Monday, 19 December 2011

Question Of The Day

18:00  Vijaykumar  6 comments


The masses of the three wires are in the ratio 1:3:5 and their lengths are in the ratio 5:3:1. then the ratio of their electrical resistance is
        [a] 1:3:5                  [b] 5:3:1
        [c]1:15:125              [d]125:15:1

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6 comments:

its option d).
Resistance=(Resisitivity*Length)/Area
We have resistivity same for each wire(as it's of copper)
We have ratio of lengths
We should get ratios of areas
For that apply Density= Mass/Volume
or Density=Mass/(Area*Length)
Density is same for the 3 wires so area is proportional to mass/length

Ratio of areas is1/5:3/3:5/1 or 1:5:25
Now ratios of areas are 1:5:25

Resistance is proportional to length/area

Ratio is5/1:3/5:1/25 or 125:15:1
So ratio is 125:15:1

I did it the same way.. But i thought, the first wire has less mass and has greater length. So it must be thin.. So it has less resistance..And 3rd wire has large mass and short length. So 3rd wire is thick and has large resistance...
So resistance of 1st is smaller than 3rd.. So,
R1<R2<R3
So... 1:25:125...
??

greater length should have higher resistance.Go through your comment again

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