JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2026
- Medium
- MCQ
To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm . The least count of scale is 1 cm . The percentage error in the ratio of EMFs is ___________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 1.65
Step-by-step Solution
Detailed explanation
\(\epsilon=\lambda \ell\)
Potential gradient
\(\epsilon_1=\lambda \ell_1\)
\(\epsilon_2=\lambda \ell_2\)
\(y =\frac{\epsilon_1}{\epsilon_2}=\frac{\ell_1}{\ell_2}\)
\(\frac{\Delta y }{ y }=\frac{\Delta \ell_1}{\ell_1}+\frac{\Delta \ell_2}{\ell_2}\)
\(\frac{\Delta y}{y}=\frac{1}{200}+\frac{1}{150}\)
\(\frac{\Delta y}{y} \times 100\left(\frac{1}{200}+\frac{1}{150}\right) \times 100\)
\(=\left(\frac{3+4}{600}\right) \times 100 \Rightarrow \frac{7}{6}=1.16 \%\)
Potential gradient
\(\epsilon_1=\lambda \ell_1\)
\(\epsilon_2=\lambda \ell_2\)
\(y =\frac{\epsilon_1}{\epsilon_2}=\frac{\ell_1}{\ell_2}\)
\(\frac{\Delta y }{ y }=\frac{\Delta \ell_1}{\ell_1}+\frac{\Delta \ell_2}{\ell_2}\)
\(\frac{\Delta y}{y}=\frac{1}{200}+\frac{1}{150}\)
\(\frac{\Delta y}{y} \times 100\left(\frac{1}{200}+\frac{1}{150}\right) \times 100\)
\(=\left(\frac{3+4}{600}\right) \times 100 \Rightarrow \frac{7}{6}=1.16 \%\)
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