Question

In a meter bridge, two balancing resistances are \(30\,\Omega\) and \(20\,\Omega\).
If the galvanometer shows zero deflection for the jockey’s contact point \(P\), find the length \(AP\).

• A. \(40\ \text{cm}\)
• B. \(30\ \text{cm}\)
• C. \(60\ \text{cm}\)
• D. \(70\ \text{cm}\)

Hint:

For a balanced meter bridge: \[ \frac{R_1}{R_2}=\frac{l_1}{l_2},\qquad l_1+l_2=100\ \text{cm} \] Use ratios directly to save time.

Answer 1

The Correct Option is C

Concept: At balance condition of a meter bridge (Wheatstone bridge), the ratio of resistances equals the ratio of the corresponding wire lengths: \[ \frac{R_1}{R_2}=\frac{AP}{PB} \]
Step 1: Apply the balance condition Given: \[ R_1=30\,\Omega,\qquad R_2=20\,\Omega \] \[ \frac{AP}{PB}=\frac{30}{20}=\frac{3}{2} \]
Step 2: Use total length of the wire Total length of meter bridge wire: \[ AP+PB=100\ \text{cm} \] Let \(AP=3x\) and \(PB=2x\). \[ 3x+2x=100 \Rightarrow 5x=100 \Rightarrow x=20 \]
Step 3: Find \(AP\) \[ AP=3x=3\times 20=60\ \text{cm} \] Final Answer: \[ \boxed{AP=60\ \text{cm}} \]