- A$5$
- ✓$6$
- C$4$
- D$3$
$=\frac{\in_0 A}{\left(\frac{d}{3}+\frac{d}{2}\right)}=\frac{6 \in_0 A}{5 d}$
$=\frac{6}{5} \times 5 \mu F =6 \mu F$
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$[1]$ The magnetic field strength may have been increased while the particle was travelling in air
$[2]$ The particle lost energy by ionising the air
$[3]$ The particle lost charge by ionising the air
Which of the following statement($s$) is(are) correct?
$(A)$ When a voltage source of $6 V$ is connected across $A$ and $B$ in both circuits, $P_1$
$(B)$ When a constant current source of $2 Amp$ is connected across $A$ and $B$ in both circuits, $P_1>P_2$.
$(C)$ When a voltage source of $6 V$ is connected across $A$ and $B$ in Circuit-$1$, $Q_1>P_1$.
$(D)$ When a constant current source of $2 Amp$ is connected across $A$ and $B$ in both circuits, $Q_2$
