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
B
5×1014N/m2
C
8×109N/m2
D
1×109N/m2

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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