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:

Question

A meter bridge with two resistances \( R_1 \) and \( R_2 \) as shown in figure was balanced (null point) at 40 cm from the point \( P \). The null point changed to 50 cm from the point \( P \), when a \( 16\,\Omega \) resistance is connected in parallel to \( R_2 \). The values of resistances \( R_1 \) and \( R_2 \) are

Answer 1

Given:

\( \ell_1 = 40 \, \text{cm}, \quad \ell_2 = 60 \, \text{cm} \)

Step 1: Resistor Equations

\[ \frac{25}{r_1} = \frac{X}{r_2} \quad \cdots \, (i) \]

\[ \frac{25}{2r_1'} = \frac{X}{2r_2'} \quad \cdots \, (ii) \]

Step 2: System Solution

From equations (i) and (ii), it is derived that: \[ l_2' = l_2 = 40 \, \text{cm} \]

Final Answer:

\[ l_2' = 40 \, \text{cm} \]

The Correct Option is D

Solution and Explanation

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