MHT CET · Physics · Oscillations
The period of oscillation of a second's pendulum on the planet whose mass and radius are twice that of earth will
- A \(\sqrt{2} \mathrm{~s}\)
- B \(\frac{1}{\sqrt{2}} \mathrm{~s}\)
- C \(\frac{1}{2} \mathrm{~s}\)
- D \(2 \sqrt{2} \mathrm{~s}\)
Answer & Solution
Correct Answer
(D) \(2 \sqrt{2} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
Gravitational acceleration on the surface of planet of mass \(M\) and radius \(R\) is given by,
\(g=\frac{G M}{R^2}\)
Gravitational acceleration on the surface of that planet with twice mass and radius will be,
\(g_p=\frac{G(2 M)}{(2 R)^2}=\frac{g_e}{2}\)
The period of oscillation of a second's pendulum on earth,
\(T_e=2 \pi \sqrt{\frac{l}{g_e}}=2 \mathrm{~s}\)
The period of oscillation of a second's pendulum on planet,
\(T_p=2 \pi \sqrt{\frac{l}{g_p}}=2 \pi \sqrt{\frac{2 l}{g_e}}=2 \sqrt{2} \mathrm{~s}\)
\(g=\frac{G M}{R^2}\)
Gravitational acceleration on the surface of that planet with twice mass and radius will be,
\(g_p=\frac{G(2 M)}{(2 R)^2}=\frac{g_e}{2}\)
The period of oscillation of a second's pendulum on earth,
\(T_e=2 \pi \sqrt{\frac{l}{g_e}}=2 \mathrm{~s}\)
The period of oscillation of a second's pendulum on planet,
\(T_p=2 \pi \sqrt{\frac{l}{g_p}}=2 \pi \sqrt{\frac{2 l}{g_e}}=2 \sqrt{2} \mathrm{~s}\)
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