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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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Shearing stress causes a change in-

Length

Breadth

Shape

Volume

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Explanation

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To break a wire, a force of $10^{6} N / m^{2}$ is required. If the density of the material is $3 \times 10^{3} k g / m^{3}$ , then the length of the wire which will break by its own weight will be -

34 m

30 m

300 m

3 m

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Explanation

(a) $L = \frac{P}{d g} = \frac{10^{6}}{3 \times 10^{3} \times 10} = \frac{100}{3} = 34 m$

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If the potential energy of a spring is V on stretching it by 2 cm, then its potential energy when it is stretched by 10 cm will be

V/25

V/5

25V

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Explanation

(d) $U = \frac{1}{2} (\frac{Y A}{L}) l^{2} ∴ U \propto l^{2} \frac{U_{2}}{U_{1}} = {(\frac{l_{2}}{l_{1}})}^{2} = {(\frac{10}{2})}^{2} = 25 \Rightarrow U_{2} = 25 U_{1}$

i.e. potential energy of the spring will be 25 V

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The work done in stretching an elastic wire per unit volume is or strain energy in a stretched string is

Stress $\times$ Strain

$\frac{1}{2} \times$ Stress $\times$ Strain

$2 \times s t r a i n \times s t r e s s$

Stress/Strain

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Explanation

(b) $s t r e s s e n e r g y = \frac{1}{2} s t r e s s \times s t r a i n$

work done by a force on a wire

$W = \frac{1}{2} (y A \frac{∆ L}{L}) \times ∆ L = \frac{1}{2} (y \frac{∆ L}{L}) (\frac{∆ L}{L}) (A L) = \frac{1}{2} s t r e s s \times s t r a i n \times v o l u m e \Rightarrow \frac{W}{v o l u m e} = \frac{1}{2} s t r e s s \times s t r a i n$

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Two wires of same diameter of the same material having the length l and 2l. If the force F is applied on each, the ratio of the work done in the two wires will be

1 : 2

1 : 4

2 : 1

1 : 1

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Explanation

(a)

$W = \frac{1}{2} F l ∴ W \propto l (F i s c o n s \tan t) ∴ \frac{W_{1}}{W_{2}} = \frac{l_{1}}{l_{2}} = \frac{l}{2 l} = \frac{1}{2}$

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A 5 metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 metre above the floor. The wire was elongated by 1 mm. The energy stored in the wire due to stretching is

Zero

0.05 joule

100 joule

500 joule

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Explanation

(b)

$W = \frac{1}{2} \times F \times l = \frac{1}{2} m g l = \frac{1}{2} \times 10 \times 10 \times 1 \times 10^{- 3} = 0.05 J$

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