In a meter bridge, two balancing resistances are $ 30 \, \Omega $ and $ 20 \, \Omega $. If the galvanometer shows zero deflection for the jockey's contact point $ P $, then find the length $ A P $.

Question

An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :

Answer 1

To solve this problem, we need to apply the principles of a meter bridge, which uses a form of a Wheatstone bridge to find unknown resistances.

In a balanced Wheatstone bridge:

$\frac{R_1}{R_2} = \frac{L_1}{L_2}$

Where:

$R_1 = 30 \, \Omega$ (resistance on one side of the bridge)
$R_2 = 20 \, \Omega$ (resistance on the other side)
$L_1$ is the length of wire $AP$
$L_2$ is the length of wire $PB$

The total length of the wire AB is 100 cm since it is a meter bridge.

Now, applying the formula:

$\frac{30}{20} = \frac{L_1}{100 - L_1}$

Cross-multiplying gives:

$30 \times (100 - L_1) = 20 \times L_1$

Expanding and simplifying:

$3000 - 30L_1 = 20L_1$

Combining like terms:

$3000 = 50L_1$

Solving for $L_1$:

$L_1 = \frac{3000}{50} = 60 \, \text{cm}$

Thus, the length $AP$ is 60 cm. Therefore, the correct answer is:

60 cm

In a meter bridge, two balancing resistances are \( 30 \, \Omega \) and \( 20 \, \Omega \). If the galvanometer shows zero deflection for the jockey's contact point \( P \), then find the length \( A P \).

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The Correct Option is C

Solution and Explanation

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