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13.Oscillations
medium
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 $g =\pi^{2}\,m / s ^{-2}$ )
A
$\frac{2}{9}\,m$
B
$\frac{4}{9}\,m$
C
$\frac{8}{9}\,m$
D
$\frac{1}{9}\,m$
(JEE MAIN-2022)
Solution
$T =2 \pi \sqrt{\frac{ L }{ g }}, \quad g ^{\prime}=\frac{ GM }{9 R ^{2}}=\frac{ g }{9}=\frac{\pi^{2}}{9}$
$2=2 \pi \sqrt{\frac{ L }{\pi^{2}} \times 9}$
$1=\pi \sqrt{ L } \times \frac{3}{\pi} \Rightarrow L =\frac{1}{9}\,m$
Standard 11
Physics
