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 $\text{T}\propto \frac{1}{\sqrt{\text{g}}}$
$\therefore \frac{{\text{T}}^{{}^{\prime }}}{\text{T}}=\sqrt{\frac{\text{g}}{{\text{g}}^{{}^{\prime }}}}$
But ${\mathrm{g}}^{\prime }=\frac{\mathrm{g}}{{\left(1+\frac{{\mathrm{h}}^2}{\mathrm{R}}\right)}^2}$ or $\frac{{\mathrm{g}}^{\prime }}{\mathrm{g}}=\frac{1}{{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)}^2}$
$\therefore \frac{{\mathrm{T}}^{\prime }}{\mathrm{T}}=\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)$
or ${\mathrm{T}}^{\prime }=\mathrm{T}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)$
$\therefore \Delta \mathrm{T}={\mathrm{T}}^{\prime }-\mathrm{T}=\mathrm{T}\left(\frac{\mathrm{h}}{\mathrm{R}}\right)$
$\therefore \Delta \text{T}={\left(\text{T}\right)}^{{}^{\prime }}-\text{T}=\text{T}\left(\frac{\text{h}}{\text{R}}\right)$
$\therefore$ Time lost in $t=1$ day is
$=\frac{\left(24\times 3600\right)\left(200\right)}{\text{6.4}\times 10^6}\text{ s}=\text{2.7 s}$