balance point for the cell of emf $E$ is at length $\dfrac{L}{5}$.

According to the potentiometer principle:

$ E = k \times \left(\frac{L}{5}\right) $

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{\text{Total P.D.}}{\text{new length}} $

But originally:

$k = \frac{\text{Total P.D.}}{L}$
$k' = \frac{\text{Total P.D.}}{\frac{3L}{2}} = \frac{2}{3} \times \frac{\text{Total P.D.}}{L} = \frac{2}{3} k$

Step 3: New Balance Length

Let the new balance length be $x$:

$ E = k' x $

But from before, $E = k \times \frac{L}{5}$, so:

$ k' x = k \frac{L}{5} $

Substitute $k' = \frac{2}{3}k$:

$ \left(\frac{2}{3}k\right)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.

Current Electricity

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