Question

Four particle each of mass $1kg$ are at the four corners of a square of side $1m$. The work one to move one of the particles to infinity

• A. $\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)G$
• B. $2\sqrt{2}G$
• C. $(2\sqrt{2}+1)G$
• D. $3G$

Answer 1

The correct option is A $\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)G$

mass of each particle, $m=1kg$

side of square, $a=1m$

diagonal of square, $d=\sqrt{2}m$

Initial potential energy of system, $U_1=\frac{4G\times m\times m}{a}+\frac{2G\times m\times m}{d}$

$U_1=\frac{4G\times 1\times 1}{1}+\frac{2G\times 1\times 1}{\sqrt{2}}$

$U_1=G\left(\frac{4\sqrt{2}+1}{\sqrt{2}}\right)$

When one mass is removed,

Potential Energy,

$U_2=\frac{2G\times 1\times 1}{1}+\frac{G\times 1\times 1}{\sqrt{2}}$

$U_2=G\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)$

Work done,

$W=U_1-U_2$

$W=G\left(\frac{4\sqrt{2}+1}{\sqrt{2}}\right)-G\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)$

$W=G\left(\frac{4\sqrt{2}-2\sqrt{2}-1}{\sqrt{2}}\right)W$

$=G\left(\frac{2\sqrt{2}+1}{\sqrt{2}}\right)$