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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Gravitation

BITSAT 2016 · Physics · Gravitation

What should be the velocity of rotation of earth due to rotation...

What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes 35\frac{3}{5} of the present weight at the equator. Equatorial radius of the earth is 6400km6400 \mathrm{ km}.

Options

  1. A8.7×10−7rad/s8.7 \times 10^{-7} \mathrm{rad} / \mathrm{s}
  2. B7.8×10−4rad/s7.8 \times 10^{-4} \mathrm{rad} / \mathrm{s}
  3. C6.7×10−4rad/s6.7 \times 10^{-4} \mathrm{rad} / \mathrm{s}
  4. D7.4×10−3rad/s7.4 \times 10^{-3} \mathrm{rad} / \mathrm{s}

Correct answer

B. 7.8×10−4rad/s7.8 \times 10^{-4} \mathrm{rad} / \mathrm{s}

Step-by-step solution

True weight at equator, W=mgW=m g Observed weight at equator, \mathrm{W}^{\prime}=mg′=35mgAtequator,latitude\mathrm{mg}^{\prime}=\frac{3}{5} \mathrm{mg}At equator, latitude\lambda=0Usingtheformula,Using the formula,mg′=mg−mRω2cos⁡2λ35mg=mg−mRω2cos⁡20=mg−mRω2⇒mRω2=mg−35mg=25mg∴ω=(25gR)1/2=(2×9.85×6.4×106)1/2=7.8×10−4rad/s\begin{array}{l}\mathrm{mg}^{\prime}=\mathrm{mg}-\mathrm{mR} \omega^{2} \cos ^{2} \lambda\frac{3}{5}\mathrm{mg}=\mathrm{mg}-\mathrm{mR} \omega^{2} \cos ^{2}0=\mathrm{mg}-\mathrm{mR} \omega^{2}\\ \Rightarrow \mathrm{mR} \omega^{2}=\mathrm{mg}-\frac{3}{5}\mathrm{mg}=\frac{2}{5}\mathrm{mg}\\ \therefore \quad \omega=\left(\frac{2}{5} \frac{\mathrm{g}}{\mathrm{R}}\right)^{1 / 2}\\ =\left(\frac{2 \times 9.8}{5 \times 6.4 \times 10^{6}}\right)^{1 / 2}=7.8 \times10^{-4} \mathrm{rad}/ \mathrm{s} \end{array}$
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