Q145 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Physics

Two metal spheres have their radii in the ratio of $4: 7$. They are put in contact and a charge $8.8 \times 10^{-7} \mathrm{C}$ is given to the system. Then they are separated so that each can exert no influence on the other. The potential due to the smaller sphere at $60 \mathrm{~m}$ from it in Volt is

  • A
    85
  • B
    76
  • Correct answer
    C
    48
  • D
    66

Solution

We have $\mathrm{q}=\mathrm{q}_1+\mathrm{q}_2$ and, $\frac{\mathrm{Kq}_1}{\mathrm{R}_1}=\frac{\mathrm{K}_{\mathrm{q}_2}}{\mathrm{R}_2} \Rightarrow \frac{\mathrm{q}_1}{4 \mathrm{R}}=\frac{\mathrm{q}_2}{7 \mathrm{R}} \Rightarrow \frac{\mathrm{q}_1}{\mathrm{q}_2}=\frac{4}{7}$ $\Rightarrow \mathrm{q}_2=\frac{7}{4} \mathrm{q}_{\mathrm{l}}$ Now, $\mathrm{q}=\mathrm{q}_{\mathrm{l}}+\frac{7}{4} \mathrm{q}_{\mathrm{l}} \Rightarrow \mathrm{q}=\frac{11}{4} \mathrm{q}_{\mathrm{l}} \Rightarrow \mathrm{q}_{\mathrm{l}}=\frac{4 \mathrm{q}}{11}$ So, $\mathrm{q}_{\mathrm{l}}=\frac{4}{11} \times 8.8 \times 10^{-7}=3.2 \times 10^{-7} \mathrm{C}$ Then, $V(60 \mathrm{~m})=\frac{9 \times 10^9 \times 3.2 \times 10^{-7}}{60}$ $=48$ Volts.

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

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