Physics Circle home Student Login

Home›Chapters›Gravitation PYQs›Acceleration Due to Gravity›Question 11

JEE Main 2026 (5 Apr, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2026 (5 Apr, Shift 1)

Q11JEE Main 2026 · 5 APR · SHIFT 1Medium

Acceleration Due to Gravity

When one moves from a point 16 km below the earth's surface to a point 16 km above the earth's surface, the change in g is approximately α%\alpha\%. The value of α\alpha is ______.

(Take radius of the earth =6400 km= 6400\text{ km}.)

Show answer and solution

Answer: (b)

  1. 16 km below: g1=g(1−166400)=g(1−0.0025)g_1 = g\left(1 - \frac{16}{6400}\right) = g(1 - 0.0025)
  2. 16 km above: g2=g(1−2×166400)=g(1−0.005)g_2 = g\left(1 - \frac{2 \times 16}{6400}\right) = g(1 - 0.005)
  3. Change =g1−g2=0.0025g= g_1 - g_2 = 0.0025g, i.e. 0.25%0.25\%
  4. So α=0.25\alpha = 0.25, answer (b).
Report a mistake

More questions on Acceleration Due to Gravity