Physics Circle home Student Login

Home›Chapters›Gravitation PYQs›Acceleration Due to Gravity›Question 104

JEE Main 2020 (5 Sep, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2020 (5 Sep, Shift 1)

Q104JEE Main 2020 · 5 SEP · SHIFT 1Medium

Acceleration Due to Gravity

The value of the acceleration due to gravity is g1 at a height h = R2\frac{R}{2} (R = radius of the earth) from the surface of the earth. It is again equal to g1 at a depth d below the surface of the earth. The ratio (dR)\left(\frac{d}{R}\right) equals:

Show answer and solution

Answer: (a)

  1. At h=R2h = \frac{R}{2}: g1=g(R1.5R)2=4g9g_1 = g\left(\frac{R}{1.5R}\right)^2 = \frac{4g}{9}
  2. At depth dd: g(1−dR)=4g9g\left(1 - \frac{d}{R}\right) = \frac{4g}{9}
  3. dR=1−49=59\frac{d}{R} = 1 - \frac{4}{9} = \frac{5}{9}
  4. So the answer is (a).
Report a mistake

More questions on Acceleration Due to Gravity