Two simple pendulum of length `1m` and `16m` respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed `n` oscillations. The value of `n` is

Explanation

As, `T=2pisqrt(l//s)`, so `Tpropsqrt(l)`. It mean time period of shorter pendulum will be smaller, i.e., it will complete more oscillations in the same time than the longer pendulum. If `T_(s)`and `T_(l)` are the time periods of shorter and longer pendulums, then as per question, the two pendulums will be in same phase for the first time, when shorter one has completes n oscillations and longer one `(n-1)` oscillations. So `nT_(s)=(n-1)T_(l) or (T_(s))/(T_(l))=((n-1))/(n)`
As, `(T_(s))/(T_(l))=sqrt((l_(s))/(l_(l)))=sqrt(10/(16)) :. sqrt((1)/(16))=((n-1))/(n)`
On solving, we get `n=(4//3)`