MCQEasyJEE 2023Variation with Altitude & Depth

JEE Physics 2023 Question with Solution

The weight of a body on the earth is 400N400 \, \text{N}. Then weight of the body when taken to a depth half of the radius of the earth will be:

  • A

    300N300 \, \text{N}

  • B

    Zero

  • C

    100N100 \, \text{N}

  • D

    200N200 \, \text{N}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Weight on the surface of the earth is W=400NW = 400 \, \text{N}. The body is taken to a depth d=R2d = \frac{R}{2}.

Find: The weight of the body at that depth.

At a depth dd inside the earth, weight is given by

W=W(1dR)W' = W\left(1 - \frac{d}{R}\right)

Substitute d=R2d = \frac{R}{2}:

W=400(1R/2R)W' = 400\left(1 - \frac{R/2}{R}\right) W=400(112)W' = 400\left(1 - \frac{1}{2}\right) W=40012=200NW' = 400 \cdot \frac{1}{2} = 200 \, \text{N}

Therefore, the weight at half the earth's radius depth is 200N200 \, \text{N}. The correct option is D.

Common mistakes

  • Using the inverse square law for depth inside the earth is incorrect because that relation applies outside the earth. Inside a uniformly dense earth, weight decreases linearly with depth. Use W=W(1dR)W' = W\left(1 - \frac{d}{R}\right) instead.

  • Substituting d=R2d = \frac{R}{2} incorrectly as dR=R2\frac{d}{R} = \frac{R}{2} is wrong because dR=R/2R=12\frac{d}{R} = \frac{R/2}{R} = \frac{1}{2}. Always simplify the ratio carefully.

  • Marking zero weight is incorrect because weight becomes zero only at the center of the earth, where d=Rd = R. Here the body is only at half the radius depth, so the weight is reduced but not zero.

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