(B) The energy stored in a fully charged capacitor is given by $U = \frac{1}{2}CV^2$.When the capacitor discharges through the coil,this electrical en

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(B) The energy stored in a fully charged capacitor is given by $U = \frac{1}{2}CV^2$.When the capacitor discharges through the coil,this electrical energy is converted into heat energy.The heat energy absorbed by the block is given by $Q = ms\Delta T$.By the principle of conservation of energy,the electrical energy stored in the capacitor equals the heat energy gained by the block:$\frac{1}{2}CV^2 = ms\Delta T$Solving for $V$:$V^2 = \frac{2ms\Delta T}{C}$$V = \sqrt{\frac{2ms\Delta T}{C}}$Therefore,the correct option is $B$.

Class 12 Physics Electric Potential and Capacitance Mix Examples - Electric Potential and Capacitance

Medium

$A$ fully charged capacitor has a capacitance $C$. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity $s$ and mass $m$. If the temperature of the block is raised by $\Delta T$,the potential difference $V$ across the capacitor is

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A
$\frac{ms\Delta T}{C}$
B
$\sqrt{\frac{2ms\Delta T}{C}}$
C
$\sqrt{\frac{2mC\Delta T}{s}}$
D
$\frac{mC\Delta T}{s}$

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Electric Potential and Capacitance

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Mix Examples - Electric Potential and Capacitance

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In the circuit shown in the figure,$K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now,the key $K_1$ is closed and at steady state,the charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is

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Four capacitors are connected in a circuit as shown in the figure. The effective capacitance in $\mu F$ between points $A$ and $B$ will be

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In the circuit shown in the figure,$K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now,the key $K_1$ is closed and at steady state,the charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is

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