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MHT CET 2023 (11 May, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2023 (11 May, Shift 1)

Q71MHT CET 2023 · 11 MAY · SHIFT 1Medium

Acceleration Due to Gravity

The radius of earth is 6400 km6400\text{ km} and acceleration due to gravity g=10 m s−2g = 10\text{ m s}^{-2}. For the weight of body of mass 5 kg5\text{ kg} to be zero on equator, rotational velocity of the earth must be (in rad/s\text{rad/s})

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Answer: (c)

  1. Weightless at equator: g−ω2R=0⇒ω=gRg - \omega^2 R = 0 \Rightarrow \omega = \sqrt{\frac{g}{R}} (mass does not matter)
  2. ω=106.4×106=1800\omega = \sqrt{\frac{10}{6.4 \times 10^6}} = \frac{1}{800} rad/s
  3. So the answer is (c) 1800\frac{1}{800}.
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