JEE Main 2022 · 25 Jul · Shift 2 · Physics Q43

PhysicsSingle correct+4 / -1Question 43 of 90

The length of a seconds pendulum at a height $h=2R$ from earth surface will be :

(Given R = Radius of earth and acceleration due to gravity at the surface of earth, $g=\pi^2$ ms $^{-2}$ )

\dfrac29

\dfrac49

\dfrac89

\dfrac19

Answer and worked solution

Official answer: D

At h = 2R $g'=\frac g9=\frac{\pi^2}9$ . Seconds pendulum ( $T=2$ s): $L=\dfrac{g'T^2}{4\pi^2}=\dfrac19$ m.

(D)

Topics:Variation of g Simple and compound pendulum

JEE Main 2022 · 25 Jul · Shift 2, Physics question 43