A bob pendulum executes simple harmonic motion with a time period of 16 s. If the  apparatus is taken to a planet where acceleration due to gravity is 4 times less than  earth, then its new time period of oscillation is

For simple pendulum, the time period of oscillation is given by  $T=2\pi \sqrt{\frac{l}{g}}$ 
The new time period is given by  $T^{\text{'}}=2\pi \sqrt{\frac{l}{\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}$
Comparing this with time period on earth, we get  $\frac{T^{\text{'}}}{T}=\frac{2\pi \sqrt{\frac{l}{\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}{2\pi \sqrt{\frac{l}{g}}}=\sqrt{4}=2$
Simplifying we get  $T^{\text{'}}=2T=2\times 16=32s$