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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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A simple pendulum has a time period $T_1$ When on earth's surface, and $T_2$ When taken to a height R above the earth's surface, where R is radius of earth. Then $\large\frac{T_2}{T_1}$ is

\[(a)\;1 \quad (b)\;\sqrt 2 \quad (c)\; 4 \quad (d)\;2\]
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1 Answer

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$\large\frac{g_2}{g_1}=\frac{{R_1}^2}{{R_2}^2}$
$R_1=R$
$R_2=2R$
$\large\frac{g_2}{g_1}=\frac{R^2}{4R^2}$
$\qquad=\large\frac{1}{4}$
$T= 2 \pi \sqrt {\large\frac{l}{g}}$
$\large\frac{T_2}{T_1}=\sqrt {\large\frac{g_1}{g_2}}$
$\qquad=\sqrt 4=2$
Hence  d is the correct answer.

 

answered Aug 10, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

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