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JEE Main 2023 (31 Jan, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2023 (31 Jan, Shift 1)

Q69JEE Main 2023 · 31 JAN · SHIFT 1Medium

Acceleration Due to Gravity

At a certain depth "d " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height 3R3R above earth surface. Where RR is Radius of earth (Take R=6400 kmR = 6400\text{ km} ). The depth dd is equal to

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Answer: (d)

  1. At height 3R3R: gh=g(R4R)2=g16g_h = g\left(\frac{R}{4R}\right)^2 = \frac{g}{16}
  2. Given gd=4gh=g4g_d = 4g_h = \frac{g}{4}
  3. g(1−dR)=g4⇒d=3R4g\left(1 - \frac{d}{R}\right) = \frac{g}{4} \Rightarrow d = \frac{3R}{4}
  4. d=34×6400=4800d = \frac{3}{4} \times 6400 = 4800 km, so the answer is (d).
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