Q111 / 125 · BITSAT 2012 · Physics
Three voltmeters $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ having resistances $\mathrm{R}, 1.5 \mathrm{R}$ and $3 \mathrm{R},$ respectively, are connected as shown. When some potential difference is applied between $\mathrm{X}$ and $\mathrm{Y},$ the voltmeter readings are $\mathrm{V}_{\mathrm{A}}, \mathrm{V}_{\mathrm{B}}$ and $\mathrm{V}_{\mathrm{C}}$ respectively. Then -


- A$V_{A} \neq V_{B}=V_{C}$
- B$V_{\mathrm{A}}=V_{\mathrm{B}} \neq V_{\mathrm{C}}$
- C$\mathrm{V}_{\mathrm{A}} \neq \mathrm{V}_{\mathrm{B}} \neq \mathrm{V}_{\mathrm{C}}$
- Correct answerD$V_{A}=V_{B}=V_{C}$
Solution
$V_{A}=I R$
$V_{B}=\left(\frac{2 I}{3}\right) 1.5 R=I R \quad V_{C}=\left(\frac{1}{3}\right) 3 R=I R$
$\therefore \quad V_{A}=V_{B}=V_{C}$
Hence, the correct answer is option: 'D'.