A simple pendulum has a time period ' $\mathrm{T}$ ' $\mathrm{in}$ air. Its time period when it is…
A simple pendulum has a time period ' $\mathrm{T}$ ' $\mathrm{in}$ air. Its time period when it is completely immersed in a liquid of density one eighth the density of - the material of bob is
$\left(\sqrt{\frac{7}{8}}\right) \mathrm{T}$
$\left(\sqrt{\frac{5}{8}}\right) \mathrm{T}$
$\left(\sqrt{\frac{3}{8}}\right) \mathrm{T}$
$\left(\sqrt{\frac{8}{7}}\right) \mathrm{T}$
Solution
Time period of simple pendulum $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$
$\Rightarrow \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}}$
Net downward force acting on the bob inside the liquid $=$ Weight of bob - Upthrust
$\Rightarrow \mathrm{Vpg}-\frac{p}{8} =\frac{7}{8} \mathrm{~V} \rho \mathrm{g}$
The value of $g$ inside the liquid will be $\frac{g}{8}$
$\begin{aligned} \therefore \quad \text { Time period in liquid } \mathrm{T}_1 & =\frac{1}{2 \pi} \sqrt{\frac{l}{\frac{7}{8}}} \\ & =\sqrt{\frac{8}{7}} \mathrm{~T}\end{aligned}$