Q131 / 160 · AP EAMCET 2024 (18 May Shift 1) · Physics

Time period of a simple pendulum in Air is T. If the pendulum is in water and executes SHM, its time period is $t$. The value of $\frac{\mathrm{T}}{t}$ is [density of bob is $\frac{5000}{3} \mathrm{~kg} \mathrm{~m}^{-3}$ ]

  • A
    $\frac{2}{5}$
  • Correct answer
    B
    $\sqrt{\frac{2}{5}}$
  • C
    $\frac{5}{2}$
  • D
    $\sqrt{\frac{5}{2}}$

Solution

For simple pendulum Time period in air, $\mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{~g}}}$ ...(i) Time period in water, $\mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{~g}\left(1-\frac{\delta}{\sigma}\right)}}$ $\therefore \mathrm{t}=2 \pi \sqrt{\frac{1}{\mathrm{~g}\left(1-\frac{1000 \times 3}{5000}\right)}}=\left(\sqrt{\left.\frac{5}{2}\right)}\right) \cdot 2 \pi \sqrt{\frac{1}{g}}$ $\Rightarrow \mathrm{t}=\sqrt{\frac{5}{2}} \mathrm{~T}[$ Using eq $(\mathrm{i})]$ $\therefore \frac{\mathrm{T}}{\mathrm{t}}=\sqrt{\frac{2}{5}}$

Hence, the correct answer is option: 'B'.

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