Question

When two resistances \(P\) and \(Q\) are connected in the left and right gaps of a meter bridge respectively, the balancing point is obtained at a distance greater than \(25\,cm\). If the resistance \(P\) is increased by \(5\Omega\), the balancing point shifts by \(10\,cm\). After that, if the resistance \(Q\) is halved, the balancing point further shifts by \(15\,cm\), then the initial value of \(P\) is:

• A. \(5\Omega\)
• B. \(15\Omega\)
• C. \(10\Omega\)
• D. \(20\Omega\)

Hint:

For meter bridge questions always use \[ \frac{R_1}{R_2} = \frac{l}{100-l} \] and write a new equation after every change in the circuit.

Answer 1

The Correct Option is C

Concept: For a meter bridge at balance, \[ \frac{P}{Q}=\frac{l}{100-l}. \] where \(l\) is the balancing length.

Step 1: Let initial balancing length be \(l\).
\[ \frac{P}{Q}=\frac{l}{100-l} \] Since \(l>25\). After increasing \(P\) by \(5\Omega\), balancing length shifts by \(10cm\). \[ \frac{P+5}{Q} = \frac{l+10}{90-l}. \]

Step 2: After halving \(Q\).
Now \[ \frac{P+5}{Q/2} = \frac{l+25}{75-l}. \] Thus, \[ \frac{2(P+5)}{Q} = \frac{l+25}{75-l}. \]

Step 3: Solve simultaneously.
Using the three equations and eliminating \(Q\) and \(l\), \[ P=10\Omega. \] Therefore, \[ \boxed{10\Omega}. \]