Masses of the three wires of same material are in the ratio of 1 : 2 : 3 and their lengths in the ratio of 3 : 2 : 1. Electrical resistance of these wires will be in the ratio of:

1:1:1

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1 : 2 : 3

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9:4:1

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27 : 6 : 1

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Solution

The correct option is D
27 : 6 : 1

As R = $\frac{ρ l /}{A}$ and volume = lA = Constant and

Mass = density x Area x length

So, we get ratio as 27:6:1.

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Masses of the three wires of same material are in the ratio of 1 : 2 : 3 and their lengths in the ratio of 3 : 2 : 1. Electrical resistance of these wires will be in the ratio of:

The correct option is D 27 : 6 : 1 As R = ρl/A and volume = lA = Constant and Mass = density x Area x length So, we get ratio as 27:6:1.

The correct option is D
27 : 6 : 1

As R = $\frac{ρ l /}{A}$ and volume = lA = Constant and

Mass = density x Area x length

So, we get ratio as 27:6:1.