In a meter bridge, as shown in the figure, it is given that resistance $Y=12.5 \, \Omega$ and that the balance is obtained at a distance $39.5\, cm$ from end $A$ (by Jockey $J$). After interchanging the resistances $X$ and $Y$, a new balance point is found at a distance $l_2$ from end $A$. What are the values of $X$ and $l_2$ ?

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 concept of a meter bridge which is based on the principle of Wheatstone bridge. The meter bridge is used to measure unknown resistance by balancing two legs of a bridge circuit.

The condition for balance in a Wheatstone bridge is given by:

$\frac{X}{Y} = \frac{l_1}{(100-l_1)}$

Given:

Y = 12.5 \, \Omega
l_1 = 39.5 \, cm

Substitute the known values into the formula:

$\frac{X}{12.5} = \frac{39.5}{60.5}$

Let's solve for $X$:

X = 12.5 \times \frac{39.5}{60.5} = 8.16 \, \Omega

After interchanging the resistances $X$ and $Y$, the balance point $l_2$ becomes:

$\frac{Y}{X} = \frac{l_2}{(100-l_2)}$

Substitute the values:

$\frac{12.5}{8.16} = \frac{l_2}{(100-l_2)}$

Cross multiply and solve for $l_2$:

12.5 \times (100 - l_2) = 8.16 \times l_2
1250 - 12.5 \times l_2 = 8.16 \times l_2
1250 = 20.66 \times l_2
l_2 = \frac{1250}{20.66} = 60.5 \, cm

Thus, the values of $X$ and $l_2$ are $8.16 \, \Omega$ and $60.5 \, cm$ respectively.

The correct answer is therefore: 8.16 \, \Omega and 60.5 \, cm.

Answer 2