Q131 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Physics
A simple pendulum of length $1 \mathrm{~m}$ and having a bob of mass $100 \mathrm{~g}$ is suspended in a car, moving on a circular track of radius $100 \mathrm{~m}$ with uniform speed $10 \mathrm{~m} / \mathrm{s}$. If the pendulum makes small oscillation in a radial direction about its equilibrium position, then its time period can be given by $\mathrm{T}=2 \pi / \alpha^{1 / 4}$. The value of $\alpha$ is [Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ]
[Take g = 10 m/s2]
- A11
- B110
- Correct answerC101
- D1100
Solution
The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car. Acceleration due to gravity $=\mathrm{g}$ Centripetal acceleration $=\mathrm{v}^2 / \mathrm{R}$ Effective acceleration will be
$$
g^{\prime}=\sqrt{g^2+\frac{v^4}{R^2}}
$$
Time period, $T=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}^{\prime}}}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{2 \pi}{2 / 4}=2 \pi \sqrt{\frac{1}{\sqrt{g^2+\frac{\mathrm{v}^4}{\mathrm{R}^2}}}} \\
& \Rightarrow \quad \alpha=\mathrm{g}^2+\frac{\mathrm{v}^4}{\mathrm{R}^2}=(10)^2+\frac{(10)^4}{(100)^2}=101
\end{aligned}
$$
Hence, the correct answer is option: 'C'.