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A metallic cube ABCD has length of side 5 cm. The lower end CD is rigidly clamped to a horizontal surface and a horizontal force F = 10 kN is applied on the top surface AB, parallel to the surface. If modulus of rigidity of metal is $10^{8} N / m^{2}$ , the angular displacement of the upper face AB relative to the lower face CD is

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0.04 rad

0.08 rad

0.25 rad

0.12 rad

answer is A.

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Detailed Solution

Shear stress $τ = \frac{F}{A} = \frac{10 x 10^{3}}{25 x 10^{- 4}} N / m^{2} = 4 x 10^{6} N / m^{2}$

Angular displacement = Shear strain, $θ = \frac{τ}{η} = \frac{4 x 10^{6}}{10^{8}} rad = 0.04 rad$

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