Q97 / 160 · JIPMER 2010 · Physics
A cube of side 40 mm has its upper face displaced by 0.1 mm by a tangential force of 8 kN . The shearing modulus of cube is
- Correct answerA$2 \times 10^9 \mathrm{Nm}^{-2}$
- B$4 \times 10^9 \mathrm{Nm}^{-2}$
- C$8 \times 10^9 \mathrm{Nm}^{-2}$
- D$16 \times 10^9 \mathrm{Nm}^{-2}$
Solution
Shearing modulus of cube
$\begin{aligned}
\eta=\frac{F l}{A \Delta l} & =\frac{8 \times 10^3 \times 40 \times 10^{-3}}{\left(40 \times 10^{-3}\right)^2 \times\left(0.1 \times 10^{-3}\right)} \\
& =2 \times 10^9 \mathrm{Nm}^{-2}
\end{aligned}$
Hence, the correct answer is option: 'A'.