Question

An accurate pendulum clock is mounted on ground floor of a high building. How much time will it lose or gain in one day if it is transferred to top storey of a building which is $\text{h = 200 m}$ higher than the ground floor? Radius of earth is $6.4 \times 10^{6} \text{m}$ .

• A. It will lose $\text{6} \text{.2 s}$
• B. It will lose $\text{2} \text{.7 s}$
• C. It will gain $\text{5} \text{.2 s}$
• D. It will gain $\text{1} \text{.6 s}$

Answer 1

Answer: (B)

$\text{T} = 2 \pi \sqrt{\frac{l }{\text{g}}}$ or $\text{T} \propto \frac{1}{\sqrt{\text{g}}}$
$∴ \, \frac{\text{T}^{'}}{\text{T}} = \sqrt{\frac{\text{g}}{\text{g}^{'}}}$
But $\mathrm{g}^{\prime}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}^2}{\mathrm{R}}\right)^2}$ or $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}$
$\therefore \quad \frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)$
or $\mathrm{T}^{\prime}=\mathrm{T}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)$
$\therefore \quad \Delta \mathrm{T}=\mathrm{T}^{\prime}-\mathrm{T}=\mathrm{T}\left(\frac{\mathrm{h}}{\mathrm{R}}\right)$
$∴ \, \Delta \text{T} = \left(\text{T}\right)^{'} - \text{T} = \text{T} \left(\frac{\text{h}}{\text{R}}\right)$
$\therefore $ Time lost in $t = 1$ day is
$= \frac{\left(2 4 \times 3 6 0 0\right) \left(2 0 0\right)}{\text{6.4} \times 1 0^{6}} \text{ s} = \text{2.7 s}$