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(C) $1$. In a magnetic field $B$, the magnetic force provides the centripetal force: $QvB = Mv^2 / R$, which gives $v = QBR / M$.$2$. In a cylindrical

Vedclass · Accepted Answer

$QB^2R^2 \ln(3) / M$

Vedclass · Answer

(C) $1$. In a magnetic field $B$, the magnetic force provides the centripetal force: $QvB = Mv^2 / R$, which gives $v = QBR / M$.$2$. In a cylindrical capacitor, the electric field $E$ at a distance $r$ is given by $E = \lambda / (2\pi \epsilon_0 r)$, where $\lambda$ is the linear charge density.$3$. The electric force provides the centripetal force: $QE = Mv^2 / R$. Substituting $E$ and $v$: $Q (\lambda / (2\pi \epsilon_0 R)) = M (QBR/M)^2 / R = Q^2 B^2 R / M$.$4$. This simplifies to $\lambda / (2\pi \epsilon_0) = Q B^2 R^2 / M$.$5$. The potential difference $V$ between electrodes of radii $r_1 = R/2$ and $r_2 = 3R/2$ is $V = \int_{r_1}^{r_2} E dr = (\lambda / 2\pi \epsilon_0) \int_{R/2}^{3R/2} (1/r) dr = (\lambda / 2\pi \epsilon_0) \ln((3R/2) / (R/2)) = (\lambda / 2\pi \epsilon_0) \ln(3)$.$6$. Substituting the value from step $4$: $V = (Q B^2 R^2 / M) \ln(3)$.

$List-I$	$List-II$
$(P)$ 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	$(4)$ $2 C_0 / 3$
	$(5)$ $2 C_0$

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