MHT CET · Physics · Current Electricity
To determine the internal resistance of a cell by using a potentiometer, the null point is at \(1 \mathrm{~m}\), when shunted by \(3 \Omega\) resistance and at a length \(1.5 \mathrm{~m}\), when cell is shunted by \(6 \Omega\) resistance. The internal resistance of the cell is
- A \(1 \Omega\)
- B \(4 \Omega\)
- C \(2 \Omega\)
- D \(6 \Omega\)
Answer & Solution
Correct Answer
(D) \(6 \Omega\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{r}=\mathrm{R}\left(\frac{\ell_1}{\ell_2}-1\right)=\mathrm{R}^{\prime}\left(\frac{\ell_1}{\ell_2^{\prime}}-1\right) \\
& \ell_2=1 \mathrm{~m}, \ell_2^{\prime}=1.5 \mathrm{~m} \mathrm{R}=3 \Omega, \mathrm{R}^{\prime}=6 \Omega \\
& \therefore 3\left(\frac{\ell_1}{1}-1\right)=6\left(\frac{\ell_1}{1.5}-1\right)
\end{aligned}
\)
Solving we get \(\ell_1=3 \mathrm{~m}\)
\(
\therefore \mathrm{r}=3\left(\frac{3}{1}-1\right)=3 \times 2=6 \Omega
\)
\begin{aligned}
& \mathrm{r}=\mathrm{R}\left(\frac{\ell_1}{\ell_2}-1\right)=\mathrm{R}^{\prime}\left(\frac{\ell_1}{\ell_2^{\prime}}-1\right) \\
& \ell_2=1 \mathrm{~m}, \ell_2^{\prime}=1.5 \mathrm{~m} \mathrm{R}=3 \Omega, \mathrm{R}^{\prime}=6 \Omega \\
& \therefore 3\left(\frac{\ell_1}{1}-1\right)=6\left(\frac{\ell_1}{1.5}-1\right)
\end{aligned}
\)
Solving we get \(\ell_1=3 \mathrm{~m}\)
\(
\therefore \mathrm{r}=3\left(\frac{3}{1}-1\right)=3 \times 2=6 \Omega
\)
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