Question:
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 \(x/10\) V.
The value of \(x\) is
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 \(x/10\) V.
The value of \(x\) is
The value of \(x\) is
Show Hint
Balancing length is proportional to emf: \(E_1/l_1=E_2/l_2\).
Updated On: Oct 1, 2026
- \(20\)
- \(25\)
- \(28\)
- \(30\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
In a potentiometer, the emf is proportional to the balancing length, since the wire has a uniform potential gradient.
Step 2: New length:
First null point: \(60\) cm for \(1.5\) V. The null point increases by \(40\) cm, so the second length is \(100\) cm.
Step 3: Calculate:
\[ \frac{E}{1.5} = \frac{100}{60} \Rightarrow E = 2.5\ \text{V} \]
The problem writes \(E = \frac{x}{10}\) V, so \(x = 25\).
Final Answer:
The value of \(x\) is \(25\), option (B). \[ \boxed{25} \]
In a potentiometer, the emf is proportional to the balancing length, since the wire has a uniform potential gradient.
Step 2: New length:
First null point: \(60\) cm for \(1.5\) V. The null point increases by \(40\) cm, so the second length is \(100\) cm.
Step 3: Calculate:
\[ \frac{E}{1.5} = \frac{100}{60} \Rightarrow E = 2.5\ \text{V} \]
The problem writes \(E = \frac{x}{10}\) V, so \(x = 25\).
Final Answer:
The value of \(x\) is \(25\), option (B). \[ \boxed{25} \]
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