Q130 / 160 · TS EAMCET 2024 (10 May Shift 1) · Physics
If a body dropped freely from a height of 20 m reaches the surface of a planet with a velocity of $31.4 \mathrm{~ms}^{-1}$, then the length of a simple pendulum that ticks seconds on the planet is
- A1 m
- B0.625 m
- Correct answerC2.5 m
- D2 m
Solution
$\mathrm{h}=20 \mathrm{~m}, \mathrm{v}=31.4 \mathrm{~ms}^{-1}$
By conservation of energy $\begin{aligned} & \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2 \\ & \Rightarrow \mathrm{~g}=\frac{\mathrm{v}^2}{2 \mathrm{~h}}=\frac{31.4 \times 31.4}{2 \times 20}=\frac{5 \pi^2}{2} \mathrm{~ms}^{-2} \end{aligned}$
For simple pendulum ticks second, $T=25$ $\begin{aligned} & \therefore \mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{~g}}} \Rightarrow \mathrm{l}=\left(\frac{\mathrm{T}}{2 \pi}\right)^2 \mathrm{~g} \\ & \therefore \quad 1=\left(\frac{2}{2 \pi}\right)^2 \times \frac{5 \pi^2}{2}=2.5 \mathrm{~m} \end{aligned}$
By conservation of energy $\begin{aligned} & \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2 \\ & \Rightarrow \mathrm{~g}=\frac{\mathrm{v}^2}{2 \mathrm{~h}}=\frac{31.4 \times 31.4}{2 \times 20}=\frac{5 \pi^2}{2} \mathrm{~ms}^{-2} \end{aligned}$
For simple pendulum ticks second, $T=25$ $\begin{aligned} & \therefore \mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{~g}}} \Rightarrow \mathrm{l}=\left(\frac{\mathrm{T}}{2 \pi}\right)^2 \mathrm{~g} \\ & \therefore \quad 1=\left(\frac{2}{2 \pi}\right)^2 \times \frac{5 \pi^2}{2}=2.5 \mathrm{~m} \end{aligned}$
Hence, the correct answer is option: 'C'.