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One of the square faces of a metal slab of side 50 cm and thickness 20 cm is rigidly fixed on a horizontal surface. If a tangential force of 30 N is applied to the top face and it is known that the shear modulus of the material is $4 \times 10^{10} N / m^{2}$ . Then the displacement (in m) of the top face is

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$4 \times 10^{- 12}$

$4 \times 10^{- 10}$

$6 \times 10^{- 10}$

$8 \times 10^{- 10}$

answer is C.

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Detailed Solution

$(6 \times 10^{- 10})$

Shear Strain, $\in = \tan θ = θ = \frac{x}{l}$ $η = \frac{a}{\in}$ $o r x = \frac{F . l}{A . η}$ Given $F = 30 N$

$A = 50 \times 50 \times 10^{- 4} η = 4 \times 10^{10} x = \frac{30 \times 20 \times 10^{- 2}}{50 \times 50 \times 10^{- 4} \times 4 \times 10^{10}} = \frac{6}{25 \times 4 \times 10^{- 3} \times 10^{10}} = 6 \times 10^{- 10} m .$

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