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JEE Main 20178 AprilOnline
Exam
JEE Main
Year
2017
Date
8 April
Mode
Online
Subject
Physics
Chapter
Gravitation

JEE Main 2017 · 8 April · Online · Physics · Gravitation

If the Earth has no rotational motion, the weight of a person on...

If the Earth has no rotational motion, the weight of a person on the equator is WW. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh 34W\dfrac{3}{4}W. The radius of the Earth is 6400km6400km and g=10ms−2g = 10m{s}^{- 2}

Options

  1. A0.63×10−3rads−10.63 \times {10}^{- 3}rad{s}^{- 1}
  2. B0.28×10−3rads−10.28 \times {10}^{- 3}rad{s}^{- 1}
  3. C1.1×10−3rads−11.1 \times {10}^{- 3}rad{s}^{- 1}
  4. D0.83×10−3rads−10.83 \times {10}^{- 3}rad{s}^{- 1}

Correct answer

A. 0.63×10−3rads−10.63 \times {10}^{- 3}rad{s}^{- 1}

Step-by-step solution

Due to the rotation of the earth,

Changed gravitational acceleration, g′=g−ω2Rcos⁡2θ{g}^{′} = g - {\omega}^{2}R\cos^{2} \theta

At the equator, angle of inclination, θ=0\theta = 0

3g4=g−ω2R\dfrac{3g}{4} = g - {\omega}^{2}R

w2R=g4{w}^{2}R = \dfrac{g}{4}

w=g4Rw = \sqrt{\dfrac{g}{4R}}

=104×6400×103= \sqrt{\dfrac{10}{4 \times 6400 \times {10}^{3}}}

=12×8×100= \dfrac{1}{2 \times 8 \times 100}

=11600=116×10−2=0.6×10−3rads−1= \dfrac{1}{1600} = \dfrac{1}{16} \times {10}^{- 2} = 0.6 \times {10}^{- 3}rad{s}^{- 1}.
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