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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
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The correct answer is:
$\left(\sqrt{\frac{8}{7}}\right) \mathrm{T}$
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
$$
\Rightarrow \mathrm{Vpg}-\mathrm{V} \frac{\rho}{\rho} \mathrm{g}=\frac{7}{8} \mathrm{~V} \rho \mathrm{g}
$$
The value of $g$ inside the liquid will be $\frac{g}{8}$
$\begin{aligned} \therefore \quad \text { Time period in liquid } \mathrm{T}_1 & =\frac{1}{2 \pi} \sqrt{\frac{l}{\frac{7}{8}}} \\ & =\sqrt{\frac{8}{7}} \mathrm{~T}\end{aligned}$
$$
\Rightarrow \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}}
$$
Net downward force acting on the bob inside the liquid $=$ Weight of bob - Upthrust
$$
\Rightarrow \mathrm{Vpg}-\mathrm{V} \frac{\rho}{\rho} \mathrm{g}=\frac{7}{8} \mathrm{~V} \rho \mathrm{g}
$$
The value of $g$ inside the liquid will be $\frac{g}{8}$
$\begin{aligned} \therefore \quad \text { Time period in liquid } \mathrm{T}_1 & =\frac{1}{2 \pi} \sqrt{\frac{l}{\frac{7}{8}}} \\ & =\sqrt{\frac{8}{7}} \mathrm{~T}\end{aligned}$
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