Masses of 3 wires of same metal are in the ratio I : 2 : 3 and their lengths are in the ratio 3 : 2 : l' The electrical resistances are in ratio

1:4:9

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9:4:1

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1:2:3

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27:6:1

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Solution

The correct option is C 27:6:1
$R = \frac{p l}{A} = \frac{d p l^{2}}{m}; R \infty \frac{l^{2}}{m}$

$[V = A l, d = \frac{m}{v} = \frac{m}{A l} \Rightarrow A = \frac{m}{d l}]$

$R_{1} : R_{2} : R_{3} \equiv\equiv \frac{9}{1} : \frac{4}{2} : \frac{1}{3} \equiv 9 : 2 : \frac{1}{3} = 27 : 6 : 1$

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Masses of 3 wires of same metal are in the ratio I : 2 : 3 and their lengths are in the ratio 3 : 2 : l' The electrical resistances are in ratio

The correct option is C 27:6:1R=plA=dpl2m;R∞l2m[V=Al,d=mv=mAl⇒A=mdl]R1:R2:R3≡≡91:42:13≡9:2:13=27:6:1

The correct option is C 27:6:1
$R = \frac{p l}{A} = \frac{d p l^{2}}{m}; R \infty \frac{l^{2}}{m}$

$[V = A l, d = \frac{m}{v} = \frac{m}{A l} \Rightarrow A = \frac{m}{d l}]$

$R_{1} : R_{2} : R_{3} \equiv\equiv \frac{9}{1} : \frac{4}{2} : \frac{1}{3} \equiv 9 : 2 : \frac{1}{3} = 27 : 6 : 1$