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JEE Main 2021 (24 Feb, Shift 2) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2021 (24 Feb, Shift 2)

Q99JEE Main 2021 · 24 FEB · SHIFT 2Medium

Acceleration Due to Gravity

A body weights 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?

[Use g=GMR2g = \frac{GM}{R^2} = 9.8 ms−2{}^{-2} and radius of earth, R = 6400 km.]

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Answer: (c)

  1. Mass m=499.8=5m = \frac{49}{9.8} = 5 kg; ω=2π86400=7.27×10−5\omega = \frac{2\pi}{86400} = 7.27 \times 10^{-5} rad/s
  2. ω2R=(7.27×10−5)2×6.4×106≈0.0338\omega^2 R = (7.27 \times 10^{-5})^2 \times 6.4 \times 10^6 \approx 0.0338 m/s2^2
  3. Weq=m(g−ω2R)=49−5×0.0338=49−0.17W_{eq} = m(g - \omega^2 R) = 49 - 5 \times 0.0338 = 49 - 0.17
  4. Weq≈48.83W_{eq} \approx 48.83 N, so the answer is (c).
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