Question

A 2.0 cm cube of some substance has its upper face displaced by 0.15 cm , by a tangential force of 0.30 N fixing its lower face. Calculate the rigidity modulus of the substance.

• A. $4\times {10}^{4}N/m^2$
• B. $1\times {10}^{4}N/m^2$
• C. $3\times {10}^{4}N/m^2$
• D. $0.5\times {10}^{4}N/m^2$

Answer 1

The correct option is B $1\times {10}^{4}N/m^2$

Given that,

Force $F=0.30N$

Area $A=4\times {10}^{-2}{m}^2$

We know that,

The shearing strain is

$\theta =\frac{\Delta L}{L}=\frac{0.15}{2.0}$

$\theta =\frac{\Delta L}{L}=0.075$

Now, the shearing stress is

$\frac{F}{A}=\frac{0.30}{4\times {10}^{-4}}$

$\frac{F}{A}=0.075\times {10}^4$

Now, the rigidity modulus is

$\eta =\frac{F}{A\times \theta }$

$\eta =\frac{0.075\times {10}^4}{0.075}$

$\eta =1\times {10}^{4}N/m^2$

Hence, the rigidity modulus of the substance is $1\times {10}^{4}N/m^2$