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Question
JEE Main
Physics

In a meter bridge experiment to determine the value of unknown resistance, first the resistances $2 \text{ }\Omega$ and $3 \text{ }\Omega$ are connected in the left and right gaps of the bridge and the null point is obtained at a distance $l \text{ cm}$ from the left. Now when an unknown resistance $x \text{ }\Omega$ is connected in parallel to $3 \text{ }\Omega$ resistance, the null point is shifted by $10 \text{ cm}$ to the right of wire. The value of unknown resistance $x$ is _______ $\Omega$.

Meter Bridge Experiment Setup

The meter bridge operates on the principle of a balanced Wheatstone bridge. The ratio of resistances in the gaps is equal to the ratio of the lengths of the wire on the meter bridge.

Initial Setup:

  • Resistance in the left gap, $R_L = 2 \text{ }\Omega$.
  • Resistance in the right gap, $R_R = 3 \text{ }\Omega$.
  • Null point distance from the left end, $l \text{ cm}$.
  • Total length of the wire = $100 \text{ cm}$.

Balancing Condition Initial State

According to the meter bridge formula:

$ \frac{R_L}{R_R} = \frac{l}{100 - l} $

Substituting the initial values:

$ \frac{2 \text{ }\Omega}{3 \text{ }\Omega} = \frac{l}{100 - l} $

Solving for $l$:

$ 2(100 - l) = 3l $

$ 200 - 2l = 3l $

$ 200 = 5l $

$ l = \frac{200}{5} = 40 \text{ cm} $

Modified Setup and Balancing

In the second scenario, an unknown resistance $x \text{ }\Omega$ is connected in parallel to the $3 \text{ }\Omega$ resistance in the right gap.

  • Resistance in the left gap remains $R_L = 2 \text{ }\Omega$.
  • The new equivalent resistance in the right gap is $R'_R$.
  • $R'_R = \frac{3 \times x}{3 + x} \text{ }\Omega$.
  • The null point shifts by $10 \text{ cm}$ to the right.
  • New null point distance, $l' = l + 10 = 40 + 10 = 50 \text{ cm}$.
  • The length of the right segment is now $100 - l' = 100 - 50 = 50 \text{ cm}$.

Calculating Unknown Resistance

Applying the balancing condition again:

$ \frac{R_L}{R'_R} = \frac{l'}{100 - l'} $

Substituting the values:

$ \frac{2 \text{ }\Omega}{\frac{3x}{3 + x} \text{ }\Omega} = \frac{50 \text{ cm}}{50 \text{ cm}} $

$ \frac{2(3 + x)}{3x} = 1 $

Solving for $x$:

$ 2(3 + x) = 3x $

$ 6 + 2x = 3x $

$ 6 = 3x - 2x $

$ x = 6 \text{ }\Omega $

Conclusion

The value of the unknown resistance $x$ is $6 \text{ }\Omega$. This value lies within the specified range of 6 to 6.

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