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  • Two isolated metallic solid spheres of radii R and 2R are charged such that both have same charge density  Th

Two isolated metallic solid spheres of radii R and 2R are charged such that both have same charge density \sigma The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is \sigma'.The ratio \frac{\sigma^{\prime}}{\sigma} is :
 

Option: 1

\frac{4}{3}


Option: 2

\frac{5}{3}


Option: 3

\frac{5}{6}


Option: 4

\frac{9}{4}


Answers (1)

best_answer

\begin{aligned} & Q_2=\sigma 4 \pi(2 \mathrm{R})^2 \\ & Q_2=\sigma 16 \pi R^2 \end{aligned}

Now,

Charge will flow until voltage of both sphere become equal so

\begin{aligned} & c=4 \pi \varepsilon_0 R \\ & v_1^{\prime}=v_2^1 \\ & \frac{Q_1}{c_1}=\frac{Q_2}{c_2} \Rightarrow \frac{Q_1^{\prime}}{4 \pi \varepsilon_0 R}=\frac{Q_2^{\prime}}{4 \pi \varepsilon_0(2 R)} \\ \Rightarrow \quad & 2Q_1^{\prime}=Q_2^{\prime} \\ & Q_1+Q_2=Q_1^{\prime}+Q_2^{\prime} \\ & \sigma 20 \pi R^2=Q_2^{\prime}+\frac{Q_2^{\prime}}{2}=\frac{3}{2} Q_2^{\prime} \Rightarrow Q_2^{\prime}=\frac{\sigma 40 \pi R^2}{3} \ldots(2) \end{aligned}

\begin{aligned} & \mathrm{Q}_2{ }^{\prime}=\frac{\sigma 40 \pi \mathrm{R}^2}{3} \\ & \text { Now } \sigma^{\prime} 4 \pi(2 R)^2=\frac{\sigma 40 \pi R^2}{3} \\ & \sigma^{\prime} 16 \pi R^2=\frac{\sigma 40 \pi R^2}{3} \\ & \frac{\sigma^{\prime}}{\sigma}=\frac{40}{3} \times \frac{1}{16}=\frac{5}{6} \\ & \end{aligned}

Posted by

vishal kumar

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