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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2024 · 4 April · Evening Shift · Physics · Gravitation

A 90 ~kg body placed at 2 R distance from surface of earth...

A 9090 kg\mathrm{~kg}bodyplacedatbody placed at2 R\mathrm{R} distance from surface of earth experiences gravitational pull of :  ( R= Radius of earth, g=10 m s−2\text{ ( } \mathrm{R}=\text{ Radius of earth, } \mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2} ) \text{ ) }

Options

  1. A100 N100 \mathrm{~N}
  2. B300 N300 \mathrm{~N}
  3. C225 N225 \mathrm{~N}
  4. D120 N120 \mathrm{~N}

Correct answer

A. 100 N100 \mathrm{~N}

Step-by-step solution

Here gs=g_s= gravitational acceleration at surface Value of  g=g5(1+hR)−2=g5(1+2)−2=g59\begin{aligned} & \text{Value\text{ of } } g=g_5\left(1+\frac{h}{R}\right)^{-2} \\ & =g_5(1+2)^{-2}=\frac{g_5}{9} \end{aligned}\text{Force }=\mathrm{mg}=90 \times \frac{\mathrm{g}_5}{9}=100 \mathrm{ N}$
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