The length of a seconds pendulum at a height h =2 R from earth surface will be: (Given: R = Radius of earth and acceleration due to gravity at the surface of earth .g =π2 ms -2)

Question

The length of a seconds pendulum at a height h =2 R from earth surface will be: (Given: R = Radius of earth and acceleration due to gravity at the surface of earth .g =π2 ms -2) (A) (2/9) m (B) (4/9) m (C) (8/9) m (D) (1/9) m

Tardigrade Admin · Accepted Answer

T =2 π √( L / g ), g prime=( GM /9 R 2)=( g /9)=(π2/9) 2=2 π √( L /π2) × 9 ⇒ 1=π √ L × (3/π) ⇒ L =(1/9) m

Q. The length of a seconds pendulum at a height $h = 2 R$ from earth surface will be :
(Given : $R =$ Radius of earth and acceleration due to gravity at the surface of earth $g = π^{2} m s^{- 2})$

$\frac{2}{9} m$

$\frac{4}{9} m$

$\frac{8}{9} m$

$\frac{1}{9} m$

$T = 2 π \frac{L}{g}, g^{'} = \frac{GM}{9 R ^{2}} = \frac{g}{9} = \frac{π ^{2}}{9}$
$2 = 2 π \frac{L}{π ^{2}} \times 9$
$\Rightarrow 1 = π L \times \frac{3}{π} \Rightarrow L = \frac{1}{9} m$

Q. 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=π2ms−2)

92​m

94​m

98​m

91​m