Q135 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Physics
A slab of side $50 \mathrm{~cm}$ and thickness $10 \mathrm{~cm}$ is subjected to a shearing force of $10^5 \mathrm{~N}$ on its narrow edge. If the lower edge is riveted to the floor and upper edge is displaced by $0.2 \mathrm{~mm}$, then shear modulus of the material of the slab is
- A6 GPa
- Correct answerB5 GPa
- C4 GPa
- D4.5 GPa
Solution
The given situation is shown in the following figure.
Shear strain $=\frac{x}{l}=\frac{0.2 \mathrm{~mm}}{50 \mathrm{~cm}}=\frac{0.2 \times 10^{-3}}{50 \times 10^{-2}}$ Shear stress $=\frac{F}{A}=\frac{10^5}{\frac{50}{100} \times \frac{10}{100}}=\frac{10^5 \times 10^4}{5 \times 10^2}=2 \times 10^6$ $\begin{aligned} \text { Modulus of rigidity } & =\frac{\text { Stress }}{\text { Strain }}=\frac{2 \times 10^6 \times 50 \times 10^{-2}}{0.2 \times 10^{-3}} \\ & =5 \times 10^9 \mathrm{~N} / \mathrm{m}^2=5 \mathrm{GPa} \end{aligned}$
Shear strain $=\frac{x}{l}=\frac{0.2 \mathrm{~mm}}{50 \mathrm{~cm}}=\frac{0.2 \times 10^{-3}}{50 \times 10^{-2}}$ Shear stress $=\frac{F}{A}=\frac{10^5}{\frac{50}{100} \times \frac{10}{100}}=\frac{10^5 \times 10^4}{5 \times 10^2}=2 \times 10^6$ $\begin{aligned} \text { Modulus of rigidity } & =\frac{\text { Stress }}{\text { Strain }}=\frac{2 \times 10^6 \times 50 \times 10^{-2}}{0.2 \times 10^{-3}} \\ & =5 \times 10^9 \mathrm{~N} / \mathrm{m}^2=5 \mathrm{GPa} \end{aligned}$
Hence, the correct answer is option: 'B'.