The metal cube of side 10 cm is subjected to a shearing stress of `10^(4) N m^(-2)`. The modulus of rigidity if the top of the cube is displaced by 0.05 cm with respect to its bottom is

Options

A. `2xx10^(6) N m^(-2)`
B. `10x^(5) Nm^(-2)`
C. `1xx10^(7) N m^(-2)`
D. `4xx 10^(5)N m^(-2)`

Answer

A

Explanation

Here, l= 10 cm ,`Deltal = 0.05cm`
Shearing stress `= 10^(4) N m^(-2)`
Shearing strain `= (Deltal)/(l) = (0.05)/(10)`
`therefore eta =("Shearing Stress")/("Shearing Strain")= (10^(4))/(0.005) = 2 xx 10^(6) N m^(-2)`