First slide

Elastic behaviour of solids

Question

A 5 cm cube has its upper face displaced by 0.2 cm by a tangential force of 8 N. The modulus of rigidity of the material of cube is

Moderate

A

$5 \times 10^{4} {Nm}^{- 2}$
B

$6 \times 10^{4} {Nm}^{- 2}$
C

$7 \times 10^{4} {Nm}^{- 2}$
D

$8 \times 10^{4} {Nm}^{- 2}$

Solution

Here $l = 5 cm = 5 \times 10^{- 2} m$
$Δl = 0 .2 cm = 0 .2 \times 10^{- 2} m, F = 8 N$
$Modulus of rigidity, η = \frac{Shearing stress}{Shearing strain} Hence, shearing stress = \frac{F}{A} = \frac{F}{l^{2}} = \frac{8}{{(5 \times 10^{- 2})}^{2}} = 3200 {Nm}^{- 2} Shearing strain = \frac{Δl}{l} = \frac{0 .2}{5} = 0 .04 ∴ η = \frac{3200}{0 .4} = 80000 {Nm}^{- 2} = 8 \times 10^{4} {Nm}^{- 2}$

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