The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh 3/5 th as much as at present weight is ( g = gravitational acceleration, R = equatorial radius of the earth) (a) √(2g/5R) (b) √(3g/5R) (c) √(5R/3g) (d) √(3/5gR)

Answer: (a). At the equator: g' = g - ω² R g - ω² R = 3/5g ⇒ ω² = 2g/5R ω = √(2g/5R) (angular speed) So the answer is (a).

The speed with which the earth would have to rotate about its axis so… | MHT CET 2024 (16 May, Shift 1)

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Acceleration Due to Gravity

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as at present weight is ( $g =$ gravitational acceleration, $R =$ equatorial radius of the earth)

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Answer: (a)

At the equator: $g' = g - \omega^2 R$
$g - \omega^2 R = \frac{3}{5}g \Rightarrow \omega^2 = \frac{2g}{5R}$
$\omega = \sqrt{\frac{2g}{5R}}$ (angular speed)
So the answer is (a).

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Acceleration Due to Gravity question from MHT CET 2024 (16 May, Shift 1)

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