Question

A potentiometer wire has length $L$ For given cell of emf $E$, the balancing length is $\frac{L}{3}$ from the positive end of the wire. If the length of potentiometer wire is increased by $50\%$, then for the same cell, the balance point is obtained at length

• A. $\frac{L}{2}$ from positive end
• B. $\frac{L}{5}$ from positive end
• C. $\frac{L}{3}$ from positive end
• D. $\frac{L}{4}$ from positive end

Answer 1

Answer: (A)

For cell of emf E balancing length
$=\frac{L}{3}=\frac{1}{3} \times L \text { (length of wire) }$
Increased length of wire $=L+\frac{L}{2}=\frac{3 L}{2}$
Since, the cell is the same, therefore the new balancing length $=\frac{1}{3}$ (new increased length)
$=\frac{1}{3} \times \frac{3}{2} L$
$=\frac{L}{2}$
Therefore, for the same cell the balance point will be as length $\frac{L}{2}$ from positive end.