Masses of 3 wires of same metal are in the ratio 1 : 2 : 3 and their lengths are in the ratio 3 : 2 : 1. The electrical resistances are in ratio
A. `1:1:1`
B. `1:2:3`
C. `9:4:1`
D. `27: 6:1`


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Correct Answer - 4
`rho` is same so `Rprop(l)/(A)`
Now m d `rArr m=Ald`
`Aprop(m)/(l) rArrR prop(l^(2))/(m)`
`R_(1):R_(2):R_(3)=(9)/(1):(4)/(2):(1)/(3)=27:6:1`

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