A) 1 m
B) 2 m
C) 0.5 m
D) 4 m
Correct Answer: C
Solution :
Key Idea: The acceleration due to gravity of earth is equal in magnitude to the force exerted by the earth on a body of unit mass. The motion of the bob is simple harmonic, hence its time period is given by \[T=2\pi \sqrt{\frac{\text{displacement}}{\text{acceleration}}}=2\pi \sqrt{\frac{l}{g}}\] Also if the periodic time of a pendulum is 2 seconds, then it is called a second's pendulum. Also, \[g=\frac{GM}{{{R}^{2}}}\]where, M is mass and R is radius. \[\therefore \] \[T=2\pi \sqrt{\frac{{{R}^{2}}l}{GM}}=2\] ?(i) Second's pendulum on other planet is \[2=2\pi \sqrt{\frac{4{{R}^{2}}l'}{G(2M)}}\] ?(ii) From Eqs. (i) and (ii), we have \[\frac{{{R}^{2}}l}{GM}=\frac{4{{R}^{2}}l'}{G(2M)}\] \[\Rightarrow \] \[l'=0.5\,m\] Hence, length of pendulum on planet is 0.5 m.
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