Q175 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
An unknown resistance wire in the left gap of a meter bridge balances a $10\ \Omega$ resistance in the right gap at a point that divides the bridge wire in a ratio $3:2$. If the length of this unknown resistance wire is $1.5\ \text{m}$, what is its resistance per unit length? (Bridge wire is uniform)
- Correct answerA$10.0\ \Omega/\text{m}$
- B$6.67\ \Omega/\text{m}$
- C$15.0\ \Omega/\text{m}$
- D$1.5\ \Omega/\text{m}$
Solution
Let the unknown resistance in the left gap be $R$ and the resistance in the right gap be $S = 10\ \Omega$.
At the balance point of the meter bridge, the ratio of resistances is equal to the ratio of the lengths of the bridge wire segments:
$\dfrac{R}{S} = \dfrac{l_1}{l_2}$
Given that the balance point divides the bridge wire in the ratio $3:2$, we have:
$\dfrac{R}{10} = \dfrac{3}{2}$
$R = 10 \times \dfrac{3}{2} = 15\ \Omega$
The length of the unknown resistance wire is $L = 1.5\ \text{m}$.
The resistance per unit length of the wire is:
$\dfrac{R}{L} = \dfrac{15}{1.5} = 10.0\ \Omega/\text{m}$
Answer: $10.0\ \Omega/\text{m}$
At the balance point of the meter bridge, the ratio of resistances is equal to the ratio of the lengths of the bridge wire segments:
$\dfrac{R}{S} = \dfrac{l_1}{l_2}$
Given that the balance point divides the bridge wire in the ratio $3:2$, we have:
$\dfrac{R}{10} = \dfrac{3}{2}$
$R = 10 \times \dfrac{3}{2} = 15\ \Omega$
The length of the unknown resistance wire is $L = 1.5\ \text{m}$.
The resistance per unit length of the wire is:
$\dfrac{R}{L} = \dfrac{15}{1.5} = 10.0\ \Omega/\text{m}$
Answer: $10.0\ \Omega/\text{m}$
Hence, the correct answer is option: 'A'.