A) 5
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
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 \[{{N}_{L}}=1\] \[\Rightarrow \] \[{{N}_{S}}=2\]
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