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7.Gravitation
hard
If a man at the equator would weigh $(3/5)^{th}$ of his weight, the angular speed of the earth is
A
$\sqrt {\frac{2}{5}\,\frac{g}{R}} $
B
$\sqrt {\frac{g}{R}} $
C
$\sqrt {\frac{R}{g}} $
D
$\sqrt {\frac{2}{5}\,\frac{R}{g}} $
Solution
At equator,
$W^{\prime}=\frac{3}{5} W$
$\mathrm{mg}^{\prime}=\frac{3}{5} \mathrm{mg} \Rightarrow \mathrm{g}^{\prime}=\frac{3}{5} \mathrm{g}$
$\because \quad \mathrm{g}^{\prime}=\mathrm{g}-\omega^{2} \mathrm{R} \Rightarrow \frac{3}{5} \mathrm{g}=\mathrm{g}-\omega^{2} \mathrm{R}$
$\Rightarrow \omega^{2} R=g-\frac{3}{5} g \Rightarrow \omega=\sqrt{\frac{2}{5} \cdot \frac{g}{R}}$
Standard 11
Physics
