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A wire of resistance 10 ohm is stretched 20% determinant its new resistance

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8 months ago

```							resistance is given by, R = ρl/A R = ρl²/(Al) = ρl²/V.....(1), where V is volume of wire. at constant volume, V resistance of wire depends on only length of wire. as it is clear in expression (1), resistance is directly proportional to square of its length. now, according to question, length is increased by 20% so, new length of wire = l + 20% of l = l + 0.2l = 1.2l now, new resistance , R' = ρ(1.2l)²/V....(2) from equations (1) and (2), R'/R = 1.44 => R' = 1.44R it is given initial resistance, R = 10ohm so, R' = 1.44R = 14.4ohm
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8 months ago
```							Dear student resistance is given by, R = ρl/AR = ρl²/(Al) = ρl²/V.....(1), where V is volume of wire.at constant volume, V resistance of wire depends on only length of wire.as it is clear in expression (1), resistance is directly proportional to square of its length.now, according to question,length is increased by 20%so, new length of wire = l + 20% of l= l + 0.2l = 1.2lnow, new resistance , R' = ρ(1.2l)²/V....(2)from equations (1) and (2),R'/R = 1.44 => R' = 1.44RGood Luck
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8 months ago
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