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Physics Question 28 – JEE-MAIN 2026

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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 %. The value of α is _______. (Take radius of the earth = 6400 km.)

Recall how the acceleration due to gravity changes with depth and height from the Earth's surface.

Step 1: Identify formulas for g at depth and height✦ Active

The acceleration due to gravity at depth d from the Earth's surface is given by gd=g(1−dR). The acceleration due to gravity at height h above the Earth's surface (for h≪R) is given by gh=g(1−2hR). We are given d=16 km, h=16 km, and the radius of the Earth R=6400 km.

Step 2: Calculate g at initial and final points○ Expand

The initial point is 16 km below the surface, so ginitial=gd. Substituting the values:

ginitial=g(1−166400)=g(1−1400)

The final point is 16 km above the surface, so gfinal=gh. Substituting the values:

gfinal=g(1−2×166400)=g(1−326400)=g(1−1200)
💡 Teacher's Secret Hint

Remember to use the appropriate approximation for g at height when h≪R.

Step 3: Calculate the percentage change in g○ Expand

The change in g is Δg=gfinal−ginitial. We are looking for the magnitude of this change as a percentage of g at the surface.

Δg=g(1−1200)−g(1−1400) Δg=g(1−1200−1+1400) Δg=g(−1200+1400)=g(−2400+1400) Δg=−g1400 The magnitude of the percentage change is |Δg|g×100%.\text{Percentage change} = \frac{|-g\frac{1}{400}|}{g} \times 100\% = \frac{1}{400} \times 100\% = \frac{1}{4}\% = 0.25\%$$ Thus, the value of α is 0.25.
💡 Teacher's Secret Hint

Pay attention to the sign of the change. Since the options are positive, the question implies the magnitude of the percentage change.

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