
When a voltmeter of resistance $150\,\Omega$ is connected across $A$ and $B$, the resistance across $A$ and $B$ will be
$R _{ ab }=\frac{150 \times 100}{150+100}=60\,\Omega$
Now the equivalent resistance of the circuit will be $R_{a c}=R_{a b}+R_{b c}=60+100=$ $160\,\Omega$
The current in the circuit will be $I=50 / 160=0.31\,A$
Thus, Potential drop across B and C will be $V _{ bc }= IR _{ bc }=(0.31)(100)=31\,V$

$1.$ When $d \approx$ a but wires are not touching the loop, it is found that the net magnetic filed on the axis of the loop is zero at a height $h$ above the loop. In that case
$(A)$ current in wire $1$ and wire $2$ is the direction $P Q$ and $R S$, respectively and $h \approx a$
$(B)$ current in wire $1$ and wire $2$ is the direction $PQ$ and $SR$, respectively and $h \approx a$
$(C)$ current in wire $1$ and wire $2$ is the direction $PQ$ and $SR$, respectively and $h \approx 1.2 a$
$(D)$ current in wire $1$ and wire $2$ is the direction $PQ$ and $RS$, resepectively and $h \approx 1.2 a$
$2.$ Consider $d \gg a$, and the loop is rotated about its diameter parallel to the wires by $30^{\circ}$ from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)
$(A)$ $\frac{\mu_0 I^2 a^2}{d}$ $(B)$ $\frac{\mu_0 I^2 a^2}{2 d}$ $(C)$ $\frac{\sqrt{3} \mu_0 I^2 a^2}{d}$ $(D)$ $\frac{\sqrt{3} \mu_0 I^2 a^2}{2 d}$
Give the answer question $1$ and $2.$