Question

In a meter bridge, when an unknown resistance ‘R’ is connected in the left gap, the null point is obtained at 25 cm from the left end of the wire. - If the resistance in the left gap is increased by 100\%, the distance of the null point from the left end of the wire increases by

• A. \(50\%\)
• B. \(40\%\)
• C. \(60\%\)
• D. \(80\%\)

Hint:

For meter bridge problems, use the formula: \[ \frac{R}{S} = \frac{l}{100 - l} \] If the resistance in one gap is changed by a factor \( k \), the new null point is found by solving: \[ \frac{k R}{S} = \frac{l'}{100 - l'} \]

Answer 1

The Correct Option is C

Step 1: Understanding the Principle of a Meter Bridge A meter bridge works on the principle of the Wheatstone bridge: \[ \frac{R}{S} = \frac{l_1}{100 - l_1} \] where: 
- \( R \) is the unknown resistance in the left gap. 
- \( S \) is the resistance in the right gap. 
- \( l_1 \) is the initial balance length (null point), given as \(25\) cm. 

Step 2: Applying the Given Condition If the resistance \( R \) in the left gap is increased by 100%, the new resistance becomes \( 2R \). The new balance length \( l_2 \) is found using: \[ \frac{2R}{S} = \frac{l_2}{100 - l_2} \] Dividing both equations: \[ \frac{l_2}{100 - l_2} \div \frac{l_1}{100 - l_1} = \frac{2R}{S} \div \frac{R}{S} \] \[ \frac{l_2}{100 - l_2} \times \frac{100 - l_1}{l_1} = 2 \] Substituting \( l_1 = 25 \): \[ \frac{l_2}{100 - l_2} \times \frac{100 - 25}{25} = 2 \] \[ \frac{l_2}{100 - l_2} \times \frac{75}{25} = 2 \] \[ \frac{l_2}{100 - l_2} \times 3 = 2 \] \[ \frac{l_2}{100 - l_2} = \frac{2}{3} \] Solving for \( l_2 \): \[ 3 l_2 = 2(100 - l_2) \] \[ 3 l_2 + 2 l_2 = 200 \] \[ 5 l_2 = 200 \] \[ l_2 = 40 \text{ cm} \] 

Step 3: Calculating Percentage Increase \[ \frac{40 - 25}{25} \times 100 = \frac{15}{25} \times 100 = 60\% \] Thus, the correct answer is \( \mathbf{(3)} \ 60\% \).