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NEET PhysicsHard

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

1.0×10−2 m1.0 \times 10^{-2} \text{ m}

B

1.0×10−1 m1.0 \times 10^{-1} \text{ m}

C

1.5×10−1 m1.5 \times 10^{-1} \text{ m}

D

1.5×10−2 m1.5 \times 10^{-2} \text{ m}

Step-by-Step Solution

Initially, P10=l1l2=32⇒P=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=l1l2⇒151=1.5l2⇒l2=0.1 m=1.0×10−1 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}.

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