The length of the seconds pendulum is decreased by $0.3 \mathrm{~cm}$ when it is shifted from place A to…

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. $975 \mathrm{~cm} / \mathrm{S}^{2}$
  2. $978 \mathrm{~cm} / \mathrm{s}^{2}$
  3. $984 \mathrm{~cm} / \mathrm{s}^{2}$
  4. $981 \mathrm{~cm} / \mathrm{s}^{2}$

Solution

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

Asked in: MHT CET 2020 (14 Oct Shift 2)

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