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A tangential force of 2100 N is applied on a surface of area $3 \times 10^{- 6} m^{2}$ which is 0.1 m from a fixed face. The force produces a shift of 7mm of upper surface with respect to bottom. Then the modulus of rigidity of the material.

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$5 \times 10^{10} N m^{- 2}$

$1 \times 10 N m^{- 2}$

$15 \times 10^{10} N m^{- 2}$

$1 \times 10^{10} N m^{- 2}$

answer is D.

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Detailed Solution

$F = 2100 N; A = 3 \times 10^{- 6} m^{2};$ $L = 0.1 m; ℓ = 7 \times 10^{- 3} m$ .

$η = \frac{F L}{A ℓ} = \frac{2100 \times 0.1}{3 \times 10^{- 6} \times 7 \times 10^{- 3}} = 1 \times 10^{10} N m^{- 2}$

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