MHT CET · Physics · Current Electricity
A battery of 6 V is connected to the ends of uniform wire 3 m long and of resistance \(100 \Omega\). The difference of potential between two points 50 cm apart on the wire is
- A 1 V
- B 2 V
- C 1.5 V
- D 3 V
Answer & Solution
Correct Answer
(A) 1 V
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{R}=\frac{\rho l}{\mathrm{~A}} \\
& \frac{100}{3}=\frac{\rho}{A}
\end{aligned}\)
\(\because \quad\) Total resistance for 50 cm wire is
\(\begin{aligned}
& \mathrm{R}^{\prime}=\frac{\rho}{\mathrm{A}} l=\frac{100}{3} \times\left(50 \times 10^{-2}\right)=\frac{50}{3} \Omega ...(i)\\
& \mathrm{I} \doteq \frac{\mathrm{~V}}{\mathrm{R}}=\frac{6}{100} \mathrm{~A}...(ii)
\end{aligned}\)
\(\therefore \quad\) The potential difference between two points 50 cm apart is
\(V=I R^{\prime}=\frac{6}{100} \times \frac{50}{3}=1 \mathrm{~V}\)
...[From (i) and (ii)]
& \mathrm{R}=\frac{\rho l}{\mathrm{~A}} \\
& \frac{100}{3}=\frac{\rho}{A}
\end{aligned}\)
\(\because \quad\) Total resistance for 50 cm wire is
\(\begin{aligned}
& \mathrm{R}^{\prime}=\frac{\rho}{\mathrm{A}} l=\frac{100}{3} \times\left(50 \times 10^{-2}\right)=\frac{50}{3} \Omega ...(i)\\
& \mathrm{I} \doteq \frac{\mathrm{~V}}{\mathrm{R}}=\frac{6}{100} \mathrm{~A}...(ii)
\end{aligned}\)
\(\therefore \quad\) The potential difference between two points 50 cm apart is
\(V=I R^{\prime}=\frac{6}{100} \times \frac{50}{3}=1 \mathrm{~V}\)
...[From (i) and (ii)]
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