Q116 / 150 · MHT CET 2026 (19 April Shift 1) · Physics
In a potentiometer circuit when two cells of e.m.f. $1.5$ volt and $1.2$ volt are connected to assist each other, balancing length is $270$ cm. What will be the balancing length in cm when these two cells are connected in opposition?
- A$90$
- B$81$
- C$60$
- Correct answerD$30$
Solution
Let the potential gradient of the potentiometer wire be $k$.
When the cells assist each other, the equivalent EMF is $E_1 + E_2$.
$E_1 + E_2 = k l_1$
$1.5 + 1.2 = k \times 270$
$2.7 = 270 k$
When the cells are connected in opposition, the equivalent EMF is $E_1 - E_2$.
$E_1 - E_2 = k l_2$
$1.5 - 1.2 = k l_2$
$0.3 = k l_2$
Dividing the two equations, we get:
$\dfrac{2.7}{0.3} = \dfrac{270}{l_2}$
$9 = \dfrac{270}{l_2}$
$l_2 = 30$ cm
Answer: $30$
When the cells assist each other, the equivalent EMF is $E_1 + E_2$.
$E_1 + E_2 = k l_1$
$1.5 + 1.2 = k \times 270$
$2.7 = 270 k$
When the cells are connected in opposition, the equivalent EMF is $E_1 - E_2$.
$E_1 - E_2 = k l_2$
$1.5 - 1.2 = k l_2$
$0.3 = k l_2$
Dividing the two equations, we get:
$\dfrac{2.7}{0.3} = \dfrac{270}{l_2}$
$9 = \dfrac{270}{l_2}$
$l_2 = 30$ cm
Answer: $30$
Hence, the correct answer is option: 'D'.