Q110 / 125 · BITSAT 2010 · Physics

Two metallic spheres of radii $1 \mathrm{~cm}$ and $3 \mathrm{~cm}$ are given charges of $-1 \times 10^{-2} \mathrm{C}$ and $5 \times 10^{-2} \mathrm{C}$, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is

  • A
    $2 \times 10^{-2} \mathrm{C}$
  • Correct answer
    B
    $3 \times 10^{-2} C$
  • C
    $4 \times 10^{-2} C$
  • D
    $1 \times 10^{-2} C$

Solution

$\quad \frac{k x}{1 \mathrm{~cm}}=\frac{k(Q-x)}{3 \mathrm{~cm}}$ $3 x=Q-x \Rightarrow 4 x=Q$ $x=\frac{Q}{4}=\frac{4 \times 10^{-2}}{4} C=1 \times 10^{-2}$ $Q^{\prime}=\mathrm{Q}-x=3 \times 10^{-2} \mathrm{C}$

Hence, the correct answer is option: 'B'.

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