Question

Figure shows a meter-bridge. Initially null point was achieved at point P as shown in the figure. When an unknown resistance "R" is connected in parallel with 3$\Omega$ the null point was shifted by 22.5 cm. Then the value of unknown resistance is :

• A. 2$\Omega$
• B. 3$\Omega$
• C. 2.5 $\Omega$
• D. 5$\Omega$

Hint:

In meter bridge problems, always identify the initial balancing length first. Remember that the total length of the wire is 100 cm. When a resistance is added in parallel, the equivalent resistance decreases, which will cause the null point to shift. Tracing the direction of the shift helps in setting up the final equation correctly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem describes a meter bridge, which is an application of a balanced Wheatstone bridge. We need to find the value of an unknown resistance 'R' based on the initial balance point and the shift in the balance point after 'R' is added in parallel to one of the known resistances.
Step 2: Key Formula or Approach:
The balancing condition for a meter bridge is given by:
\[ \frac{R_1}{R_2} = \frac{l}{100 - l} \] where \(l\) is the balancing length in cm.
When two resistors \(R_a\) and \(R_b\) are in parallel, their equivalent resistance \(R_{eq}\) is:
\[ R_{eq} = \frac{R_a \times R_b}{R_a + R_b} \] Step 3: Detailed Explanation:
Initial State:
Let the initial balancing length from the left end be \(l_1\). The resistances are 2$\Omega$ and 3$\Omega$.
Using the balancing condition:
\[ \frac{2}{3} = \frac{l_1}{100 - l_1} \] \[ 2(100 - l_1) = 3l_1 \] \[ 200 - 2l_1 = 3l_1 \] \[ 5l_1 = 200 \implies l_1 = 40 \text{ cm} \] The initial null point P is at 40 cm from the left end.
Final State:
An unknown resistance R is connected in parallel with the 3$\Omega$ resistor. The new equivalent resistance \(R_{eq}\) is:
\[ R_{eq} = \frac{3 \times R}{3 + R} \] The null point shifts by 22.5 cm. The new balancing length \(l_2\) is \(40 + 22.5 = 62.5\) cm. The other length is \(100 - 62.5 = 37.5\) cm.
The new balancing condition is:
\[ \frac{2}{R_{eq}} = \frac{l_2}{100 - l_2} = \frac{62.5}{37.5} \] We know that \(\frac{62.5}{37.5} = \frac{5}{3}\).
\[ \frac{2}{\frac{3R}{3+R}} = \frac{5}{3} \] \[ \frac{2(3+R)}{3R} = \frac{5}{3} \] \[ 6(3+R) = 5(3R) \] \[ 18 + 6R = 15R \] \[ 18 = 9R \] \[ R = 2 \Omega \] Step 4: Final Answer:
The value of the unknown resistance is 2$\Omega$.