Question

When two resistors are connected in the two gaps of a meter bridge, the balancing point is obtained at \(25\,\text{cm}\) from the left end of the bridge wire. When a \(48\,\Omega\) resistor is connected in series to the smaller of the two resistors, the balancing point is obtained at \(75\,\text{cm}\) from the left end of the bridge wire. Then the initial values of the resistors connected in the left and right gaps of the bridge respectively are

• A. \(18\,\Omega,\;54\,\Omega\)
• B. \(6\,\Omega,\;18\,\Omega\)
• C. \(12\,\Omega,\;36\,\Omega\)
• D. \(24\,\Omega,\;72\,\Omega\)

Hint:

For a balanced meter bridge, \[ \boxed{ \frac{R_1}{R_2} = \frac{l}{100-l}. } \] If resistance is added in series to one gap, simply replace that resistance by its new value and apply the balance condition again.

Answer 1

The Correct Option is B

Step 1: Use the meter bridge balance condition. Let the resistors in the left and right gaps be \(R_L\) and \(R_R\). For a meter bridge, \[ \frac{R_L}{R_R} = \frac{l}{100-l}, \] where \(l\) is the balancing length from the left end. Initially, \[ l=25\,\text{cm}. \] Hence, \[ \frac{R_L}{R_R} = \frac{25}{75} = \frac13. \] Therefore, \[ R_R=3R_L. \] Since the smaller resistor is \[ R_L, \] the \(48\,\Omega\) resistor is connected in series with it.

Step 2: Use the new balancing condition. Now, \[ l=75\,\text{cm}. \] Therefore, \[ \frac{R_L+48}{R_R} = \frac{75}{25} = 3. \] Substituting \[ R_R=3R_L, \] \[ \frac{R_L+48}{3R_L}=3. \] Hence, \[ R_L+48=9R_L, \] \[ 48=8R_L, \] \[ R_L=6\,\Omega. \] Thus, \[ R_R=3\times6=18\,\Omega. \] Hence, \[ \boxed{R_L=6\,\Omega,\qquad R_R=18\,\Omega.} \] Therefore, the correct option is \(\boxed{(B)}\).