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 \( A P \).
Show Hint
In a meter bridge, the ratio of resistances is equal to the ratio of the lengths of the bridge. This principle is useful for determining unknown resistances.
Step 1: Using the formula for meter bridge.
In a meter bridge, the ratio of the two resistances is equal to the ratio of the lengths on the bridge. Let the length of the bridge between points \( A \) and \( P \) be \( x \), and the total length of the bridge is 100 cm. The other segment of the bridge will be \( 100 - x \). The relation between the resistances and the lengths is given by:
\[
\frac{R_1}{R_2} = \frac{x}{100 - x}
\]
where \( R_1 = 30 \, \Omega \) and \( R_2 = 20 \, \Omega \).
Step 2: Substitute the values.
Substituting \( R_1 = 30 \, \Omega \), \( R_2 = 20 \, \Omega \) into the equation:
\[
\frac{30}{20} = \frac{x}{100 - x}
\]
\[
\frac{3}{2} = \frac{x}{100 - x}
\]
Step 3: Solve for \( x \).
Cross multiply to solve for \( x \):
\[
3(100 - x) = 2x
\]
\[
300 - 3x = 2x
\]
\[
300 = 5x
\]
\[
x = 60 \, \text{cm}
\]
Thus, the length \( A P \) is 60 cm.
