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  • A 0.05 m cube has its upper face displaced by 0.2 cm . by a tangential force of 8 N the modulus of rigidity of the cube is Option: 1 <img alt="8

A 0.05 m cube has its upper face displaced by 0.2 cm . by a tangential force of 8 N the modulus of rigidity of the cube is 

Option: 1

8 \times 10 ^3 N/m^2


Option: 2

8 \times 10 ^4 N/m^2


Option: 3

8 \times 10 ^5 N/m^2


Option: 4

8 \times 10 ^6 N/m^2


Answers (1)

best_answer

As we have learned

Modulus of Rigidity -

Ratio of tangential stress to the shearing stress.

- wherein

it is denoted by G

 

 shearing strain = \frac{\Delta l}{l} = \frac{0.2}{5} = 0.04

shearing stress = \frac{\Delta F}{l^2} = \frac{8}{0.05\times 0.05} = \frac{8}{25} \times 10^4modulus of rigiidty 

G = 8 \times 10 ^4 N/m^2

Posted by

HARSH KANKARIA

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