Verified Solution • 2025–26 Session moderate

QNEET-2020-Ph-743: NEET 2020 Physics - Class 12 NEET-2020-Ph-743 Solution: A resistance wire connected in the left gap of a metre bridge…

A resistance wire connected in the left gap of a metre bridge balances a 10Ω10\Omega resistance in the right gap at a point which divides the bridge wire in the ratio 3:2. If the length of the resistance wire is 1.5 m then the length of 1Ω1\Omega of the resistance wire is: 1.0×10−1 m1.0 \times 10^{-1}\ \text{m} 1.5×10−1 m1.5 \times 10^{-1}\ \text{m} 1.5×10−2 m1.5 \times 10^{-2}\ \text{m} 1.0×10−2 m1.0 \times 10^{-2}\ \text{m}

Multiple Choice Options

  • A 1.0×10−1m1.0 \times {10^{ - 1}}m 1.0 _ 1 0 _ 1 m
  • B 1.5×10−1m1.5 \times {10^{ - 1}}m 1.5 _ 1 0 _ 1 m
  • C 1.5×10−2m1.5 \times {10^{ - 2}}m 1.5 _ 1 0 _ 2 m
  • D 1.0×10−2m1.0 \times {10^{ - 2}}m 1.0 _ 1 0 _ 2 m

Step-by-Step Verified Solution

Initially, P 10 = l 1 l 2 = 3 2 _ P = 30 2 = 15 � Now Resistance, R = _ l A R 1 R 2 = l 1 l 2 _ 15 1 = 1.5 l 2 l 2 = 0.1 m = 1.0 _ 10 _ 1 m

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