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Q.

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60^o latitude becomes zero is (Radius of earth = 6400 km. At the poles $g = 10 {ms}^{- 2})$

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a

$5 .0 \times 10^{- 1} rad / s$

b

$2 .5 \times 10^{- 3} rad / s$

c

$10 \times 10^{1} rad / s$

d

$7 .8 \times 10^{- 2} rad / s$

answer is A.

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Detailed Solution

$\begin{array}{l} g^{'} = g - ω^{2} {Rcos}^{2} λ \Rightarrow 0 = g - ω^{2} {Rcos}^{2} 60^{\circ} \\ 0 = g - \frac{ω^{2} R}{4} \Rightarrow ω = 2 \sqrt{\frac{g}{R}} = \frac{1}{400} \frac{rad}{\sec} = 2 .5 \times 10^{- 3} \frac{rad}{\sec} \end{array}$

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