Solveeit Logo
Login

Question

Question: The metal cube of side 10 cm is subjected to a shearing stress of \(10^{4}\text{ N }\text{m}^{- 2}.\...

The metal cube of side 10 cm is subjected to a shearing stress of 104 N m−2.10^{4}\text{ N }\text{m}^{- 2}. The modulus of rigidity if the top of the cube is displaced by 0.05 cm with respect to its bottom is

A

2×106 N m−22 \times 10^{6}\text{ N }\text{m}^{- 2}

B

105 N m−210^{5}\text{ N }\text{m}^{- 2}

C

1×107 N m−21 \times 10^{7}\text{ N }\text{m}^{- 2}

D

4×105 N m−24 \times 10^{5}\text{ N }\text{m}^{- 2}

Answer

2×106 N m−22 \times 10^{6}\text{ N }\text{m}^{- 2}

Explanation

Solution

: Here, l=10cmΔl=0.05cml = 10cm\Delta l = 0.05cm

Shearing stress =104Nm−2= 10^{4}Nm^{- 2}

Shearing strain =Δll=0.0510= \frac{\Delta l}{l} = \frac{0.05}{10}

∴η=ShearingstressShearingstrain=1040.005=2×106Nm−2\therefore\eta = \frac{Shearingstress}{Shearingstrain} = \frac{10^{4}}{0.005} = 2 \times 10^{6}Nm^{- 2}