JEE Mains · Physics · STD 12 - 3. current electricity
In an experiment to find \(emf\) of a cell using potentiometer, the length of null point for a cell of emf \(1.5\,V\) is found to be \(60\,cm\). If this cell is replaced by another cell of \(emf\; E\). the length-of null point increases by \(40\,cm\). The value of \(E\) is \(\frac{x}{10} V\). The value of \(x\) is \(............\)
- A \(24\)
- B \(25\)
- C \(23\)
- D \(22\)
Answer & Solution
Correct Answer
(B) \(25\)
Step-by-step Solution
Detailed explanation
\(\frac{E_1}{E_2}=\frac{l_1}{l_2}\) \(\frac{1.5}{E_2}=\frac{60}{60+40}=\frac{6}{10}=\frac{3}{5}\) \(E_2=\frac{5}{2}=\frac{x}{10}\) \(x=25\)
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