JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 6)
At what distance above and below the surface of the earth a body will have same weight. (take radius of earth as $$R$$.)
$$\frac{\sqrt{3} R-R}{2}$$
$$\frac{R}{2}$$
$$\frac{\sqrt{5} R-R}{2}$$
$$\sqrt{5} R-R$$
Explanation
_29th_January_Morning_Shift_en_6_1.png)
$$\begin{aligned} & g_p=\frac{g R^2}{(R+h)^2} \\ & g_q=g\left(1-\frac{h}{R}\right) \\ & g_p=g_q \\ & \frac{g}{\left(1+\frac{h}{R}\right)^2}=g\left(1-\frac{h}{R}\right) \\ & \left(1-\frac{h^2}{R^2}\right)\left(1+\frac{h}{R}\right)=1 \end{aligned}$$
Take $$\frac{\mathrm{h}}{\mathrm{R}}=\mathrm{x}$$
So
$$\begin{aligned} & \mathrm{x}^3-\mathrm{x}+\mathrm{x}^2=0 \\ & \mathrm{x}=\frac{\sqrt{5}-1}{2} \\ & \mathrm{~h}=\frac{\mathrm{R}}{2}(\sqrt{5}-1) \end{aligned}$$
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