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Question

A metal cube of side length $$8.0 cm$$ has its upper surface displaced with respect to the bottom by $$0.10 mm$$ when a tangential force of 4 x 10$$^{9}$$ N is applied at the top with bottom surface fixed. The rigidity modulus of the material of the cube is :                    


A
4×109N/m2
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B
5×1014N/m2
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C
8×109N/m2
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D
1×109N/m2
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Solution

The correct option is B $$5 \times 10^{14} N/m^{2}$$
Given, 

Side length, $$L= 8 cm$$

$$\Delta L= 0.1 mm$$

Force, $$F=4\times 10^9 N$$

Shearing strain $$=\dfrac{0.01}{8}$$

Stress $$= \dfrac{4\times 10^{9}}{8\times 8\times 10^{-4}}  N/m^{2}$$

Rigidity Modulus, $$y=\dfrac{\dfrac{F}{A}}{\frac{\Delta X}{X}}=\dfrac{\dfrac{4\times 10^{9}}{9\times 8\times 10^{-4}}}{\dfrac{0.01}{8}}$$

$$y=\dfrac{1}{2}\times 10^{14}  N/m^{2}$$

$$=5\times 10^{14} N/m^{2}$$

Physics

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