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Byju's Answer

Standard XII

Chemistry

Nernst Equation

Two spherical...

Question

Two spherical conductors each of capacity C are charged to potential V and these are then connected by means of fine wire. The loss of energy is:

zero

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\frac{1}{2} {C V}^{2}

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{C V}^{2}

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2 {C V}^{2}

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Solution

The correct option is C ${C V}^{2}$
Using the formula.
Loss of energy , $Δ U = \frac{C_{1} C_{2} (V_{1} + V_{2})^{2}}{2 (C_{1} + C_{2}})$

here, $C_{1}$ and $C_{2}$ is the capacitance of capacitors that is connected by wire and $V_{1}$ and $V_{2}$ is their respective potential.

Given, $C_{1} = C_{2} = C$ and $V_{1} = V_{2} = V$

Hence, $Δ U = \frac{C \times C (V + V)^{2}}{2 (C + C)} = C V^{2}$

So, option $(C)$ is correct.

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Two spherical conductors each of capacity C are charged to potential V and these are then connected by means of fine wire. The loss of energy is:

The correct option is C CV2Using the formula.Loss of energy , ΔU=C1C2(V1+V2)22(C1+C2)here, C1 and C2 is the capacitance of capacitors that is connected by wire and V1 and V2 is their respective potential.Given, C1=C2=C and V1=V2=V Hence, ΔU=C×C(V+V)22(C+C)=CV2So, option (C) is correct.