1. Use the meter bridge balance condition

For a meter bridge at balance,

$\frac{R_{\text{left}}}{R_{\text{right}}}=\frac{\text{resistance of left wire segment}}{\text{resistance of right wire segment}}$

If resistance per unit length of the wire is uniform, then resistance is proportional to length. Hence,

$\frac{X}{25}=\frac{40}{60}=\frac{2}{3}$

So,

$X=25\cdot \frac{2}{3}=\frac{50}{3}\ \Omega$

2. Now replace the wire

The new wire has resistance per centimeter $2r$ instead of $r$.

Let the new balancing length from the left end be $l$ cm.

Then resistance of left segment of wire is

$2r\, l$

and resistance of right segment is

$2r(100-l)$

At balance,

$\frac{X}{25}=\frac{2r l}{2r(100-l)}=\frac{l}{100-l}$

But we already know

$\frac{X}{25}=\frac{2}{3}$

Therefore,

$\frac{l}{100-l}=\frac{2}{3}$

Solving:

$3l=2(100-l)$ $3l=200-2l$ $5l=200$ $l=40\text{ cm}$

3. Conclusion

The balancing length remains unchanged because both wire segments are multiplied by the same resistance-per-unit-length factor.

So the new balancing length is:

$\boxed{40\text{ cm}}$

Hence, Option C is correct.