Question

With resistance \(x\) in the left gap and resistance \(9\ \Omega\) in right gap of a meter bridge, the balance point is obtained at \(40\) cm from left end. In what way and to which resistance \(3\ \Omega\) resistance be connected to obtain the balance point at \(50\) cm from the left end?

• A. In series with \(9\ \Omega\)
• B. Parallel to \(x\ \Omega\)
• C. In series with \(x\ \Omega\)
• D. Parallel to \(9\ \Omega\)

Hint:

First find x from the 40 cm balance. A balance at 50 cm needs equal resistances in the two gaps, so decide where the extra resistance must go.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A meter bridge balances when \(\frac{x}{9}=\frac{l}{100-l}\).

Step 2: Find x
\(\frac{x}{9}=\frac{40}{60}\), so \(x=6\ \Omega\).

Step 3: Condition for 50 cm
At the middle, \(\frac{l}{100-l}=1\), so the two gap resistances must be equal.

Step 4: Decide the Connection
Left needs to rise from \(6\ \Omega\) to \(9\ \Omega\). A series \(3\ \Omega\) gives \(6+3=9\ \Omega\).

Final Answer:
Connect the 3 ohm resistor in series with x. \[ \boxed{\text{(C) }\text{In series with }x\ \Omega} \]