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  • The length of a seconds pendulum at a height from earth surface will be : (Given <img alt="\mathrm{\mathrm{R}=}" src

The length of a seconds pendulum at a height \mathrm{h=2 R} from earth surface will be :
(Given \mathrm{\mathrm{R}=} Radius of earth and acceleration due to gravity at the surface of earth, \mathrm{g}=\pi^{2} \mathrm{~ms}^{-2})

Option: 1

\frac{2}{9}\mathrm{m}


Option: 2

\frac{4}{9}\mathrm{m}


Option: 3

\frac{8}{9}\mathrm{m}


Option: 4

\frac{1}{9}\mathrm{m}


Answers (1)

best_answer

For Seconds pendulam ,

\mathrm{T=2 s=2 \pi \sqrt{\frac{l}{9}} \rightarrow(1) }

\mathrm{g_n=\frac{g R^2}{(R+h)^2}=\frac{9}{9} }

\mathrm{T^{\prime}=2 \pi \sqrt{\frac{l^{\prime}}{g_h}}=2 s \rightarrow(2) }

From eq (1) & (2)    

\mathrm{2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{l'}{g/9}}}

\mathrm{l^{\prime}=\frac{l}{9} }

\mathrm{(\because l=1 m) }

\mathrm{\therefore l^{\prime}=\frac{1}{9} m}

Hence (4) is correct option

 

Posted by

shivangi.bhatnagar

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