Question

Three particles each having a mass of 100 g are placed on the vertices of an equilateral triangle of side 20 cm. The work done in increasing the side of this triangle to 40 cm is

$(G=6.67\times {10}^{-11}\frac{N-m^2}{k{g}^2})$

• A. 5.0×10^-12J
• B. 2.25×10^-10J
• C. 4.0×10^-11J
• D. 6.0×10^-15J

Answer 1

The correct option is A

5.0×10^-12J

W = $U_f-U_i$

=$=3[-\left\{\frac{G\left(m\right)\left(m\right)}{r_f}\right\}-\left\{\frac{-G\left(m\right)\left(m\right)}{r_i}\right\}]$

Here, G = $6.67\times {10}^{-11}\frac{N-m^2}{k{g}^2}$.

M = 0.1 kg, $r_f=0.4m$ and $r_i=0.2m$

Substituting the values, we get

W = $5.0\times {10}^{-12}J$