Q133 / 160 · AP EAMCET 2018 (23 Apr Shift 1) · Physics
A rubber cube of side $5 \mathrm{~cm}$ has one face fixed, while a tangential force $1800 \mathrm{~N}$ is applied on its opposite face. If modulus of rigidity of rubber is $2.4 \times 10^6 \mathrm{Nm}^{-2}$, then the lateral displacement of the strained face is ________
- A3 mm
- B5 mm
- Correct answerC15 mm
- D1.5 mm
Solution
Shear stress
$$
=\frac{\text { Tangential force }}{\text { Area }}=\frac{1800 \mathrm{~N}}{(0.05)^2}
$$
Shear strain $=\tan \theta$
$$
\begin{aligned}
& =\frac{x}{h}=\frac{\text { Lateral displacement }}{\text { Height of cube }}=\frac{x}{5} \mathrm{~cm} \\
& =\frac{x \mathrm{~m}}{0.05 \mathrm{~m}}
\end{aligned}
$$
Modulus of rigidity,
$$
\eta=\frac{\text { Shear stress }}{\text { Shear strain }}
$$
$$
2.4 \times 10^6 \frac{\mathrm{N}}{\mathrm{m}^2}=\frac{1800}{25 \times 10^{-4}} \times \frac{0.05}{x} \mathrm{~N} / \mathrm{m}^2
$$
or
$$
\begin{aligned}
x & =\frac{1800 \times 0.05}{25 \times 10^{-4} \times 2.4 \times 10^6} \\
& =15 \times 10^{-3} \mathrm{~m} \\
& =15 \mathrm{~mm}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.