The work done is bringing three particles each of mass 10 g from large distances to the vertices of an equilateral triangle of side 10 cm

The correct option is 2 x 10⁻¹³ J.

To find the work done in assembling the system of three particles at the vertices of an equilateral triangle, we calculate the magnitude of the gravitational potential energy of the system.

1. Identify the given values:
- Mass of each particle, $m=10\text{g}=10\times {10}^{-3}\text{kg}={10}^{-2}\text{kg}$
- Side of the equilateral triangle (distance between each pair of particles), $r=10\text{cm}=0.1\text{m}={10}^{-1}\text{m}$
- Universal gravitational constant, $G=6.67\times {10}^{-11}{\text{N m}}^2{\text{/kg}}^2$

2. Formula:
The work done by an external agent to bring the particles from infinity to form the configuration is equal to the change in the potential energy of the system. For a three-particle system, there are three distinct interacting pairs. The potential energy $U$ of the system is given by:

$U=-\left(\frac{G{m}^2}{r}+\frac{G{m}^2}{r}+\frac{G{m}^2}{r}\right)=-\frac{3G{m}^2}{r}$

$W=\frac{3G{m}^2}{r}$

3. Calculation:
Substitute the given values into the formula:

$W=\frac{3\times 6.67\times {10}^{-11}\times {\left({10}^{-2}\right)}^2}{{10}^{-1}}$

$W=\frac{20.01\times {10}^{-11}\times {10}^{-4}}{{10}^{-1}}$

$W=\frac{20.01\times {10}^{-15}}{{10}^{-1}}$

$W=20.01\times {10}^{-14}\text{J}$

$W=2.001\times {10}^{-13}\text{J}\approx 2\times {10}^{-13}\text{J}$