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When a pressure of 100 atmosphere is applied on a spherical ball, then its volume reduces by 0.01%. The bulk modulus of the material of the rubber in $d y n e / c m^{2}$ is:

$10 \times 10^{12}$

$100 \times 10^{12}$

$1 \times 10^{12}$

$20 \times 10^{12}$

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Explanation

(c) $∆ V = \frac{0.01}{100} V K = \frac{p}{\frac{∆ V}{V}} = \frac{100}{0.01 / 100} = 10^{6} a t m = 10^{11} N / m^{2} = 10^{12} d y n e / c m^{2}$

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A uniform cube is subjected to volume compression. If each side is decreased by 1%, then bulk strain is

0.01

0.06

0.02

0.03

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Explanation

(d) If side of the cube is L then $V = L^{3} \Rightarrow \frac{d V}{V} = 3 \frac{d L}{L} ∴ % c h a n g e i n v o l u m e = 3 \times (% c h a n g e i n l e n g t h) = 3 \times 1 % = 3 % ∴ B u l k s t r a i n \frac{∆ V}{V} = 0.03$

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A ball falling in a lake of depth 200 m shows 0.1% decrease in its volume at the bottom. What is the bulk modulus of the material of the ball

$19.6 \times 10^{8} N / m^{2}$

$19.6 \times 10^{- 10} N / m^{2}$

$19.6 \times 10^{10} N / m^{2}$

$19.6 \times 10^{- 8} N / m^{2}$

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Explanation

(a) $B = \frac{∆ p}{∆ V / V} = \frac{h p g}{0.1 / 100} = \frac{200 \times 10^{3} \times 9.8}{1 / 1000} = 19.6 \times 10^{8} N / m^{2}$

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The Bulk modulus for an incompressible liquid is

Zero

Unity

Infinity

Between 0 to 1

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Explanation

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The pressure applied from all directions on a cube is P. How much its temperature should be raised to maintain the original volume ? The volume elasticity of the cube is $β$ and the coefficient of volume expansion is $α$ -

$\frac{P}{α β}$

$\frac{P α}{β}$

$\frac{P β}{α}$

$\frac{α β}{P}$

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Explanation

(a) If coefficient of volume expansion is $α$ and rise in temperature is $∆ θ t h e n$ $∆ V = V α ∆ θ \Rightarrow \frac{∆ V}{V} = α ∆ θ V o l u m e e l a s t i c i t y β = \frac{P}{∆ V / V} = \frac{P}{α ∆ θ} \Rightarrow ∆ θ = \frac{P}{α β}$

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The ratio of lengths of two rods A and B of same material is 1 : 2 and the ratio of their radii is 2 : 1, then the ratio of modulus of rigidity of A and B will be

4 : 1

16 : 1

8 : 1

1 : 1

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Explanation

(d) Modulus of rigidity is the property of material.

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Two wires A and B of same length and of the same material have the respective radii $r_{1}$ and $r_{2}$ . Their one end is fixed with a rigid support, and at the other end equal twisting couple is applied. Then the ratio of the angle of twist at the end of A and the angle of twist at the end of B will be

$\frac{r_{1}^{2}}{r_{2}^{2}}$

$\frac{r_{2}^{2}}{r_{1}^{2}}$

$\frac{r_{2}^{4}}{r_{1}^{4}}$

$\frac{r_{1}^{4}}{r_{2}^{4}}$

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Explanation

If material and length of the wires A and B are equal and equal twisting couple are applied then

$θ \propto \frac{1}{r^{4}} ∴ \frac{θ_{1}}{θ_{2}} = {(\frac{r_{2}}{r_{1}})}^{4}$

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When a spiral spring is stretched by suspending a load on it, the strain produced is called-

Shearing

Longitudinal

Volume

Transverse

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Explanation

Due to weight (mg), transverse stress will be produced.

So, due to this stress, shear strain will be developed.

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The Young's modulus of the material of a wire is $6 \times 10^{12} N / m^{2}$ and there is no transverse strain in it, then its modulus of rigidity will be

$3 \times 10^{12} N / m^{2}$

$2 \times 10^{12} N / m^{2}$

$10^{12} N / m^{2}$

None of the above

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Explanation

(a)

$Y = 2 η (1 + σ) F o r n o t r a n s v e r s e s t r a i n (σ = 0) Y = 2 η \Rightarrow η = \frac{Y}{2} = 3 \times 10^{12} N / m^{2}$

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If the Young's modulus of the material is 3 times its modulus of rigidity, then its volume elasticity will be

Zero

Infinity

$2 \times 10^{10} N / m^{2}$

$3 \times 10^{10} N / m^{2}$

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Explanation

(b) $Y = 2 η (1 + σ) \Rightarrow 3 η = 2 η (1 + σ) \Rightarrow σ = \frac{3}{2} - 1 = \frac{1}{2}$

Now substituting the value of $σ$ in the following expression.

$Y = 3 K (1 - 2 σ) \Rightarrow K = \frac{Y}{3 (1 - 2 σ)} = \infty$

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