Q105 / 150 · MHT CET 2026 (17 April Shift 1) · Physics
The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\dfrac{3}{5}$th as much as at present is ($g$ = gravitational acceleration, $R$ = equatorial radius of the earth.)
- A$\sqrt{\dfrac{3}{5}gR}$
- Correct answerB$\sqrt{\dfrac{2g}{5R}}$
- C$\sqrt{\dfrac{3g}{5R}}$
- D$\sqrt{\dfrac{5R}{2g}}$
Solution
The apparent acceleration due to gravity at the equator when the Earth rotates with angular speed $\omega$ is given by:
$g' = g - R\omega^2$
The present weight of a person at the equator is approximately $mg$ (neglecting the small effect of the Earth's current rotation). The new weight is given as $\dfrac{3}{5}$ of the present weight.
$mg' = \dfrac{3}{5}mg$
$g' = \dfrac{3}{5}g$
Substituting the expression for $g'$:
$g - R\omega^2 = \dfrac{3}{5}g$
$R\omega^2 = g - \dfrac{3}{5}g$
$R\omega^2 = \dfrac{2}{5}g$
$\omega^2 = \dfrac{2g}{5R}$
$\omega = \sqrt{\dfrac{2g}{5R}}$
Answer: $\sqrt{\dfrac{2g}{5R}}$
$g' = g - R\omega^2$
The present weight of a person at the equator is approximately $mg$ (neglecting the small effect of the Earth's current rotation). The new weight is given as $\dfrac{3}{5}$ of the present weight.
$mg' = \dfrac{3}{5}mg$
$g' = \dfrac{3}{5}g$
Substituting the expression for $g'$:
$g - R\omega^2 = \dfrac{3}{5}g$
$R\omega^2 = g - \dfrac{3}{5}g$
$R\omega^2 = \dfrac{2}{5}g$
$\omega^2 = \dfrac{2g}{5R}$
$\omega = \sqrt{\dfrac{2g}{5R}}$
Answer: $\sqrt{\dfrac{2g}{5R}}$
Hence, the correct answer is option: 'B'.