
- ✓

- B

- C

- D







at $t =5 \mu s$
$V _{0}( t )=5\left(1- e ^{-\frac{5 \times 10^{-6}}{10^{3} \times 10 \times 10^{-9}}}\right)$
$=5\left(1- e ^{-0.5}\right)=2 V$
Now $V_{\text {in }}=0$ means discharging
$V _{0}( t )=2 e ^{-\frac{ t }{ RC }}=2 e ^{-0.5}$
$=1.21 V$
Now for next $5 \mu s$
$V _{0}( t )=5-3.79 e ^{-\frac{t}{ RC }}$
after $5 \mu s$ again
$V _{0}( t )=2.79 Volt \approx 3 V$
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Statement $1 :$ Self inductance of a long solenoid of length $L,$ total number of turns $N$ and radius $r$ is less than $\frac{{\pi {\mu _0}{N^2}{r^2}}}{L}$
Statement $2:$ The magnetic induction in the solenoid in Statement $1$ carrying current $I$ is $\frac{{{\mu _0}NI}}{L}$ in the middle of the solenoid but becomes less as we move towards its ends.