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NEET PhysicsGeneralHard

Question

A resistance wire connected in the left gap of a metre bridge balances a 10 Ω10 \ \Omega resistance in the right gap at a point which divides the bridge wire in the ratio 3:23 : 2. If the length of the resistance wire is 1.5 m1.5 \text{ m}, then the length of 1 Ω1 \ \Omega of the resistance wire is :

A

1.0×102 m1.0 \times 10^{-2} \text{ m}

B

1.0×101 m1.0 \times 10^{-1} \text{ m}

C

1.5×101 m1.5 \times 10^{-1} \text{ m}

D

1.5×102 m1.5 \times 10^{-2} \text{ m}

Step-by-Step Solution

Initially, P10=l1l2=32P=302=15 Ω\frac{P}{10} = \frac{l_1}{l_2} = \frac{3}{2} \Rightarrow P = \frac{30}{2} = 15 \ \Omega. Now Resistance, R=ρlAR = \frac{\rho l}{A}. R1R2=l1l2151=1.5l2l2=0.1 m=1.0×101 m\frac{R_1}{R_2} = \frac{l_1}{l_2} \Rightarrow \frac{15}{1} = \frac{1.5}{l_2} \Rightarrow l_2 = 0.1 \text{ m} = 1.0 \times 10^{-1} \text{ m}.

Exam Context & Concepts Covered

This question aligns with the NEET Physics syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

Physicsresistanceconnectedbridgebalancesresistance

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