Time period of a simple pendulum in Air is T. If the pendulum is in water and executes SHM, its time period…
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 $\frac{5000}{3} \mathrm{~kg} \mathrm{~m}^{-3}$ ]
$\frac{2}{5}$
$\sqrt{\frac{2}{5}}$
$\frac{5}{2}$
$\sqrt{\frac{5}{2}}$
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 $(\mathrm{i})]$
$\therefore \frac{\mathrm{T}}{\mathrm{t}}=\sqrt{\frac{2}{5}}$