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Standard XII

Physics

Flux as the Flow

Figure shows ...

Question

Figure shows a solid metal sphere of radius a surrounded by a concentric thin metal shell of radius $2 a$ . Initially both are having charges $Q$ each. When the two are connected by a conducting wire ,then amount of heat produced in this process will be:

\frac{K Q^{2}}{2 a}

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\frac{K Q^{2}}{4 a}

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\frac{K Q^{2}}{6 a}

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\frac{K Q^{2}}{8 a}

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Solution

The correct option is B $\frac{K Q^{2}}{4 a}$
Heat produced $= U_{i} - U_{r}$
$= (U_{1} + U_{2}) - U_{2} = U_{1}$
$= \frac{q^{2}}{2 c} = \frac{q^{2}}{2 {4 π ε_{0} {\frac{a \times 2 a}{2 a - a}}}} = \frac{k q^{2}}{4 a} = \frac{k Q^{2}}{4 a}$

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Figure shows a solid metal sphere of radius a surrounded by a concentric thin metal shell of radius 2a. Initially both are having charges Q each. When the two are connected by a conducting wire ,then amount of heat produced in this process will be:

The correct option is B KQ24aHeat produced =Ui−Ur=(U1+U2)−U2=U1=q22c=q22{4πε0{a×2a2a−a}}=kq24a=kQ24a

Figure shows a solid metal sphere of radius a surrounded by a concentric thin metal shell of radius $2 a$ . Initially both are having charges $Q$ each. When the two are connected by a conducting wire ,then amount of heat produced in this process will be:

The correct option is B $\frac{K Q^{2}}{4 a}$
Heat produced $= U_{i} - U_{r}$
$= (U_{1} + U_{2}) - U_{2} = U_{1}$
$= \frac{q^{2}}{2 c} = \frac{q^{2}}{2 {4 π ε_{0} {\frac{a \times 2 a}{2 a - a}}}} = \frac{k q^{2}}{4 a} = \frac{k Q^{2}}{4 a}$