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MHT CET · Physics · Oscillations

The length of the seconds pendulum is decreased by \(0.3 \mathrm{~cm}\) when it is shifted from
place A to place B. If the acceleration due to gravity at place \(\mathrm{A}\) is \(981 \mathrm{~cm} / \mathrm{s}^{2}\), the
acceleration due to gravity at place \(\mathrm{B}\) is \(\left(\right.\) Take \(\pi^{2}=10\) )

  1. A \(975 \mathrm{Cm} / \mathrm{S}^{2}\)
  2. B \(978 \mathrm{~cm} / \mathrm{s}^{2}\)
  3. C \(984 \mathrm{~cm} / \mathrm{s}^{2}\)
  4. D \(981 \mathrm{Cm} / \mathrm{s}^{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(978 \mathrm{~cm} / \mathrm{s}^{2}\)

Step-by-step Solution

Detailed explanation

For seconds pendulum \(\mathrm{T}=26\)
\(T=2 \pi \sqrt{\frac{1}{g}} \quad \therefore \quad 2=2 \pi \sqrt{\frac{\ell}{g}}\)
of \(1=\pi \sqrt{\frac{l}{g}}\)
gquaring: \(1=\pi^{2} \frac{\ell}{g}\)
\(t=\frac{9}{\pi^{2}}=\frac{981}{10}=98.1 \mathrm{~cm}\)
At place B, \(\ell^{\prime}=98.1-0.3=97.8 \mathrm{~cm}\)
Again, \(2=2 \pi \sqrt{\frac{\ell^{\prime}}{\mathrm{g}^{\prime}}}\)
\(1=\pi^{2} \frac{\mathscr{L}}{\mathrm{g}^{\prime}}\)
\(\therefore g^{\prime}=\pi^{2} \ell^{\prime}=10 \times 97.8=978 \mathrm{~cm} / \mathrm{s}^{2}\)