Question 502: Current Electricity

Question

The length of a potentiometer wire is ' $L$ ' . A cell of e.m.f. $E$ is balanced at a length $\frac{L}{5}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$, the same cell will give balance point at distance ' $x$ '. The value of ' $x$ ' is

Accepted Answer

Let the potentiometer wire have length $L$ and let its potential gradient (potential drop per unit length) be $k$.
Step 1: Balance Point for Original Wire
The balance point for the cell of emf $E$ is at length $\dfrac{L}{5}$.
According to the potentiometer principle:
$ E = k 	imes \left(\frac{L}{5}ight) $
So,
$ k = \frac{E}{L/5} = \frac{5E}{L} $
Step 2: Wire Length is Increased
Now, the length of the wire becomes:
$ L' = L + \frac{L}{2} = \frac{3L}{2} $
The total potential difference (P.D.) across the wire
remains the same
(since the driver cell is unchanged), so the potential gradient decreases.
Let $k'$ be the new potential gradient:
$ k' = \frac{	ext{Total P.D.}}{	ext{new length}} $
But originally:
$k = \frac{	ext{Total P.D.}}{L}$
$k' = \frac{	ext{Total P.D.}}{\frac{3L}{2}} = \frac{2}{3} 	imes \frac{	ext{Total P.D.}}{L} = \frac{2}{3} k$
Step 3: New Balance Length
Let the new balance length be $x$:
$ E = k' x $
But from before, $E = k 	imes \frac{L}{5}$, so:
$ k' x = k \frac{L}{5} $
Substitute $k' = \frac{2}{3}k$:
$ \left(\frac{2}{3}kight)x = k \frac{L}{5} $
$ \frac{2x}{3} = \frac{L}{5} $
$ 2x = \frac{3L}{5} $
$ x = \frac{3L}{10} $
Final answer:
The value of $x$ is
$ \boxed{\frac{3L}{10}} $
So,
Option C
is correct.

Answer

$\frac{5 \mathrm{~L}}{12}$

Answer

$\frac{4 \mathrm{~L}}{15}$

Answer

$\frac{3 \mathrm{~L}}{10}$

Answer

$\frac{2 \mathrm{~L}}{15}$

Current Electricity

The length of a potentiometer wire is ' L ' . A cell of e.m.f. E is balanced at a length \frac{L}{5} from the positive end of the wire. If the length of the wire is increased by \frac{L}{2}, the same cell will give balance point at distance ' x '. The value of ' x ' is

Choose the correct answer: