Three particles each of mass 100 gm are brought from a very large distance to the vertices of an equilateral triangle whose side is 20 cm in length. The work done will be

The correct option/answer is: -1.00 x 10⁻¹¹ Joule.

To find the work done in bringing the three particles from a very large distance (infinity) to the vertices of an equilateral triangle, we calculate the change in the potential energy of the system.

Initially, the particles are at an infinite distance from each other, which means the initial potential energy of the system ($U_i$) is zero:
$U_i=0$

Let the three particles have masses $m_1=m_2=m_3=m=100\text{gm}=0.1\text{kg}$.
The side length of the equilateral triangle is $r=20\text{cm}=0.2\text{m}$.
The universal gravitational constant is $G\approx 6.67\times {10}^{-11}{\text{N m}}^2{\text{/kg}}^2$.

The final gravitational potential energy ($U_f$) of the system consisting of three interacting pairs of masses is:
$U_f=-3\times \frac{G{m}^2}{r}$

Substituting the given values into the formula:
$U_f=-3\times \frac{(6.67\times {10}^{-11})\times {\left(0.1\right)}^2}{0.2}$

Let's simplify the expression:
$U_f=-3\times \frac{6.67\times {10}^{-11}\times 0.01}{0.2}$
$U_f=-3\times 6.67\times {10}^{-11}\times 0.05$
$U_f=-0.15\times 6.67\times {10}^{-11}$
$U_f\approx -1.00\times {10}^{-11}\text{Joule}$

The work done ($W$) by the gravitational force in assembling the system is equal to the negative change in potential energy, or the work done by an external agent against gravity is equal to the change in potential energy:
$W=U_f-U_i=-1.00\times {10}^{-11}\text{Joule}$