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7.Gravitation
normal
Consider two configurations of a system of three particles of masses $m, 2m$ and $3m.$ The work done by external agent in changing the configuration of the system from figure $(i)$ to figure $(ii)$ is
A
zero
B
$ - \frac{{6G{m^2}}}{a}\left( {1 + \frac{1}{{\sqrt 2 }}} \right)$
C
$ - \frac{{6G{m^2}}}{a}\left( {1 - \frac{1}{{\sqrt 2 }}} \right)$
D
$ - \frac{{6G{m^2}}}{a}\left( {2 - \frac{1}{{\sqrt 2 }}} \right)$
Solution
$U_{1}=0-\frac{G m 2 m}{a}-\frac{G m 3 m}{a}-\frac{G 3 m 2 m}{a}$
$=-\frac{5 G m^{2}}{a}-\frac{3 \sqrt{2} m^{2}}{a}$
$=-\frac{G m^{2}}{a}(5+3 \sqrt{2})$
$U_{2}=0-\frac{G m 3 m}{a}-\frac{G m 2 m}{a}-\frac{G 3 m 2 m}{a}$
$=-\frac{11 G m^{2}}{a}$
$\Delta W=\Delta U=\frac{G m^{2}}{a}(11-5-3 \sqrt{2})$
$=\frac{6 G m^{2}}{a}\left(1-\frac{1}{\sqrt{2}}\right)$
Standard 11
Physics
