MHT CET · Physics · Electrostatics
A metallic sphere ' A ' isolated from ground is charged to \(+50 \mu \mathrm{C}\). This sphere is brought in contact with other isolated metallic sphere ' \(B\) ' of half the radius of sphere ' A '. Then the charge on the two isolated spheres \(\mathrm{A} \& \mathrm{~B}\) are in the ratio
- A \(1: 2\)
- B \(2: 1\)
- C \(4: 1\)
- D \(1: 1\)
Answer & Solution
Correct Answer
(B) \(2: 1\)
Step-by-step Solution
Detailed explanation
Charge will flow from sphere A to B till the potential becomes same
\(\begin{aligned}
& \therefore \quad \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \text {. } \\
& \therefore \quad \frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{A}}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{B}}} \\
& \therefore \quad \frac{q_A}{q_B}=\frac{r_A}{r_B} \\
& \therefore \quad \frac{q_A}{q_B}=\frac{2 r_B}{r_B} \\
& \ldots .\left(\because r_B=\frac{r_A}{2}\right) \\
& \therefore \quad \frac{q_A}{q_B}=\frac{2}{1}
\end{aligned}\)
\(\begin{aligned}
& \therefore \quad \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \text {. } \\
& \therefore \quad \frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{A}}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{B}}} \\
& \therefore \quad \frac{q_A}{q_B}=\frac{r_A}{r_B} \\
& \therefore \quad \frac{q_A}{q_B}=\frac{2 r_B}{r_B} \\
& \ldots .\left(\because r_B=\frac{r_A}{2}\right) \\
& \therefore \quad \frac{q_A}{q_B}=\frac{2}{1}
\end{aligned}\)
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