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Mechanical Properties of Solids MCQs for NEET — Physics Questions with Answers

Practice free Mechanical Properties of Solids (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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The extension of a wire by the application of load is 3 mm. The extension in a wire of the same material and length but half the radius by the same load is -

12 mm

0.75 mm

15 mm

6 mm

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Explanation

(a) $l = \frac{F L}{A Y} \Rightarrow l \propto \frac{1}{r^{2}} (F, L a n d Y a r e c o n s \tan t) \frac{l_{2}}{l_{1}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(2)}^{2} \Rightarrow l_{2} = 4 l_{1} = 4 \times 3 = 12 m m$

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The isothermal elasticity of a gas is equal to

Density

Volume

Pressure

Specific heat

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Explanation

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The adiabatic elasticity of a gas is equal to

$γ \times$ density

$γ \times$ volume

$γ \times$ pressure

$γ \times$ specific heat

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Explanation

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The specific heat at constant pressure and at constant volume for an ideal gas are $C_{p}$ and $C_{v}$ and its adiabatic and isothermal elasticities are $E_{\emptyset}$ and $E_{θ}$ respectively. The ratio of $E_{\emptyset}$ to $E_{θ}$ is

$C_{v} / C_{p}$

$C_{p} / C_{v}$

$C_{p} C_{v}$

$1 / C_{p} C_{v}$

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Explanation

(b) Ratio of adiabatic and isothermal elasticities

$\frac{E_{\emptyset}}{E_{θ}} = \frac{γ P}{P} = γ = \frac{C_{p}}{C_{}}$

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The compressibility of water is $4 \times 10^{- 5}$ per unit atmospheric pressure. The decrease in volume of 100 cubic centimeter of water under a pressure of 100 atmosphere will be -

0.4 cc

$4 \times 10^{- 5} c c$

0.025 cc

0.004 cc

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Explanation

(a) $C = \frac{1}{K} = \frac{∆ V / V}{∆ P} \Rightarrow ∆ V = C \times ∆ P \times V = 4 \times 10^{- 5} \times 100 \times 100 = 0.4 c c$

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If a rubber ball is taken at the depth of 200 m in a pool, its volume decreases by 0.1%. If the density of the water is $1 \times 10^{3} k g / m^{3}$ and $g = 10 m / s^{2}$ , then the volume elasticity in will be

$10^{8}$

$2 \times 10^{8}$

$10^{9}$

$2 \times 10^{9}$

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Explanation

(d) $K = \frac{∆ P}{∆ V / V} = \frac{h p g}{∆ V / V} = \frac{200 \times 10^{3} \times 10}{0.1 / 100} = 2 \times 10^{9}$

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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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