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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 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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Modulus of rigidity of a liquid

Non zero constant

Infinite

Zero

Can not be predicted

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Explanation

So, the modulus of rigidity for water is zero.

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A cube of aluminium of sides 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be

0.02

0.1

0.005

0.002

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Explanation

(d) Shearing strain $\emptyset = \frac{x}{L} = \frac{0.02 c m}{10 c m} ∴ \emptyset = 0.002$

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The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30°. Then angle of shear is

$12 °$

$0.12 °$

$1.2 °$

$0.012 °$

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Explanation

(b) Angle of shear $\emptyset = \frac{r θ}{L} = \frac{4 \times 10^{- 1}}{100} \times 30 ° = 0.12 °$

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A rod of length l and radius r is joined to a rod of length l/2 and radius r/2 of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of $θ$ , the twist angle at the joint will be

$θ / 4$

$θ / 2$

$5 θ / 6$

$8 θ / 9$

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