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: Test Option (C):
Try $R=45\ \Omega$. First balance: $\dfrac{l_1}{100-l_1}=\dfrac{15}{45}=\dfrac13\Rightarrow l_1=25$ cm.

Step 2: After Shunting:
New right-hand resistance is $22.5\ \Omega$. Then $\dfrac{l}{100-l}=\dfrac{15}{22.5}=\dfrac23\Rightarrow l=40$ cm.

Step 3: Compare:
$40/25=1.6$, exactly the given factor. Other values such as 25 or 30 ohm give factors different from 1.6. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } 45\ \Omega} \]