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Assertion : The difference in the value of acceleration due to gravity at pole and equator is proportional to square of angular velocity of earth.
Reason : The value of acceleration due to gravity is minimum at the equator and maximum at the pole.
Options:
Reason : The value of acceleration due to gravity is minimum at the equator and maximum at the pole.
Solution:
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Verified Answer
The correct answer is:
If both assertion and reason are true but reason is not the correct explanation of assertion
Acceleration due to gravity,
$$
g^{\prime}=g-R_e \omega^2 \cos ^2 \lambda
$$
At equator, $\lambda=0^{\circ}, \quad \therefore \cos 0^{\circ}=1$
$$
\therefore \mathrm{g}_e=\mathrm{g}-R_e \omega^2
$$
At poles, $\lambda=90^{\circ}, \quad \therefore \cos 90^{\circ}=0 \quad \therefore g_p=g$
Thus, $g_p-g_e=g-g+R_e\left(\omega^2=R_e\left(\omega^2\right.\right.$
Also, the value of $g$ is maximum at poles and minimum at equator.
$$
g^{\prime}=g-R_e \omega^2 \cos ^2 \lambda
$$
At equator, $\lambda=0^{\circ}, \quad \therefore \cos 0^{\circ}=1$
$$
\therefore \mathrm{g}_e=\mathrm{g}-R_e \omega^2
$$
At poles, $\lambda=90^{\circ}, \quad \therefore \cos 90^{\circ}=0 \quad \therefore g_p=g$
Thus, $g_p-g_e=g-g+R_e\left(\omega^2=R_e\left(\omega^2\right.\right.$
Also, the value of $g$ is maximum at poles and minimum at equator.
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