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One face CD of a brass cube ABCD having length of side 10 cm is fixed and a tangential force is applied on the free face A. As a result the angular displacement of the other face is 0.18^o. If shear modulus of brass is $4 \times 10^{10} N / m^{2}$ , the elastic potential energy stored in the cube is: $(π^{2} \approx 10)$

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150 J

300 J

200 J

50 J

answer is C.

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Detailed Solution

Shear strain, $θ = 0 {.18}^{o} = \frac{π}{180} \times 0 .18 rad = \frac{π}{1000} rad$

Strain energy stored per unit volume is given by:

$u = \frac{1}{2} {ηθ}^{2} = \frac{1}{2} \times 4 \times 10^{10} \times {(\frac{π}{1000})}^{2} \frac{J}{m^{3}}$

$= 2 \times 10^{10} \times \frac{10}{10^{6}} \frac{J}{m^{3}} = 2 \times 10^{5} \frac{J}{m^{3}}$

$∴$ Total strain energy stored = volume x energy density

$= {(\frac{10}{100})}^{3} \times 2 \times 10^{5} J = 200 J$

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