Q111 / 150 · MHT CET 2026 (13 April Shift 2) · Physics

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

  • A
    $20$
  • Correct answer
    B
    $25$
  • C
    $28$
  • D
    $30$

Solution

According to the principle of potentiometer, the emf of a cell is directly proportional to its balancing length.

$E \propto l \Rightarrow \dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}$

Given $E_1 = 1.5$ V and $l_1 = 60$ cm.

When the cell is replaced by another cell of emf $E$, the balancing length increases by $40$ cm.

$l_2 = 60 + 40 = 100$ cm.

Substituting the values:

$\dfrac{1.5}{E} = \dfrac{60}{100}$

$E = \dfrac{1.5 \times 100}{60} = 2.5$ V.

Given $E = \dfrac{x}{10}$ V.

$\dfrac{x}{10} = 2.5 \Rightarrow x = 25$

Answer: $25$

Hence, the correct answer is option: 'B'.

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