JEE Main 2022 · 26 Jun · Shift 1 · Physics Q38

← JEE Main 2022 · 26 Jun · Shift 1

PhysicsSingle correct+4 / -1Question 38 of 90

Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with $\dfrac g6$ vertically upwards then the time period will be :

(Where g = acceleration due to gravity)

\sqrt{\dfrac65}\,T

\sqrt{\dfrac56}\,T

\sqrt{\dfrac67}\,T

\sqrt{\dfrac76}\,T

Answer and worked solution

Official answer: C

Effective gravity in a lift accelerating up: $g+\frac g6=\frac{7g}6$ .

Period. $T'=T\sqrt{\dfrac g{7g/6}}=\sqrt{\dfrac67}\,T$ .

(C)

Topics:Simple and compound pendulum Pseudo forces and non-inertial frames

← Previous question Next question →

JEE Main 2022 · 26 Jun · Shift 1, Physics question 38