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Three particles each having a mass 100 gm are placed on the verticles of an equilateral triangle of side 20 cm. The work done in increasing the side of their triangle to 40 cm is $(G= 6.67 \times 10^{-11}\;Nm^2/kg^2)$

\[(a)\;5.0 \times 10^{-12}\;J \quad (b)\;2.25 \times 10^{-10}\;J \quad (c)\;4.0 \times 10^{-11}\;J \quad (d)\;6.0 \times 10^{-15}\;J \]

1 Answer

Work done= Final gravitational potential -Initial gravitational potential
$\qquad=v_f -v_i =3\bigg[\bigg(\large\frac{-Gmm}{r_f}\bigg)-\bigg(\frac{-Gmm}{r_i}\bigg)\bigg]$
$ m= 0.1 \quad r_f=0.4\;m \quad r_i=0.2\; m$
Substituting these values we get
$W= 3\bigg[-Gmm \bigg(\large\frac{1}{.4}-\frac{1}{.2}\bigg)\bigg]$
$W= 5.0 \times 10^{-12}J$
Hence a is the correct answer. 
answered Aug 26, 2013 by meena.p
edited Jul 3, 2014 by lmohan717

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