JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 7)
T is the time period of simple pendulum on the earth's surface. Its time period becomes $$x$$ T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of $$x$$ will be :
4
$$\frac{1}{2}$$
2
$$\frac{1}{4}$$
Explanation
At surface of earth time period
$$ \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} $$
At height $\mathrm{h}=\mathrm{R}$
$$ \begin{aligned} & \mathrm{g}^{\prime}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}=\frac{\mathrm{g}}{4} \\\\ & \therefore \,\mathrm{xT}=2 \pi \sqrt{\frac{\ell}{(\mathrm{g} / 4)}} \\\\ & \Rightarrow \mathrm{xT}=2 \times 2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\\\ & \Rightarrow \mathrm{xT}=2 \mathrm{~T} \Rightarrow \mathrm{x}=2 \end{aligned} $$
$$ \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} $$
At height $\mathrm{h}=\mathrm{R}$
$$ \begin{aligned} & \mathrm{g}^{\prime}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}=\frac{\mathrm{g}}{4} \\\\ & \therefore \,\mathrm{xT}=2 \pi \sqrt{\frac{\ell}{(\mathrm{g} / 4)}} \\\\ & \Rightarrow \mathrm{xT}=2 \times 2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\\\ & \Rightarrow \mathrm{xT}=2 \mathrm{~T} \Rightarrow \mathrm{x}=2 \end{aligned} $$
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