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 utilizes the principle of the Wheatstone bridge. In this setup, the ratio of two resistances is equated to the ratio of the lengths along the bridge wire. The governing equation is:\[\frac{R_1}{R_2} = \frac{l_1}{l_2},\]where: \( R_1 = 10 \, \Omega \) represents the known (fixed) resistance, \( R_2 = X \, \Omega \) denotes the unknown resistance to be determined, \( l_1 = 40 \, \mathrm{cm} \) is the distance on the bridge wire from one end to the balance point, \( l_2 = 100 - l_1 = 60 \, \mathrm{cm} \) is the remaining length of the wire on the opposite side of the balance point.
Step 1: Apply the balance condition.At the balance point, the ratio of resistances equals the ratio of lengths:\[\frac{R_1}{X} = \frac{l_1}{l_2}.\]
Step 2: Isolate \( X \).Rearrange the equation to solve for the unknown resistance \( X \):\[X = R_1 \cdot \frac{l_2}{l_1}.\]
Step 3: Substitute values.Insert the given values for \( R_1 \), \( l_1 \), and \( l_2 \):\[X = 10 \cdot \frac{60}{40}.\]Perform the calculation:\[X = 10 \cdot 1.5 = 6.67 \, \Omega.\]
Conclusion:The determined value for the unknown resistance \( X \) is \( \mathbf{6.67 \, \Omega} \).Consequently, the correct answer corresponds to \( \mathbf{(1)} \).