A tunnel is dug in the Earth across one of it | vedclass

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(C) Let the velocity of mass $m$ be $u$ and mass $2m$ be $u$ when they reach the center of the Earth. By conservation of energy, $u = \sqrt{gR} = \sqr

Vedclass · Accepted Answer

$R/3$

Vedclass · Answer

(C) Let the velocity of mass $m$ be $u$ and mass $2m$ be $u$ when they reach the center of the Earth. By conservation of energy, $u = \sqrt{gR} = \sqrt{\frac{GM}{R}}$.At the center, the masses collide and stick. By conservation of linear momentum: $(2m)u - m(u) = (3m)v$, where $v$ is the velocity of the combined mass $3m$ after collision.$mu = 3mv \Rightarrow v = u/3$.The potential energy of the combined mass at distance $x$ from the center is $U(x) = \frac{1}{2} (3m) \omega^2 x^2$, where $\omega^2 = \frac{g}{R} = \frac{GM}{R^3}$.Using conservation of energy for the combined mass: $\frac{1}{2} (3m) v^2 = \frac{1}{2} (3m) \omega^2 A^2$.$v^2 = \omega^2 A^2 \Rightarrow A = \frac{v}{\omega} = \frac{u/3}{\sqrt{GM/R^3}} = \frac{\sqrt{GM/R}/3}{\sqrt{GM/R^3}} = \frac{R}{3}$.

$\text{LIST-I}$	$\text{LIST-II}$
$A$. Gravitational constant	$I$. $[LT^{-2}]$
$B$. Gravitational potential energy	$II$. $[L^2 T^{-2}]$
$C$. Gravitational potential	$III$. $[ML^2 T^{-2}]$
$D$. Acceleration due to gravity	$IV$. $[M^{-1} L^3 T^{-2}]$

$A$ tunnel is dug in the Earth across one of its diameters. Two masses $m$ and $2m$ are dropped from the ends of the tunnel. The masses collide, stick to each other, and perform $S.H.M.$ The amplitude of the $S.H.M.$ will be:

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