A) 1 m
B) 2 m
C) 0.5 m
D) 4 m
Correct Answer: C
Solution :
\[T=2\pi \sqrt{\frac{l}{g}}\] \[\Rightarrow \] \[2=2\pi \sqrt{\frac{{{R}^{2}}l}{GM}}\] ??..(i) Here,\[l=1m\]) and on given planet, \[2=2\pi \sqrt{\frac{R{{'}^{2}}l'}{GM'}}\] \[\Rightarrow \] \[2=2\pi \sqrt{\frac{4{{R}^{2}}l'}{G\times 2M}}\] ?..(ii) From Eqs. (i) and (ii), \[2l'=l\] \[\Rightarrow \] \[l'=1/2=0.5\,m\]You need to login to perform this action.
You will be redirected in
3 sec