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Current Electricity question

2013 · Q125
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Current Electricity question

2013 · Q125

NEETPhysicsCurrent ElectricityMCQ+4 / −1
A 12 cm wire is given a shape of a right angled triangle ABC having sides 3 cm, 4 cm and 5 cm as shown in the figure. The resistance between two ends (AB, BC, CA) of the respective sides are measuread one by one ratio
NEET 2013 (Karnataka) Physics - Current Electricity Question 91 English
  1. A
    9 : 16 : 25
  2. B
    27 : 32 : 35
  3. C
    21 : 24 : 25
  4. D
    3 : 4 : 5
View written solutionFree

Correct answer: B

Resistance is directly proportional to length

1RAB=13+14+5=(4+5)+3(3)(4+5){1 \over {{R_{AB}}}} = {1 \over 3} + {1 \over {4 + 5}} = {{\left( {4 + 5} \right) + 3} \over {\left( 3 \right)\left( {4 + 5} \right)}}RAB​1​=31​+4+51​=(3)(4+5)(4+5)+3​

RAB=3×(4+5)3+(4+5)=2712{R_{AB}} = {{3 \times \left( {4 + 5} \right)} \over {3 + \left( {4 + 5} \right)}} = {{27} \over {12}}RAB​=3+(4+5)3×(4+5)​=1227​

Similarly,

RBC=4×(3+5)4+(3+5)=3212{R_{BC}} = {{4 \times \left( {3 + 5} \right)} \over {4 + \left( {3 + 5} \right)}} = {{32} \over {12}}RBC​=4+(3+5)4×(3+5)​=1232​

RAC=5×(3+4)5+(3+4)=3512{R_{AC}} = {{5 \times \left( {3 + 4} \right)} \over {5 + \left( {3 + 4} \right)}} = {{35} \over {12}}RAC​=5+(3+4)5×(3+4)​=1235​

∴\therefore∴ RAB : RBC : RAC = 27 : 32 : 35

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