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A tangential force of 0.25 N is applied ...

A tangential force of 0.25 N is applied to a 5 cm cube to displace its upper surface with respect to the bottom surface. The shearing stress is

A

`10 N//m^(2)`

B

`50 N//m^(2)`

C

`75 N//m^(2)`

D

`100 N//m^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
Shearing stress `= (F)/(A) = (25 xx 10^(-2))/(25 xx 10^(-4)) = 100 N//m^(2)`
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