MHT CET · Physics · Oscillations
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
- A \(\left(\sqrt{\frac{7}{8}}\right) \mathrm{T}\)
- B \(\left(\sqrt{\frac{5}{8}}\right) \mathrm{T}\)
- C \(\left(\sqrt{\frac{3}{8}}\right) \mathrm{T}\)
- D \(\left(\sqrt{\frac{8}{7}}\right) \mathrm{T}\)
Answer & Solution
Correct Answer
(D) \(\left(\sqrt{\frac{8}{7}}\right) \mathrm{T}\)
Step-by-step Solution
Detailed explanation
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 \(\frac{g}{8}\)
\(\begin{aligned} \therefore \text { Time period in liquid } \mathrm{T}_1 & =\frac{1}{2 \pi} \sqrt{\frac{l}{\frac{7}{8}}} \\ & =\sqrt{\frac{8}{7}} \mathrm{~T}\end{aligned}\)
\(
\Rightarrow \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}}
\)
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 \(\frac{g}{8}\)
\(\begin{aligned} \therefore \text { Time period in liquid } \mathrm{T}_1 & =\frac{1}{2 \pi} \sqrt{\frac{l}{\frac{7}{8}}} \\ & =\sqrt{\frac{8}{7}} \mathrm{~T}\end{aligned}\)
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