Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The height above the surface of the earth where acceleration due to gravity becomes $\left(\frac{g}{9}\right)$ is ( $R$ is radius of the earth, $g$ is acceleration due to gravity)
PhysicsGravitationMHT CETMHT CET 2022 (11 Aug Shift 1)
Options:
  • A 2R
  • B $\frac{R}{3}$
  • C $\frac{R}{\sqrt{2}}$
  • D $\sqrt{2} R$
Solution:
1151 Upvotes Verified Answer
The correct answer is: 2R
$\begin{aligned} & g^{\prime}=\frac{g}{\left(1+\frac{h}{R}\right)^2} \\ & \frac{g}{9}=\frac{g}{\left(1+\frac{h}{R}\right)^2} \\ & h=2 R\end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.