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Current Electricity question

2020 · Q128
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Current Electricity question

2020 · Q128

NEETPhysicsCurrent ElectricityMCQ+4 / −1
A resistance wire connected in the left gap of a metre bridge balances a 10 Ω\OmegaΩ 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 1 Ω\OmegaΩ of the resistance wire is :
  1. A
    1.0×10−1m1.0 \times {10^{ - 1}}m1.0×10−1m
  2. B
    1.5×10−1m1.5 \times {10^{ - 1}}m1.5×10−1m
  3. C
    1.5×10−2m1.5 \times {10^{ - 2}}m1.5×10−2m
  4. D
    1.0×10−2m1.0 \times {10^{ - 2}}m1.0×10−2m
View written solutionFree

Correct answer: A

NEET 2020 Phase 1 Physics - Current Electricity Question 33 English Explanation


Initially, P10=l1l2=32{P \over {10}} = {{{l_1}} \over {{l_2}}} = {3 \over 2}10P​=l2​l1​​=23​
⇒\Rightarrow⇒ P=302=15ΩP = {{30} \over 2} = 15\Omega P=230​=15Ω
Now Resistance, R=ρlAR = {{{\rho l}} \over A}R=Aρl​
R1R2=l1l2{{{R_1}} \over {{R_2}}} = {{{l_1}} \over {{l_2}}}R2​R1​​=l2​l1​​
⇒\Rightarrow⇒ 151=1.5l2{{15} \over 1} = {{1.5} \over {{l_2}}}115​=l2​1.5​
l2=0.1m{l_2} = 0.1ml2​=0.1m
=1.0×10−1m = 1.0 \times {10^{ - 1}}m=1.0×10−1m

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