${F_1} = \frac{{{\mu _0}}}{{4\pi }}\frac{{2IiL}}{{(L/2)}} = \frac{{{\mu _0}Ii}}{\pi }$
acting towards $X Y$ in the plane of loop. Force on arm $C D$ due to current in conductor $X Y$ is
${F_2} = \frac{{{\mu _0}}}{{4\pi }}\frac{{2IiL}}{{3(L/2)}} = \frac{{{\mu _0}Ii}}{{3\pi }}$
acting away from $X Y$ in the plane of loop.
$\therefore$ Net force on the loop $=F_{1}-F_{2}$
$=\frac{\mu_{0} I i}{\pi}\left[1-\frac{1}{3}\right]=\frac{2}{3} \frac{\mu_{0} I i}{\pi}$
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Match the quantities mentioned in $List-I$ with their values in $List-II$ and choose the correct option.
| $List-I$ | $List-II$ |
| ($P$) The capacitance between $S_1$ and $S_4$, with $S_2$ and $S_3$ not connected, is | ($1$) $3 C_0$ |
| ($Q$) The capacitance between $S_1$ and $S_4$, with $S_2$ shorted to $S_3$, is | ($2$) $C_0 / 2$ |
| ($R$) The capacitance between $S_1$ and $S_3$, with $S_2$ shorted to $S_4$, is | ($3$) $C_0 / 3$ |
| ($S$) The capacitance between $S_1$ and $S_2$, with $S_3$ shorted to $S_1$, and $S_2$ shorted to $S_4$, is | ($4$) $2 C_0 / 3$ |
| ($5$) $2 C_0$ |
$(A)$ $R=\frac{h^2+r^2}{2 h}$
$(B)$ $R=\frac{3 r^2}{2 h}$
$(C)$ Apparent depth of the bottom of the beaker is close to $\frac{3 H }{2}\left(1+\frac{\omega^2 H }{2 g }\right)^{-1}$
$(D)$ Apparent depth of the bottom of the beaker is close to $\frac{3 H }{4}\left(1+\frac{\omega^2 H}{4 g }\right)^{-1}$