MHT CET · Physics · Oscillations
The length of seconds pendulum is \(1 \mathrm{~m}\) on the earth. If the mass and diameter of
the planet is double than that of the earth, then the length of the seconds pendulum
on the planet will be
- A \(0.2 \mathrm{~m}\)
- B \(0 \cdot 4 \mathrm{~m}\)
- C \(0 \cdot 3 \mathrm{~m}\)
- D \(0.5 \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(0.5 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(\frac{T_{e}}{T_{p}}=\sqrt{\frac{\ell_{e}}{g_{e}} \times \sqrt{\frac{g_{p}}{\ell_{p}}}} \quad\) But \(T_{e}=T_{p}\)
\(\therefore \quad 1=\sqrt{\frac{\ell_{e} \times g_{p}}{g_{e} \times \ell_{p}}}\)
But \(g_{p}=\frac{G \times 2 M}{(2 R)^{2}}=\frac{G M \times 2}{4 R^{2}}=\frac{g}{2}\)
\(g_{e} \ell_{p}=\ell_{e} g_{p}\)
\(\ell_{p}=\ell_{e} \frac{g_{p}}{g_{e}}=1 \times \frac{1}{2}=0.5 \mathrm{~m}\)
\(\therefore \quad 1=\sqrt{\frac{\ell_{e} \times g_{p}}{g_{e} \times \ell_{p}}}\)
But \(g_{p}=\frac{G \times 2 M}{(2 R)^{2}}=\frac{G M \times 2}{4 R^{2}}=\frac{g}{2}\)
\(g_{e} \ell_{p}=\ell_{e} g_{p}\)
\(\ell_{p}=\ell_{e} \frac{g_{p}}{g_{e}}=1 \times \frac{1}{2}=0.5 \mathrm{~m}\)
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