Question 328: Gravitation

Question

Time period of simple pendulum on earth's surface is '$$\mathrm{T}$$'. Its time period becomes '$$\mathrm{xT}$$' when taken to a height $$\mathrm{R}$$ (equal to earth's radius) above the earth's surface. Then the value of '$$x$$' will be

Accepted Answer

$$\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$$
At a height '$$h$$' from earth's surface,
$$\begin{aligned}
\mathrm{xT} & =2 \pi \sqrt{\frac{l}{\mathrm{~g}_{\mathrm{h}}}} \
	herefore \quad \mathrm{x} & =\sqrt{\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}} \quad 	ext{..... (i)}
\end{aligned}$$
Now, $$\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}$$
$$	herefore \quad \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{4 \mathrm{R}^2}\quad$$ .....($$\because$$ h = R)
$$	herefore \quad \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{4}\quad$$ .... (ii)
$$	herefore \quad$$ From equations (i) and (ii),
$$x=\sqrt{\frac{g}{g / 4}}=\sqrt{4}=2$$

Answer

4

Answer

2

Answer

$$\frac{1}{2}$$

Answer

$$\frac{1}{4}$$

Gravitation

Time period of simple pendulum on earth's surface is '$\mathrm{T}'. Its time period becomes '\mathrm{xT}' when taken to a height \mathrm{R} (equal to earth's radius) above the earth's surface. Then the value of 'x$' will be

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