A) \[1\,\,m\]
B) \[2\,\,m\]
C) \[0.5\,\,m\]
D) \[4\,\,m\]
Correct Answer: C
Solution :
The motion of the bob is simple harmonic, hence its time period is given by \[T=2\pi \sqrt{\frac{displacement}{acceleration}}=2\pi \sqrt{\frac{l}{g}}\] Also if the periodic time of a pendulum is\[2\,\,s\], then it is called a seconds pendulum. Also, \[g=\frac{GM}{{{R}^{2}}}\] where, \[M\]is mass, \[R\] is radius \[\therefore \] \[T=2\pi \sqrt{\frac{{{R}^{2}}l}{GM}}=2\] ? (1) Seconds pendulum on other planet is \[2=2\pi \sqrt{\frac{4{{R}^{2}}l}{G(2M)}}\] ? (2) From Eqs. (1) and (2), we have \[\frac{{{R}^{2}}l}{GM}=\frac{4{{R}^{2}}l}{G(2M)}\] \[\Rightarrow \] \[l=0.5m\] Hence, length of pendulum on planet is 0.5 m.
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