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(D) When a body is suspended by a string at a position $P$ with latitude $\lambda$,the body rotates with the angular velocity $\omega$ of the Earth. T

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$\frac{G M m}{R^2}-m \omega^2 R$

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(D) When a body is suspended by a string at a position $P$ with latitude $\lambda$,the body rotates with the angular velocity $\omega$ of the Earth. The effective force acting on the body is the gravitational force minus the centripetal force required for circular motion.The tension $T$ in the string is given by:$T = mg - mr\omega^2 \cos \lambda$Since $g = \frac{GM}{R^2}$ and the radius of the circular path at latitude $\lambda$ is $r = R \cos \lambda$,we substitute these into the equation:$T = m \left( \frac{GM}{R^2} \right) - m(R \cos \lambda) \omega^2 \cos \lambda$$T = \frac{GMm}{R^2} - mR\omega^2 \cos^2 \lambda$At the equator,the latitude $\lambda = 0^{\circ}$.Substituting $\lambda = 0^{\circ}$ and $\cos 0^{\circ} = 1$:$T = \frac{GMm}{R^2} - mR\omega^2 (1)^2$$T = \frac{GMm}{R^2} - mR\omega^2$

Consider a particle of mass $m$ suspended by a string at the equator. Let $R$ and $M$ denote the radius and mass of the Earth. If $\omega$ is the angular velocity of rotation of the Earth about its own axis,then the tension on the string will be $(\cos 0^{\circ}=1)$

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