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 \(=6400 \mathrm{~km}, \mathrm{~g}=10 \frac{\mathrm{m}}{\mathrm{s}^{2}}, \cos 0^{\circ}=1\) )
- A \(3 \cdot 5 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}\)
- B \(7 \cdot 91 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}\)
- C \(6 \cdot 5 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}\)
- D \(2 \cdot 5 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}\)
Answer & Solution
Correct Answer
(B) \(7 \cdot 91 \times 10^{-4} \frac{\mathrm{rad}}{\mathrm{s}}\)
Step-by-step Solution
Detailed explanation
\(g' = g - R\omega^2\) \(\frac{3}{5}g = g - R\omega^2\)
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