Q 11-07-179JEE MainJEE Main 2019 (10 Apr, Shift 1)Easy
The value of acceleration due to gravity at Earth's surface is $9.8\ \text{m s}^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\ \text{m s}^{-2}$, is close to: (Radius of earth $= 6.4\times10^6\ \text{m}$)
Answer: (B) $2.6\times10^6\ \text{m}$
$$\frac{g_h}{g} = \left(\frac{R}{R+h}\right)^2 = \frac12 \;\Rightarrow\; R + h = \sqrt2R$$
$$h = (\sqrt2-1)R = 0.414\times6.4\times10^6 \approx 2.6\times10^6\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics