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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.

Solution:

(a) \(\frac{\pi}{6}\)

Hint:

Question 2.

Solution:

(c) \(\frac{5}{2}\)

Hint:

Question 3.

Solution:

(c) 0

Hint:

Question 4.

Solution:

(d) \(\frac{2}{3}\)

Hint:

It is an even function

Question 5.

(b) 2π

(c) 3π

(d) 4π

Solution:

(d) 4π

Hint:

Question 6.

(a) 4

(b) 3

(c) 2

(d) 0

Solution:

(c) 2

Hint:

Question 7.

(a) cos x – x sin x

(b) sin x + x cos x

(c) x cos x

(d) x sin x

Solution:

(c) x cos x

Hint:

Question 8.

The area between y^{2} = 4x and its latus rectum is ………

Solution:

(c) \(\frac{8}{3}\)

Hint:

Question 9.

Solution:

(b) \(\frac{1}{10100}\)

Question 10.

Solution:

(a) \(\frac{\pi}{2}\)

Question 11.

(a) 10

(b) 5

(c) 8

(d) 9

Solution:

(d) 9

Hint:

Question 12.

Solution:

(b) \(\frac{2}{9}\)

Hint:

Question 13.

Solution:

\(\frac{3 \pi}{8}\)

Hint:

Question 14.

Solution:

(d) \(\frac{2}{27}\)

Hint:

Question 15.

(a) 4

(b) 1

(c) 3

(d) 2

Solution:

(d) 2

Hint:

Question 16.

The volume of solid of revolution of the region bounded by y^{2} = x(a – x) about x-axis is ……..

Solution:

(d) \(\frac{\pi a^{3}}{6}\)

Hint:

Question 17.

(a) 3

(b) 6

(c) 9

(d) 5

Solution:

(c) 9

Hint:

Question 18.

Solution:

(d) \(\frac{\pi^{2}}{4}-2\)

Hint:

Question 19.

Solution:

(b) \(\frac{3 \pi a^{4}}{16}\)

Hint:

Question 20.

Solution:

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

Hint:

### Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 Additional Problems

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.

The area bounded by the line y = x, the x – axis, the ordinates x = 1,x = 2 is …….

Solution:

(a) \(\frac{3}{2}\)

Hint:

Question 2.

The area of the region bounded by the graph of y = sin x and y = cos x between x = 0 and x = \(\frac{\pi}{4}\) is ……..

Solution:

(b) \(\sqrt{2}-1\)

Hint:

Question 3.

The area between the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 and its auxiliary circle is …….

(a) πb(a – b)

(b) 2πa(a – b)

(c) πa(a – b)

(d) 2πb(a – b)

Solution:

(c) πa(a – b)

Hint:

Question 4.

The area bounded by the parabola y^{2} = x and its latus rectum is ……..

Solution:

(b) \(\frac{1}{6}\)

Hint:

Question 5.

The volume of the solid obtained by revolving \(\frac{x^{2}}{9}+\frac{y^{2}}{16}\) = 1 about the minor axis is …….

(a) 48π

(b) 64π

(c) 32π

(d) 128π

Solution:

(b) 64π

Hint:

Question 6.

The volume, when the curve y = \(\sqrt{3+x^{2}}\) from x = 0 to x = 4 is rotated about x – axis is ……

Solution:

(c) \(\frac{100}{3} \pi\)

Hint:

Question 7.

The volume generated when the region bounded by y = x, y = 1, x = 0 is rotated about y – axis is ……….

Solution:

(c) \(\frac{\pi}{3}\)

Hint:

Question 8.

Volume of solid obtained by revolving the area of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 about major and minor axes are in the ratio …….

(a) b^{2} : a^{2}

(b) a^{2} : b^{2}

(c) a : b

(d) b : a

Solution:

(d) b : a

Hint:

Question 9.

The volume generated by rotating the triangle with vertices at (0, 0), (3, 0) and (3, 3) about x-axis is …….

(a) 18π

(b) 2π

(c) 36π

(d) 9π

Solution:

(d) 9π

Hint:

Question 10.

The length of the arc of the curve is …….

(a) 48

(b) 24

(c) 12

(d) 96

Solution:

(a) 48

Hint:

Length of the arc of the curve = 6a

∴ Required length = 6a = 6 × 8 = 48 units.

Question 11.

The surface area of the solid of revolution of the region bounded by y = 2x, x = 0 and x = 2 about x-axis is ……

Solution:

(a) \(8 \sqrt{5} \pi\)

Hint:

Question 12.

The curved surface area of a sphere of radius 5, intercepted between two parallel planes of distance 2 and 4 from the centre is ……

(a) 20π

(b) 40π

(c) 10π

(d) 30π

Solution:

(a) 20π

Hint:

The curved surface area of a sphere of radius r intercepted between two parallel planes at a distance a and b from the centre of the sphere is 2πr (b – a)

Given radius, r = 5; a = 2; b = 4

Required surface area = 2πr (b – a)

= 2π × 5 × (4 – 2) = 20π sq. units