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

- A meter bridge works on the principle of a Wheatstone bridge: \[ \frac{R}{S} = \frac{l_1}{100 - l_1} \]
- Initially:
- \( l_1 = 25 \, cm \)
- So: \[ \frac{R}{S} = \frac{25}{75} = \frac{1}{3} \]
- Now resistance is increased by 100%, so: \[ R' = 2R \]
- New balance condition: \[ \frac{2R}{S} = \frac{l_2}{100 - l_2} \]
- Substitute \( \frac{R}{S} = \frac{1}{3} \): \[ \frac{2}{3} = \frac{l_2}{100 - l_2} \]
- Cross multiply: \[ 3l_2 = 2(100 - l_2) \]
- Solve: \[ 3l_2 = 200 - 2l_2 \] \[ 5l_2 = 200 \] \[ l_2 = 40 \, cm \]
- Percentage change in balance length: \[ \frac{40 - 25}{25} \times 100 = 60\% \]
- Hence, the percentage increase is: 60%