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MHT CET · Physics · Motion In Two Dimensions

What should be the velocity of earth due to rotation about its own axis so that the weight at equator becomes \(\left(\frac{3}{5}\right)^{\text {th }}\) of initial value? (Radius of earth on equator \(=6400 \mathrm{~km}, \mathrm{~g}=10 \frac{\mathrm{m}}{\mathrm{s}^{2}}, \cos 0^{\circ}=1\) )

  1. A \(3 \cdot 5 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}\)
  2. B \(7 \cdot 91 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}\)
  3. C \(6 \cdot 5 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}\)
  4. D \(2 \cdot 5 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(7 \cdot 91 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}\)

Step-by-step Solution

Detailed explanation

\(g' = g - R\omega^2\) \(\frac{3}{5}g = g - R\omega^2\)