnangle formed ¹⁵C₃
= \(\frac{15 \times 14 \times 13}{3 \times 2 \times 1}\) = 455

Question 24.
Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1?
Answer:
Givc sum of the intercepts = 1
⇒ when x intercept a then y intercept = 1 – a
Equation of the line is \(\frac{x}{a}\) + \(\frac{y}{1-a}\) = 1
The line passes through (8, 3) ⇒ \(\frac{8}{a}\) + \(\frac{3}{1-a}\) = 1
(i.e) 8 (1 – a) + 3a = a(1 – a)
8 – 8a + 3a = a – a²
a² – 6a + 8 =0
(a – 2)(a – 4) = 0 ⇒a = 2 or 4
1. When a = 2 equation of the line is \(\frac{x}{2}\) + \(\frac{y}{-2}\) = 1 (i.e) \(\frac{x}{2}\) – y = 1 ⇒ x – 2y = 2
2.When a = 4 equation of the line is \(\frac{x}{4}\) + \(\frac{y}{1-4}\) = 1 (i.e) \(\frac{x}{4}\) – \(\frac{y}{3}\) = 1 ⇒ 3x – 4y = 12

Question 25.
Find the values of p, q, r, & s if \(\left[\begin{array}{ccc}
p^{2}-1 & 0 & -31-q^{3} \\
7 & r+1 & 9 \\
-2 & 8 & s-1
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & -4 \\
7 & \frac{3}{2} & 9 \\
-2 & 8 & -\pi
\end{array}\right]\)
Answer:
When two matrices (of some order) are equal then their correspondings entries are equal.

Question 26.
Find |\(\vec { a } \) × \(\vec { b } \)| where \(\vec { a } \) = 3\(\vec { i } \) + 4\(\vec { j } \) and \(\vec { b } \) = \(\vec { i } \) + \(\vec { j } \) + \(\vec { k } \)
Answer:

Question 27.
At the given point x₀ discover whether the given function is continous or discontinous citing the reasons for your answer?
Answer:

Question 28.
Evaluate y = xe^-x2
Answer:

Question 29.
Evaluate ∫\(\frac{1}{x logx}\) dx?
Answer:

Question 30.
Evaluate [((256)^-1/2)^-1/4]³
Answer:

PART – III

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

Question 31.
Find the largest possible domain for the real valued functionsf defined by f(x) = \(\sqrt { x^{ 2 }-5x+6 } \)

Question 32.
Show that tan 75° + cot 75° = 4?

Question 33.
There are 10 bulbs in a room. Each one of them can be operated independently. Find the number of ways in which the room can be illuminated?

Question 34.
Find the \(\sqrt [ 3 ]{ 126 } \) approximately to two decimal places?

Question 35.
Find the equation of the line passing through the point of intersection 2x + y = 5 and x + 3y + 8 = O and parallel to the line 3x + 4y = 7?

Question 36.
If \(\left|\begin{array}{ccc}
a & b & a \alpha+b \\
b & c & b \alpha+c \\
a \alpha+b & b \alpha+c & 0
\end{array}\right|\) = 0 prove that a, b, c are in G.P. or a is a root of ax² + 2bx + c =0

Question 37.
Evaluate \(\underset { x\rightarrow 3 }{ lim } \)\(\frac { x^{ 2 }-9 }{ x-3 } \) ¡f it exists by finding f(3^–) and f(3⁺)

Question 38.
Find the derivative of tan^-1 (1 + x²) with respect x² + x + 1?

Question 39.
Evaluate: ∫x⁵e^{x2 dx}

Question 40.
Prove that the line segment joining the mid points of the adjacent sides of a quadrilateral from parllelogram?

PART – IV

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

Question 41.
(a) Graph the functions f(x) = x³ and g(x) = \(\sqrt [ 3 ]{ x } \) on the same coordinate plane. Find fog and graph it on the plane as well. Explain your results.

[OR]

(b) 1f x = -2 is one root of x³ – x² – 17x = 22 then find the other roots of the equation?

Question 42.
(a) lf A + B + C= it prove that cosA + cos B + cosC = 1 + 4 sin (\(\frac{A}{2}\)) sin (\(\frac{B}{2}\) sin (\(\frac{C}{2}\))

[OR]

(b) If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) show that A² – 4A – 5I = O

Question 43.
(a) If ⁿ⁺¹C₈ : ^(n-3)P₄ = 57 : 16, find the value of n?

[OR]

(b) If the letters of the word IITJEE arc permuted in all possible ways and the strings thus formed are arranged in the lexicographic order, find the rank of the word IITJEE?

Question 44.
(a) The line \(\frac{x}{a}\) + \(\frac{y}{b}\)= 1 moves in such a way that \(\frac { 1 }{ a^{ 2 } } \) + \(\frac { 1 }{ b^{ 2 } } \) = \(\frac { 1 }{ c^{ 2 } } \) where c is a constant. Find the locus of the foot of the perpendicular from the origin on the given line?

[OR]

(b) Show that the equation 4x² + 4xy + y² – 6x – 3y – 4 = 0 represents a pair of parallel lines. Find the distance between them?

Question 45 (a).
Prove that \(\left|\begin{array}{lll}
1 & x^{2} & x^{3} \\
1 & y^{2} & y^{3} \\
1 & z^{2} & z^{3}
\end{array}\right|\) = (x – y) (y – z) (z – x) (xy + yz + zx)

(b) Evaluate \(\underset { x\rightarrow \infty }{ lim } \) \(\frac{3}{x-2}\) – \(\frac{2 x+11}{x^{2}+x-6}\)

Question 46.
(a) Evaluate \(\frac{1}{6 x-7-x^{2}}\)

(b) Evaluate ∫e^tan-1x (\(\frac { 1+x+x^{ 2 } }{ 1+x^{ 2 } } \)) dx

Question 47 (a).
Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?

[OR]

(b) Firm manufactures PVC pipes in three plants viz. X, Y and Z. The daily production volumes from the three firms X, Y and Z are respectively 2000 units, 3000 units and 5000 units. It is known from the past experience that 3% of the output from plant X, 4% from plant Y and 2% from plant Z are defective. A pipe is selected at random from a day’s total production,

1. find the probability that the selected pipe is a defective one.
2. if the selected pipe ¡s a defective, then what is the probability that it was produced by plant Y?

Students can Download Tamil Nadu 11th Physics Model Question Paper 1 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 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: 3 Hours
Maximum Marks: 70

PART – I

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

Question 1.
A cyclist moving on a circular track of radius 40 m completes half a revolution in 40 sec average velocity is ………………..
(a) 0
(b) 2 m/s
(c) 4 m/s
(d) 2π m/s
Hint:
Displacement of the cyclist in half revolution is
d = diameter of the circular track
i.e., d= 80 m
Time taken, t = 40 s
Average velocity, V =  = \(\frac{80}{40}\)
V = 2 m/s
Answer:
(b) 2 m/s

Question 2.
A wheel has angular acceleration of 3.0 rad/s² and an initial angular speed of 2.00 rad/s. In a time of 2 seconds it has rotated through an angle of (in radian) ………………..
(a) 10
(b) 12
(c) 4
(d) 6
Answer:
(a) 10

Question 3.
If the origin of co-ordinate system lies at the centre of mass. The sum of the moments of the masses of the system about the centre of mass is …………………
(a) May be greater than zero
(b) May be less than zero
(c) May be equal to zero
(d) Always zero
Answer:
(d) Always zero

Question 4.
Dimensional formula for co-efficient of viscousity
(a) ML^-2 T^-2
(b) ML^-2 T^-1
(c) ML^-1 T^-1
(d) M^-1 L^-1 T^-1
Answer:
(c) ML^-1 T^-1

Question 5.
Action and reaction …………………
(a) Acts on same object
(b) Acts on two different objects
(c) Have resultant not zero
(d) Acts on the same direction
Answer:
(b) Acts on two different objects

Question 6.
A spring is stretched by applying load to its free end. The strain produced in the spring is …………………
(a) Volumetric
(b) Shear
(c) Longitudinal
(d) Longitudinal and shear
Answer:
(d) Longitudinal and shear

Question 7.
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?
(a) 0.25 rad s^-2
(b) 25 rad s^-2
(c) 5 ms^-2
(d) 25 ms^-2

Hint:
τ = F × r
Iα = F × r
MR² × α = 30 × \(\frac{40}{100}\); \(\frac { 3\times 40\times 40\times \alpha }{ 100\times 100 } \) = 12
\(\frac { 3\times 16\times \alpha }{ 100 } \) = 12; α = 25 rad/s²
Answer:
(b) 25 rad s^-2

Question 8.
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (E is total energy) ………………
(a) \(\frac{2}{3}\) E
(b) \(\frac{1}{3}\) E
(c) \(\frac{1}{4}\) E
(d) \(\frac{1}{2}\) E
Hint:
PE = \(\frac{1}{2}\) kx²
⇒ \(P E_{V_{2}}=\frac{1}{2} K\left(\frac{A}{2}\right)^{2}=\frac{1}{4}\left(\frac{1}{2} K A^{2}\right)\)
Answer:
(c) \(\frac{1}{4}\) E

Question 9.
A particle executes simple harmonic motion with an angular velocity and maximum acceleration of 3.5 rad/s and 7.5 m/s² respectively. Amplitude of the oscillation is ………………
(a) 0.36
(b) 0.28
(c) 0.61
(d) 0.53
Hint:
x = A sin ωt
∴ a = \(\frac { d^{ 2 }x }{ dt^{ 2 } } \) = -Aω² sinωt
∴Maximum acceleration |a_max| = Aω²
Now Aω² = 7.5
A = \(\frac { 7.5 }{ \omega ^{ 2 } } \) = \(\frac { 7.5 }{ (3.5)^{ 2 } } \) = 0.61
Answer:
(c) 0.61

Question 10.
If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved, then its fundamental frequency will become ………………….
(a) \(\frac{n}{4}\)
(b) \(\sqrt{2n}\)
(c) n
(d) \(\frac { n }{ \sqrt { 2 } } \)
Hint:

Answer:
(c) n

Question 11.
The theory of refrigerator is based on …………….
(a) Joule-Thomson effect
(b) Newton’s particle theory
(c) Joule’s effect
(d) None of the above
Answer:
(d) None of the above

Question 12.
Work done by 0.1 mole of a gas at 27°C to double its volume at constant pressure is ………………..
(a) 54 cal
(b) 60 cal
(c) 546 cal
(d) 600 cal
Hint:
Workdone (W) = – p.dv = nRT
= 0.1 × (0.2 cal) × (273 + 27) = 0.1 × 2 × 300
W = 60 cal
Answer:
(b) 60 cal

Question 13.
When a lift is moving upwards with acceleration a, then time period of simple pendulum in it will ………………..
(a) 2π\(\sqrt { \frac { 1 }{ g+a } } \)
(b) 2π\(\sqrt { \frac { g+a }{ l } } \)
(c) \(\frac{1}{2π}\)\(\sqrt { \frac { 1 }{ g+a } } \)
(d) \(\frac{1}{2π}\)\(\sqrt { \frac { g+a }{ l } } \)
Answer:
(a) 2π\(\sqrt { \frac { 1 }{ g+a } } \)

Question 14.
A disc is rotating with angular speed ω. If a child sits on it, what is conserved?
(a) Linear momentum
(b) Angular momentum
(c) Kinetic energy
(d) Potential energy
Answer:
(b) Angular momentum

Question 15.
The vectors \(\vec { A } \) and \(\vec { B } \) are such that |\(\vec { A } \) + \(\vec { B } \)| = |\(\vec { A } \) – \(\vec { B } \)|. The angle between the two vector is ………………..
(a) 45°
(b) 60°
(c) 75°
(d) 90°
Hint:
The angle between two vector is always 90°.
Answer:
(d) 90°

PART – II

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

Question 16.
Velocity – time graph of a moving object is shown below. What is the acceleration of the object? Also draw displacement – time graph for the motion of the object?
Answer:
The given graph shows that the velocity of the object is constant. That is, the velocity of the object is not changing, so the acceleration of the object is zero. Since the acceleration of an object is given by


Displacement – time graph for the motion of the object is shown in the figure above.

Question 17.
Can a body subjected to a uniform acceleration always move in a straight line?
Answer:
It will be a straight line in one dimensional motion but not applicable for two dimensional motion because the projectile has a parabolic path but it has a uniform acceleration.

Question 18.
Calculate the viscous force on a ball of radius 1mm moving through a liquid of viscosity 0.2 Nsm^-2 at a speed of 0.07 ms^–
Answer:
Radius of the ball (a) = 1mm = 1 × 10^-3m
Co-effecient of viscosity of liquid (η) = 0.2 Nsm^-2
Speed of the ball (v) = 0.07 ms^-1
According to Stoke’s law
Viscous force F = 6 π η av
= 6 × 3.14 × 1 × 10^-3 × 0.2 × 0.07
= 0.26376 × 10^-3 = 2.64 × 10^-4N

Question 19.
Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10ms^-2)
Answer:
Given data: F = 30 N, load (m) 2 kg; height = 10m, g = 10 ms^-2
Gravitational forcc F = mg = 30 N
The distance moved h = 10 m
Work done on the object W = Fh = 30 × 10 = 300 J

Question 20.
Why are shockers used in automobiles like car?
Answer:
In the event of jump or jerk, the lime of action of force increases. Since the product of force aid time is constant in a given situation. therefore the force decreases.

Question 21.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighted 250 N on the surface?
Answer:
As g_d = g (1 – \(\frac{d}{R}\)) ⇒ mg_d = mg(1 – \(\frac{d}{R}\))
Here d = \(\frac{R}{2}\)
∴mg_d = (250) × (1 – \(\frac { R/2 }{ R } \)) = 250 × \(\frac{1}{2}\) = 125N

Question 22.
How do you deduce that two vectors are perpendicular?
Answer:
If two vectors \(\vec { A } \) and \(\vec { B } \) are perpendicular to each other than their scalar product \(\vec { A } \).\(\vec { B } \) = O because cos 90° = 0. Then the vectors \(\vec { A } \) and \(\vec { B } \) are said to be mutually orthogonal.

Question 23.
An air bubble of radius r in water is at a depth of h below the water surface at some instant. If P is atmospheric pressure and d and T are the density and surface tension of ater respectively. Calculate the pressure P inside the bubble?
Answer:
Excess ot pressure inside the air bubble in water = \(\frac{2T}{r}\)
∴ Total pressure inside the air bubble
= atmospheric pressure + pressure due to liquid column + Excess pressure due to surface tension
= P + hρg + \(\frac{2T}{r}\)

Question 24.
Define beats?
Answer:
Formation of beats: When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency. Number of beats per second.
n = |f₁ – f₂| per second.

PART – III

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

Question 25.
Define centripetal acceleration and give any two examples?
Answer:
The acceleration that is directed towards the centre of the circle along the radius and perpendicular to the velocity of the particle is known as centripetal or radial or normal acceleration.
Example:-

1. In the case of planets revolving round the Sun or the moon revolving round the earth, the centripetal force is provided by the gravitational force of attraction between them.
2. For an electron revolving round the nucleus in a circular path, the electrostatic force of attraction between the electron and the nucleus provides the necessary centripetal force.

Question 26.
Write any six properties of vector product of two vectors?
Answer:
(I) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors \(\vec { A } \) and \(\vec { B } \), even though the vectors \(\vec { A } \) and \(\vec { B } \) may or may not be mutually orthogonal.

(II) The vector product of two vectors is not commutative, i.e., \(\vec { A } \) × \(\vec { B } \) ≠ \(\vec { B } \) × \(\vec { A } \). But, \(\vec { A } \) × \(\vec { B } \) = – \(\vec { B } \) × \(\vec { A } \) .
Here it is worthwhile to note that |\(\vec { A } \) × \(\vec { B } \)| = |\(\vec { B } \) × \(\vec { A } \)| = AB sin θ i.e., in the case of the product vectors \(\vec { A } \) × \(\vec { B } \) and \(\vec { B } \) × \(\vec { A } \), the magnitudes are equal but directions are opposite to each other.

(III) The vector product of two vectors will have maximum magnitude when sin θ = 1, i.e., θ = 90° i.e., when the vectors \(\vec { A } \) and \(\vec { B } \) are orthogonal to each other.
(\(\vec { A } \) × \(\vec { B } \))_max = AB\(\hat { n } \)

(IV) The vector product of two non-zero vectors will be minimum when sin θ = 0, i.e., θ = 0° or θ = 180°
(\(\vec { A } \) × \(\vec { B } \))_min = 0
i.e., the vector product of two non-zero vectors vanishes, if the vectors are either parallel or antiparallel.

(V) The self-cross product, i.e., product of a vector with itself is the null vector
\(\vec { A } \) × \(\vec { A } \) = AA sin 0° \(\hat { n } \) = \(\vec { 0 } \)
In physics the null vector \(\vec { 0 } \) is simply denoted as zero.

(VI) The self-vector products of unit vectors are thus zero.
\(\hat { i } \) × \(\hat { i } \) = \(\hat { j } \) × \(\hat { j } \) = \(\hat { k } \) × \(\hat { k } \) = 0

Question 27.
Show that the pressure of the gas is equal to two third of mean kinetic energy per unit volume?
Answer:
The internal energy of the gas is given by
U = \(\frac{3}{2}\) NkT
The above equation can also be written as
U = \(\frac{3}{2}\) PV
Since PV = NkT
P = \(\frac{2}{3}\) \(\frac{U}{V}\) = \(\frac{2}{3}\) u
From the equation (1), we can state that the pressure of the gas is equal to two thirds of internal energy per unit volume or internal energy density (u = \(\frac{U}{V}\))
Writing pressure in terms of mean kinetic energy density using equation.
P = \(\frac{1}{3}\) nm\(\overline { V^{ 2 } } \) = \(\frac{1}{3}\) ρ\(\overline { V^{ 2 } } \)
where ρ = nm = mass density (Note n is number density)
Multiply and divide R.H.S of equation (2) by 2, we get
P = \(\frac{2}{3}\)(\(\frac{ρ}{2}\) \(\overline { V^{ 2 } } \))
P = \(\frac{2}{3}\) \(\overline { KE } \)
From the equation (3), pressure is equal to \(\frac{2}{3}\) of mean kinetic energy per unit volume.

Question 28.
Derive an expression for gravitational potential energy?
Answer:
The gravitational tbrce is a conservative force and hence we can define a gravitational potential energy associated with this conservative force field.
Two masses m₁ and m₂, are initially separated by a distance r’. Assuming m₁ to be fixed in its position. work must be done on m₂ to move the distance from r’ to r.


To move the mass m₂, through an infinitesimal displacement d\(\vec { r } \) from \(\vec { r } \) to \(\vec { r } \) + d\(\vec { r } \) , work has to be done externally. This infinitesimal work is given by
dW = \(\vec { F } \)_ext . d\(\vec { r } \) ……………….. (1)
The work is done against the gravitational force, therefore,
\(\vec { F } \)_ext = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \) ………………. (2)
Substituting equation (2) in (1), we get
dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \).d\(\vec { r } \) ………………… (3)
d\(\vec { r } \) = dr \(\hat { r } \) ⇒ dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \).(dr \(\hat { r } \))
\(\hat { r } \).\(\hat { r } \) = 1 (Since both are unit vectors)
∴ dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \) dr …………….. (4)
Thus the total work done for displacing the particle from r’ to r is

This work done W gives the gravitational potential energy difference of the system of masses and m₁ and m₂ when the separation between them are r and r’ respectively.

Question 29.
A satellite orbiting the Earth in a circular orbit of radius 1600 km above the surface of the Earth. What is the acceleration experienced by satellite due to Earth’s gravitational force?
Answer:
g’ = g(1 – \(\frac { 2h }{ R_{ e } } \))
= g(\(\frac { 1-2\times 1600\times 10^{ 3 } }{ 6400\times 10^{ 3 } } \)) = g(1 – \(\frac{2}{4}\))
g’ = g (1 – \(\frac{1}{2}\)) = \(\frac{g}{2}\)
g’ = g (1- \(\frac{1}{2}\)) = \(\frac{g}{2}\)
g’ = \(\frac{8}{2}\) (or) g’ = \(\frac{9.8}{2}\) = 4.9ms^-2

Question 30.
Explain the v ariation of a g with latitude?
Answer:
When an object is on the surface fo 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.
This centrifugal force is given by mωR’.
\(\mathrm{OP}_{z}, \cos \lambda=\frac{\mathrm{PZ}}{\mathrm{OP}}=\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\)
R’ = R cos λ
where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
\(a_{\mathrm{PQ}}=\omega^{2} \mathrm{R} \cos \lambda=\omega^{2} \mathrm{R} \cos ^{2} \lambda\)
Since R’ = R cos λ

Therefore, g = g – ω²R cos² λ
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; g’ = g, it is maximum. At the equator, g’ is minimum.

Question 31.
A bullet of mass 50g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take g = 10 ms^-2
Answer:
m₁ = 50 g = 0.05 kg; m₂ = 450 g = 0.45 kg
The speed of the bullet is u₁ The second body is at rest (u₂ = 0). Let the common velocity of the bullet and the object after the bullet is embedded into the object is v.
v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\)
v = \(\frac{0.05 u_{1}+(0.45 \times 0)}{(0.05+0.45)}\) = \(\frac{0.05}{0.50}\)u₁
The combined velocity is the initial velocity for the vertical upward motion of the combined bullet and the object. From second equation of motion,
v = \(\sqrt{2gh}\)
v = \(\sqrt{2 \times 10 \times 1.8}\) = \(\sqrt{36}\)
v = 6 ms^-1
Substituting this in the above equation, the value of u₁ is
6 = \(\frac{0.05}{0.50}\)u₁ or u₁ = \(\frac{0.05}{0.50}\) × 6 = 10 × 6
u₁ = 60ms^-1

Question 32.
If the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? If not, why?
Answer:
When the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly. In this intermediate stage the ideal gas equation be not valid. Because this equation can be relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Question 33.
Calculate how many times more intense is 90 dB sound compared to 40 dB sound?
Answer:
Given Data:
L = log \(\frac { I }{ I_{ 0 } } \) = log I – log I₀
We get 90 dB = 9 B = log I₁ – log I₀ ……………….. (1)
40 dB = 4 B = logI₂ – logI₀ ………………….. (2)
Subtract (2) from (1)
50 dB = 5B = log I₁ – logI₂
5 = log₁₀ (\(\frac{I_{1}}{I_{2}}\))
\(\frac{I_{1}}{I_{2}}\) = 10⁵

PART – IV

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

Question 34 (a)
Obtain an expression for the time period T of a simple pendulum. The time period T depends on

1. Mass of the bob(m)
2. Length of the pendulum (l)
3. Acceleration due to gravity (g) at the place where the pendulum is suspended, (constant k = 2π)

Answer:
Example:
An expression for the time period T of a simple pendulum can be obtained by using this method as follows.
Let true period T depend upon

1. Mass m of the bob
2. Length l of the pendulum and
3. Acceleration due to gravity g at the place where the pendulum is suspended. Let the constant involved is k = 2π.

Solution:

Here k is the dimensionless constant. Rewriting the above equation with dimensions.

Comparing the powers of M, L and T on both sides, a = 0, b + c = 0, -2c = 1
Solving for a, b and c ⇒ a = 0, b = 1/2, and c = -1/2
From the above equation
T = \(\mathrm{k} \mathrm{m}^{0} l^{1}=g^{-1}-2\)
T = \(k\left(\frac{1}{g}\right)^{1}\) = \(k \sqrt{1 / g}\)
Experimentally k = 2π, hence
T = \(2 \pi \sqrt{1 / g}\)

[OR]

(b) Obtain an expression for the escape speed in detail?
Answer:
Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed v_i the initial total energy of the object is
\(\mathrm{E}_{i}=\frac{1}{2} \mathrm{M} v_{i}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) ………………. (1)
where, M_E is the mass of the Earth and R_E the radius of the Earth.
The term \(-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) is the potential energy of the mass M.
When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be nonzero.
E_F = 0
E_i = E_f ……………….. (2)
Substituting (1) in (2) we get,

Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace v_i with v_e i.e..

From equation (4) the escape speed depends on two factors acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g (9.8 ms^-2) and R_e = 6400 km, the escape speed of the Earth is v_e = 11.2 kms^-1. The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity.

Question 35 (a).
Derive an expression for loss of kinetic energy in perfect inelastic collision?
Answer:
In perfectly inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, heat, light etc. Let KE_i be the total kinetic energy before collision and KE_f be the total kinetic energy after collision.
Total kinetic energy before collision,
\(\mathrm{KE}_{i}=\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}\) …………….. (1)
Total kinetic energy after collision
KE_f = \(\frac{1}{2}\) (m₁ + m₂)v² …………….. (2)
Then the loss of kinetic energy is
Loss of KE, ∆Q = KE_f – KE_i = \(\frac{1}{2}\) (m₁ + m₂)v² – \(\frac{1}{2}\) m₁ u₁² – \(\frac{1}{2}\) m₂ u₂² ………………. (3)
Substituting equation v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\) in equation (3), and on simplying (expand v by using the algebra) (a + b)² = a² + b² + 2ab, we get
Loss of KE, ∆Q = \(\frac{1}{2}\) \(\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)\) (u₁ – u₂)²

(b) A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. Calculate the initial velocity of the shell?
Answer:
Given Data :
m = 200 gm = 0.2 kg; M = 4 kg.
Energy generated = 1.05 KJ = 1.05 × 10³ J
According to law of Conservation of linear momentum
mv = Mv’
∴v’ = (\(\frac{m}{M}\)) v
Total K.E of the gun and bullet

[OR]

(c) State parallel axis theorem?
Answer:
Parallel axis theorem: Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.

If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is given by the relation,
I = I_C + Md²

(d) Calculate the moment of inertia of uniform circular disc of mass 500 G radius 10 cm about

1. The diameter of the disc
2. The axis, tangent to the disc and parallel to its diameter
3. The axis through the centre of the disc and perpendicular to its plane

Answer:
1. Given Data: M = 500 g = 0.5 kg. R = 10 cm = 10 × 10^-2 m
Moment of inertia of disc about diameter = Id = \(\frac{1}{4}\) MR²
Id = \(\frac{1}{4}\) × 0.5 × 0.1 kg m² = 0.0125 kg m²

2. Apply a parallel axes theorem, moment of inertia of the disc about a tangent to the disc and parallel to the diameter of the disc
= \(\frac{1}{4}\) MR² + MR² = \(\frac{5}{4}\) MR² = \(\frac{5}{4}\) × 0.5 × 1
= 0.0625 kgm²

3. Moment of inertia of the disc about an axis passing through the centre of disc and perpendicular to the plane of the disc
= \(\frac{1}{2}\) MR² = \(\frac{1}{2}\) × 0.5 × 0.1 = 0.025 kgm²

Question 36 (a).
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it?
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton’s second and third laws, we can derive the law of conservation of total linear momentum.

When two particles interact with each other, they exert equal and opposite forces on each other. The particle 1 exerts force \(\vec { F } \)₁₂ on particle 2 and particle 2 exerts an exactly equal and opposite force \(\vec { F } \)₁₂ on particle 1 according to Newton’s third law.
\(\vec { F } \)₂₁ = –\(\vec { F } \)₁₂ …………… (1)
In terms of momentum of particles, the force on each particle (Newton’s second law) can be written as
\(\vec { F } \)₁₂ = \(\frac{d \bar{p}_{1}}{d t}\) and \(\vec { F } \)₂₁ = \(\frac{d \vec{p}_{2}}{d t}\) ……………… (2)

Here \(\vec { P } \)₁ is the momentum of particle 1 which changes due to the force \(\vec { F } \)₁₂ exerted by particle 2. Further \(\vec { P } \)₂ is the momentum of particle 2. This changes due to \(\vec { F } \)₂₁ exerted by particle 1.
Substitute equation (2) in equation (1)

It implies that \(\vec { P } \)₁ + \(\vec { P } \)₂ = (constant vector always)

\(\vec { P } \)₁ + \(\vec { P } \)₂ is the total linear momentum of the two particles ( \(\vec { P } \)_tot = \(\vec { P } \)₁ + \(\vec { P } \)₂). It is also called as total linear momentum of the system. Flere, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.

If there are no external forces acting on the system, then the total linear momentum of the system ( \(\vec { P } \)_tot) is always a constant vector. In other words, the total linear momentum of the system is conserved in time. Here the word ‘conserve’ means that \(\vec { P } \)₁ and \(\vec { P } \)₂ can vary,
in such a way that \(\vec { P } \)₁ + \(\vec { P } \)₂ is a constant vector.

The forces \(\vec { F } \)₁₂ and \(\vec { F } \)₁₂ are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.

To find the recoil velocity of a gun when a bullet is fired from it:
Consider the firing of a gun. Here the system is Gun + bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec { P } \)₁ be the momentum of the bullet and \(\vec { P } \)₂ the momentum of the gun before firing. Since initially both are at rest.

\(\vec { P } \)₁ = 0, \(\vec { P } \)₂ = 0.

Total momentum before firing the gun is zero, \(\vec { P } \)₁ + \(\vec { P } \)₂ = 0

According to the law of conservation of linear momentum, total linear momemtum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec { P } \)₁ + \(\vec { P } \)₂. To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec { P } \)₂ to \(\vec { P } \)₂. Due to the conservation of linear momentum, \(\vec { P } \)₁+ \(\vec { P } \)₂‘= 0. It implies that \(\vec { P } \)₁‘ = –\(\vec { P } \)₂ the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum \(\vec { P } \)₂. It is called ‘recoil momentum’. This is an example of conservation of total linear momentum.

[OR]

(b) Derive an expression for escape speed?
Answer:
Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed v;, the initial total energy of the object is
\(E_{i}=\frac{1}{2} M v_{i}^{2}-\frac{G M M_{E}}{R_{E}}\) ………………. (1)
where, M_E is the mass of the Earth and RE the radius of the Earth. The term \(-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) is the potential energy of the mass M.

When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be nonzero.
E_f = 0

According to the law of energy conservation,
E_i – E_f = 0 …………….. (2)

Substituting (1) in (2) we get,
\(\frac{1}{2} \mathrm{M} v_{i}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{F}}}=0\)
\(\frac{1}{2} \mathrm{M} v_{i}^{2}=\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) …………….. (3)

Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace v_i with v_e, i.e.,

From equation (4) the escape speed depends on two factors acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g (9.8 ms^-2) and R_e = 6400 km, the escape speed of the Earth is v_e = 11.2 kms^-1. The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity.

Question 37 (a).
Explain in detail Newton’s law of cooling?
Answer:
Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a 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 of T 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 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)

Integrating equation (5) on both sides,

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

[OR]

(b) Derive an expression for pressure exerted by the gas on the wall of the container?
Answer:
Expression for pressure exerted by a gas:
Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.
The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.

The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision. the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.

A molecule of mass in moving with a velocity \(\vec { v } \) having components (v_x v_y, v_z) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with sanie speed and its x-component is reversed. This is shown in the figure. The components of velocity of the molecule after collision are (-v_x, v_y, v_z)
The x-component of momentum of the molecule bêfore collision = mv_x
The x-component of momentum of the molecule after collision = -mv_x
The change in momentum of the molecule in x direction
= Final momentum – initial momentum = -mv_x – mv_x = -2mv_x
According to law of conservation of linear momentum, the change in momentum of the wall = 2mv_x

The number of molecules hitting the right side wall in a small interval of time ∆t.

The molecules within the distance of v_x ∆t from the right side wall and moving towards the right will hit the wall in the time interval &. The number of molecules that will hit the right side wall in a time interval ∆t is equal to the product of volume (Av_x∆t) and number density of the molecules n). Here A is area of the wall and ii is number of molecules per
unit volume \(\frac{N}{V}\) We have assumed that the number density is the same throughout the cube.

Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side.

Te no.of molecules that hit the right side wall in a time interval ∆t

= \(\frac{n}{2} \mathrm{A} v_{x} \Delta t\) ……………….. (1)
In the same interval of time ∆t, the total momentum transferred by the molecules
\(\Delta \mathrm{P}=\frac{n}{2} \mathrm{A} v_{x} \Delta t \times 2 m v_{x}=\mathrm{A} v_{x}^{2} m n \Delta t\) ……………….. (2)

From Newton’s second law, the change in momentum in a small interval of time gives rise to force.
The force exerted by the molecules on the wall (in magnitude)

F = \(\frac{\Delta p}{\Delta t}=n m \mathrm{A} v_{x}^{2}\) ……………….. (3)
Pressure P = force divided by the area of the wall

P = \(\frac{F}{A}\) = nmv²_x …………………. (4)
Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term v²_x by the average \(\bar{v}_{x}^{2}\) in equation (4)

P = nm\(\bar{v}_{x}^{2}\) ………………… (5)

Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, \(\bar{v}_{x}^{2}\) = \(\bar{v}_{y}^{2}\) = \(\bar{v}_{z}^{2}\). The mean square speed is written as

\(\bar{v}^{2}\) = \(\bar{v}_{x}^{2}\) + \(\bar{v}_{y}^{2}\) + \(\bar{v}_{z}^{2}\) = 3\(\bar{v}_{x}^{2}\)
\(\bar{v}_{x}^{2}\) = \(\frac{1}{3}\) \(\bar{v}^{2}\)
Using this in equation (5), we get
P = \(\frac{1}{3}\)nm \(\bar{v}^{2}\) or P = \(\frac{1}{3}\) \(\frac{N}{V}\) m\(\bar{v}^{2}\) as [n = \(\frac{N}{V}\)] ……………… (6)

Question 38 (a).
Explain with graphs the difference between work done by a constant force and by a variable force. Arrive at an expression for power and velocity. Give some examples for the same?
Answer:
Work done by a constant force: When a constant force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation,
dW = (F cos θ) dr ………………. (1)
The total Work done in producing a displacement from initial position r_i to final position r_f is,

The graphical representation of the work done by a constant force is shown in figure given below. The area under the graph shows the work done by the constant force.

Work done by a variable force:
When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation.

dW = F cos θ dr [F cos θ is the component of the variable force F]
where, F and θ are variables. The total work done for a displacement from initial position r_i to final position r_f is given by the relation,
W = \(\int_{r_{i}}^{r_{f}} d \mathrm{W}=\int_{r_{i}}^{r_{f}} \mathrm{F} \cos \theta d r\) ………………. (4)

A graphical representation of the work done by a variable force is shown in figure given below. The area under the graph is the work done by the variable force.

Expression for power and velocity

The work done by a force \(\vec { F } \) for a displacement \(\bar { dr } \) is
W = ∫\(\vec { F } \).\(\vec { dr } \) ……………. (1)
Left hand side of the equation (1) can be written as
W = ∫dW = ∫\(\frac{dW}{dt}\) (multiplied and divided by dt) ………………… (2)
Since, velocity is \(\vec { v } \) = \(\frac{d \vec{r}}{d t}\); \(\vec { dr } \) = \(\vec { v } \) dt. Right hand side of the equation (I) can be written as

Substituting equation (2) and equation (3) in equation (1), we get

This relation is true for any arbitrary value of di. This implies that the term within the bracket must be equal to zero, i.e.,
\(\frac{dW}{dt}\) – \(\vec { F } \).\(\vec { v } \) = 0 Or \(\frac{dW}{dt}\) = \(\vec { F } \).\(\vec { v } \)
Hence power P = \(\vec { F } \).\(\vec { v } \)

[OR]

(b) Explain in detail Newton’s law of cooling?
Answer:
Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{dQ}{dt}\) ∝ (T – T_s)
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 a positive constant.
From equation (3) and (4)
– a(T – T_s) = ms \(\frac{dT}{dt}\)
\(\frac{d T}{T-T_{s}}\) = -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,
\(\mathrm{T}=\mathrm{T}_{s}+b_{2} e^{\frac{-a}{m s}}\) ………………… (6)
Here b₂ = e^b1 = constant

Students can download 6th Social Science Term 3 Civics Chapter 2 Local Bodies: Rural and Urban 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 Civics Solutions Term 3 Chapter 2 Local Bodies: Rural and Urban

Samacheer Kalvi 6th Social Science Local Bodies: Rural and Urban Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
……………… is set up with several village panchayats
(a) Panchayat Union
(b) District Panchayat
(c) Taluk
(d) Revenue village
Answer:
(a) Panchayat Union

Question 2.
_______ is National panchayat Raj day
a) January 24
(b) July 24
(c) November 24
(d) April 24
Answer:
(d) April 24

Question 3.
The oldest urban local body in India is ………………
(a) Delhi
(b) Chennai
(c) Kolkata
(d) Mumbai
Answer:
(b) Chennai

Question 4.
_________ District has the highest number of Panchayat Unions.
(a) Vellore
(b) Thiruvallore
(c) Villupuram
(d) Kanchipuram
Answer:
(c) Villupuram

Question 5.
The head of a corporation is called a ………………
(a) Mayor
(b) Commissioner
(c) Chair Person
(d) President
Answer:
(a) Mayor

II. Fill in the blanks

1. ……………… is the first state in India to introduce town Panchayat.
2. The Panchayat Raj Act was enacted in the year ………………
3. The tenure of the local body representative is ……………… years.
4. ……………… is the first municipality in Tamil Nadu.

Answer:

1. Tamil Nadu
2. 1992
3. 5
4. Walajahphet Municipality

III. Match the following Answer

Answer:
1. – d
2. – c
3. – a
4. – b

IV. Answer the following

Question 1.
Is there any corporation in your district? Name it.
Answer:
Yes, Tirunelveli Corporation.

Question 2.
What is the need for local bodies?
Answer:
In order to fulfill the requirements of the people and to involve them directly in governance, there is a need for an effective system of local bodies.

Question 3.
What are the divisions of a rural local body?
Answer:
The rural bodies are categorized into village panchayats, Panchayat Unions, and District panchayats.

Question 4.
What are the divisions of an Urban local body?
Answer:
The Urban local bodies are Categorized into City Municipal Corporations, Municipalities and Town Panchayats.

Question 5.
Who are the representatives elected in a Village Panchayat?
Answer:
The Elected Representatives in a Village Panchayat

1. Panchayat president
2. Ward members
3. Councillor
4. District Panchayat Ward Councillor

Question 6.
List out a few functions of corporations.
Answer:

1. Drinking-Water Supply
2. Street Light
3. Maintenance of Clean Environment
4. Primary Health facilities
5. Corporation Schools
6. Birth and Death registration etc.

Question 7.
List out a few means of revenue of village Panchayats.
Answer:

1. House tax
2. Professional tax
3. Tax on shops
4. Water charges
5. Specific fees for property tax Funds from Central and State Governments etc.

Question 8.
When are Grama Sabha meetings convened? What are the special on those days?
Answer:

1. The Grama Sabha meetings are Convened on January 26, May 1, August 15, and October 2.
2. These days are celebrated as National festival days every year.

Question 9.
What are the special features of the Panchayat Raj system?
Answer:

1. Special features of Panchayat Raj
2. Grama Sabha
3. Three-tier local body governance
4. Reservations
5. Panchayat Elections
6. Tenure
7. Finance Commission
8. Account and Audit etc.

Question 10.
What is the importance of Grama Sabha?
Answer:

1. Grama Sabha is essential for the effective functioning of Village Panchayat.
2. It enhances public participation in the planning and implementation of schemes for social benefit.

V. HOTS

Question 1.
Local bodies play an important role in the development of villages and cities. How?
Answer:

1. India is a vast nation. It is very difficult for a single government to run the entire Country.
2. Our Constitution has provided for three separate levels of government.
3. Union government
4. State government
5. Local government
6. The Local government takes care of the local administration of cities and villages.
7. The main jobs of these bodies are
8. Keeping an area clean
9. Construction of roads and schools
10. arrangements for water and electricity etc.

VI. Activities

1. Prepare a questionnaire to interview a local body representative.
2. Discuss; If there is a contribution to the improvement of your school by local body representatives.
3. If I were a local body representative, I would
4. Find out the number of local bodies in your district and list them.
   

Question 1.
Prepare …………… representative
1. What is your plan to bring in a good drainage system?
2. When will the bridgework Connecting Maharaja Nagar and Thyagaraja Nagar be completed?
3. How many months it will take to bring street lights in our area?

Question 2.
Discuss: If there representatives:
1. Local body members of our area have Contributed much to the upliftment of our school.
2. They met the VIPS, the Common public, and the old Students to meet out the needs of the school.
3. From the Sponsorship they supplied things like Computer Laboratory equipment and books for the library.

Question 3.
If I were …………… them.
1. I would take steps to eradicate Dengu and all sorts of infectious diseases.
2. I would try to maintain a proper drainage system.
3. I would bring in proper roads and lightings etc.

Samacheer Kalvi 6th Social Science Local Bodies: Rural and Urban Additional Important Questions and Answers

I. Fill in the blanks Answer

1. There are ………………. Corporations in Tamil Nadu.
2. The Chennai Corporation was founded in ……………….
3. ………………. District has the most number of municipalities
4. A ………………. Panchayat is between a Village and a city
5. The ………………. Panchayats are the local bodies of Villages.
6. The ………………. and ………………. Districts have the lowest number of Panchayat Unions.
7. The Constituencies are also called……………….

Answer:

1. Twelve
2. 1688
3. Kanchipuram
4. Town
5. Village
6. Nilgris and Perambalur
7. Wards

II. Choose the Correct answer

Question 1.
advocated Panchayat Raj as the foundation of India’s Political System ……………….
(a) Jawaharlal Nehru
(b) Mahatma Gandhi
(c) Rajendra Prasad
Answer:
(b) Mahatma Gandhi

Question 2.
A Bio element Officer (BDQ) is the administrative head of a __________
(a) Village Panchayat
(b) District Panchayat
(c) Panchayat Union
(d) Town Panchayat
Answer:
(c) Panchayat Union

Question 3.
In the 2011 Local Bodies election, percent Seats were won by women ……………….
(a) 38
(b) 28
(c) 48
Answer:
(a) 38

Question 4.
Discretionary function of a Village Panchayat __________
(a) Cleaning roads
(b) Libraries
(c) Water supply
(d) Street lighting
Answer:
(b) Libraries

Question 5.
The Tamil Nadu State Election Commission is situated in ……………….
(a) Chennai
(b) Coimbatore
(c) Trichy
Answer:
(a) Chennai

III. Answer the following briefly

Question 1.
Write about the officials of Municipal Corporation.
Answer:

1. A City Municipal Corporation has a Commissioner, who is an IAS Officer.
2. Government Officials are deputed as Commissioners for municipalities.
3. The administrative officer of a Municipality is an Executive officer (EO)

Question 2.
Explain briefly about Panchayat Union.
Answer:

1. Many Village Panchayats join to form a Panchayat Union.
2. A Councillor is elected from each Panchayat.
3. The Councillors will elect a Panchayat union Chairperson among themselves.
4. A Vice-Chairperson is also elected.
5. A Block Development Officer (BDO) is the administrative head, of a Panchayat Union.
6. The Services are provided on the Panchayat Union level.

Question 3.
List out the Strength of local bodies in Tamil Nadu.
Answer:
Tamil Nadu:

1. Village Panchayats – 12, 524
2. Panchayat Unions – 385
3. District Panchayats – 31
4. Town Panchayats – 561
5. Municipalities – 125
6. City Municipal Corporation – 12

IV. Answer the following in detail

Question 1.
Discuss the District Panchayat
Answer:

1. A District Panchayat is formed in every district.
2. A district is divided into wards on the basis of a 50,000 population.
3. The ward members are elected by the village panchayats.
4. The members of the District Panchayat elect the District Panchayat Committee Chairperson.
5. They provide essential services and facilities to the rural population.
6. They also provide planning and execution of development programmes for the district.

Question 2.
List out the Works carried out by local bodies during natural disasters and outbreaks of diseases.
Answer:

1. Rescuing the public and settle them in a safer places.
2. Arranging food packets and pure drinking water.
3. Assisting them with medical Aids.
4. Creating awareness about the clean environment to the public.
5. Keeping the medicines in the upto date conditions.
6. Preventing them from getting panic against the diseases and make the situation, calm.

V. Mind map