Question

In a meter bridge experiment, a resistance of \( 10 \, \Omega \) is balanced by a resistance \( X \) with a balance point at \( 40 \, \mathrm{cm} \). What is the value of \( X \)?

• A. \( 6.67 \, \Omega \)
• B. \( 8.00 \, \Omega \)
• C. \( 10.00 \, \Omega \)
• D. \( 15.00 \, \Omega \)

Hint:

In a meter bridge, the principle is based on the Wheatstone bridge's law, where the balance point helps determine the unknown resistance using the formula \( \frac{R_1}{R_2} = \frac{l_1}{l_2} \), with \( l_1 \) and \( l_2 \) being the lengths on either side of the meter bridge.

Answer 1

The Correct Option is A

The meter bridge experiment is based on the principle of the **Wheatstone bridge**. In a Wheatstone bridge, the ratio of two resistances is balanced by the ratio of the lengths of the bridge wire. The relationship is given by the equation: \[ \frac{R_1}{R_2} = \frac{l_1}{l_2}, \] where: \( R_1 = 10 \, \Omega \) is the known resistance (in this case, it is the fixed resistance), \( R_2 = X \, \Omega \) is the unknown resistance (which we need to find), \( l_1 = 40 \, \mathrm{cm} \) is the distance on the bridge wire where the balance point is obtained, \( l_2 = 100 - l_1 = 60 \, \mathrm{cm} \) is the remaining length of the wire on the other side of the bridge.
Step 1: Use the balanced condition. We know that the ratio of the resistances is equal to the ratio of the lengths at the balance point: \[ \frac{R_1}{X} = \frac{l_1}{l_2}. \]
Step 2: Rearrange for \( X \). Rearranging the equation to solve for the unknown resistance \( X \): \[ X = R_1 \cdot \frac{l_2}{l_1}. \]
Step 3: Substitute the known values. Substitute the values for \( R_1 \), \( l_1 \), and \( l_2 \): \[ X = 10 \cdot \frac{60}{40}. \] Simplify the equation: \[ X = 10 \cdot 1.5 = 6.67 \, \Omega. \]
Conclusion: Thus, the value of the unknown resistance \( X \) is \( \mathbf{6.67 \, \Omega} \). Therefore, the correct answer is \( \mathbf{(1)} \).