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Mechanical Properties Of Solids

Question

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A cube of side 40 mm has its upper face displaced by 0.1 mm by a tangential force of 8 kN. The shearing modulus of cube is

2 × 10⁹ Nm^-2

4 × 10⁹ Nm^-2

8 × 10⁹ Nm^-2

16 × 10⁹ Nm^-2

Solution

A.

2 × 10⁹ Nm^-2

Shearing modulus of cube

$η = \frac{F l}{A ∆ l}$

η = $\frac{8 \times 10^{3} \times 40 \times 10^{- 3}}{{(40 \times 10^{- 3})}^{2} \times (0.1 \times 10^{- 3})}$

η = 2 × 10⁹ N m^-2

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