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JEE Main 2023 (25 Jan, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2023 (25 Jan, Shift 1)

Q72JEE Main 2023 · 25 JAN · SHIFT 1Difficult

Acceleration Due to Gravity

Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately)

(Take g = 10 m s−2-2, radius of earth = 6400 km)

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Answer: (b)

  1. Inside earth g′=grRg' = g\frac{r}{R}, so a=−gRra = -\frac{g}{R}r: SHM with ω=gR\omega = \sqrt{\frac{g}{R}} (mass does not matter)
  2. T=2πRg=2π6.4×10610=2π×800T = 2\pi\sqrt{\frac{R}{g}} = 2\pi\sqrt{\frac{6.4 \times 10^6}{10}} = 2\pi \times 800 s
  3. T≈5024T \approx 5024 s ≈84\approx 84 min =1= 1 h 2424 min
  4. So the answer is (b).
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