c
(c) If \(t\) is the time taken by pendulums to come in same phase again first time after \(t = 0\).
and \({N_S} = \) Number of oscillations made by shorter length pendulum with time period \({T_S}\).
\({N_L} = \) Number of oscillations made by longer length pendulum with time period \({T_L}\).

Then \(t = {N_S}{T_S} = {N_L}{T_L}\)

\( \Rightarrow {N_S}2\pi \sqrt {\frac{5}{g}} = {N_L} \times 2\pi \sqrt {\frac{{20}}{g}} \) (\(\because T = 2\pi \sqrt {\frac{l}{g}} \))

\( \Rightarrow {N_S} = 2{N_L}\) i.e. if \({N_L} = 1\)

\( \Rightarrow {N_S} = 2\)