Question

Two resistors are connected in series in a meter bridge. The balance point is obtained at 20 cm from the left end. When a 15 ohm resistor is connected in series with the smaller resistor, the null point shifts to 40 cm. What is the value of the larger resistor?

• A. \( 9 \Omega \)
• B. \( 18 \Omega \)
• C. \( 27 \Omega \)
• D. \( 36 \Omega \)

Hint:

In meter bridge problems, always remember the balance condition: \[ \frac{R_1}{R_2} = \frac{L_1}{L_2} \] This helps to calculate the unknown resistances based on the given lengths.

Answer 1

The Correct Option is D

Let the resistance of the smaller resistor be \( r_1 \) and that of the larger resistor be \( r_2 \).
The balance point in a meter bridge is given by the formula: \[ \frac{l}{100 - l} = \frac{r_1}{r_2} \] where \( l \) is the balance point and \( r_1, r_2 \) are the resistances of the two resistors.
Initially, the balance point is 20 cm, so: \[ \frac{20}{80} = \frac{r_1}{r_2} \] This simplifies to: \[ \frac{r_1}{r_2} = \frac{1}{4} \quad \text{(1)} \]
When a 15 ohm resistor is connected in series with \( r_1 \), the new balance point shifts to 40 cm. The total resistance of the smaller side becomes \( r_1 + 15 \). The new equation is: \[ \frac{40}{60} = \frac{r_1 + 15}{r_2} \] This simplifies to: \[ \frac{r_1 + 15}{r_2} = \frac{2}{3} \quad \text{(2)} \]
Now, solve equations (1) and (2) together: \[ \frac{r_1}{r_2} = \frac{1}{4} \quad \text{and} \quad \frac{r_1 + 15}{r_2} = \frac{2}{3} \] From equation (1), \( r_1 = \frac{r_2}{4} \). Substitute this into equation (2): \[ \frac{\frac{r_2}{4} + 15}{r_2} = \frac{2}{3} \] Simplifying: \[ \frac{r_2 + 60}{4r_2} = \frac{2}{3} \] Cross-multiply: \[ 3(r_2 + 60) = 8r_2 \] \[ 3r_2 + 180 = 8r_2 \] \[ 180 = 5r_2 \] \[ r_2 = \frac{180}{5} = 36\ \Omega \]