Q. In the following four periods (i) Time of revolution of a satellite just above the earth’s surface $(T_{st})$ (ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth $(T_{ma})$ (iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform field of $9.8N/kg(T_{sp})$ (iv) Period of an infinite length simple pendulum in the earth’s real gravitational filed $(T_{is})$
A) $T_{st} > T_{ma}$
B) $T_{ma} > T_{st}$
C) $T_{sp} > T_{is}$
D) $T_{st} = T_{ma} = T_{sp} = T_{is}$
Explanation
(c) (i) $T_{st} = 2\pi\sqrt{\frac{(R + h)^{3}}{GM}} = 2\pi\sqrt{\frac{R}{g}}$ [As $h < < R$ and $GM = gR^{2}$]
(ii) $T_{ma} = 2\pi\sqrt{\frac{R}{g}}$
(iii) $T_{sp} = 2\pi\sqrt{\frac{1}{g\left( \frac{1}{l} + \frac{1}{R} \right)}} = 2\pi\sqrt{\frac{R}{2g}}$ [As $l = R$]
(iv) $T_{is} = 2\pi\sqrt{\frac{R}{g}}$ [As $l = \infty$]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.