Question

In a meter bridge experiment, the balance point is obtained at length '\(l_1\)' cm from left hand when resistances in the left gap and right gap are \(15 \Omega\) and \(R \Omega\) respectively. When the resistance \(R\) is shunted with equal resistance the new balance point is at \((1.6\,l_1)\). The resistance \(R\) in ohm is

• A. \(25\)
• B. \(30\)
• C. \(45\)
• D. \(60\)

Hint:

Use the balance condition for both situations and eliminate l1.

Answer 1

The Correct Option is C

Step 1: Balance Condition:
For a meter bridge, \(\dfrac{P}{Q}=\dfrac{l}{100-l}\).
First case: \(\dfrac{15}{R}=\dfrac{l_1}{100-l_1}\).

Step 2: After Shunting:
Shunting \(R\) with an equal resistance gives \(\dfrac R2\). Now \(\dfrac{15}{R/2}=\dfrac{30}R=\dfrac{1.6l_1}{100-1.6l_1}\).

Step 3: Divide the Two Equations:
\[ \frac{30/R}{15/R}=2=\frac{1.6l_1(100-l_1)}{l_1(100-1.6l_1)}\Rightarrow2(100-1.6l_1)=1.6(100-l_1) \]
\[ 200-3.2l_1=160-1.6l_1\Rightarrow1.6l_1=40\Rightarrow l_1=25\ \text{cm} \]

Step 4: Find R:
\(\dfrac{15}R=\dfrac{25}{75}=\dfrac13\Rightarrow R=45\ \Omega\). Check: new balance \(1.6\times25=40\) cm, and \(\dfrac{30}{45}=\dfrac{40}{60}\). True. So (C) is correct.

Final Answer:
The resistance is 45 ohm, option (C). \[ \boxed{\text{(C) } 45\ \Omega} \]