Physics Question on Meter Bridge

Question

The resistance per centimeter of a meter bridge wire is $r$ , with $X \, \Omega$ resistance in the left gap. Balancing length from the left end is at 40 cm with 25 $\Omega$ resistance in the right gap. Now the wire is replaced by another wire of $2r$ resistance per centimeter. The new balancing length for the same settings will be at:

Answer 1

Solution

The balanced condition for a meter bridge is based on the principle of a Wheatstone bridge. When balanced, the ratio of the resistances in the two gaps is equal to the ratio of the lengths of the wire on either side of the balance point.

Initially, we are given that the resistance in the left gap is $X \, \Omega$ and in the right gap is $25 \, \Omega$ . The balancing length from the left end is 40 cm, meaning that the remaining length from the balance point to the right end is 60 cm (since the wire is 100 cm in total length). Using the balanced condition:

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

Solving for $X$ , we get:

$X = 25 \times \frac{40}{60} = 16.67 \, \Omega$

Now, if the wire is replaced by another wire with twice the resistance per unit length, i.e., $2r$ instead of $r$ , each segment’s resistance will be scaled proportionally by the same factor. The new resistance per unit length affects both segments equally, so the ratio of resistances remains the same as the original setup.

Since the balance depends only on the ratio of resistances in the two arms of the bridge, the balancing length remains unaffected by the change in resistance per unit length.

Therefore, the balancing length will remain:

$40 \, \text{cm}$