BITSAT 2015 Paper

Initial potential energy of the system = $\frac{1}{4\pi\epsilon_0} \left[ \frac{q^2}{a} + \frac{q^2}{a} + \frac{q^2}{a} \right]$ = $\frac{1}{4\pi\epsilon_0} \left( \frac{3q^2}{a} \right)$ = 9 × $10^9$ $\left( \frac{3 \times (0.1)^2}{1} \right)$ = 27 × $10^7$ J Let charge at A is moved to mid-point O, Then final potential energy of the system $U_f$ = $\frac{1}{4\pi\epsilon_0} \left[ \frac{2q^2}{\frac{a}{2}} + \frac{q^2}{a} \right]$ = 5 × $\frac{1}{4\pi\epsilon_0} \left( \frac{q^2}{a^2} \right)$ = 45 × $10^7$ J Work done = $U_f - U_i$ = 18 × $10^7$ J Also, energy supplied per sec = 1000 J (given) Time required to move one of the mid-point of the line joining the other two t = $\frac{18 \times 10^7}{1000}$ = 18 × $10^4$ s = 50 h