Question

Wire having resistance of 10 ohm is stretched so that its length is increased by 20% calculate new resistance.

Answer 1

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

Answer 2

R1 = 14.4ohm

Explanation:

Given :

Resistance (R) = 10 ohm

Find:

New resistance (R1) = ?

Computation:

We know that,

R = ρ(L/A)

R = ρl²/(AL) =

R = ρl²/V................Eq1

If length is increased by 20%

then, new length of wire = L + 20% of L

new length of wire =  1.2L

New resistance (R1) = ρ(1.2L)²/V.........Eq2

from Eq2 / Eq1

R1 = 1.44R

so, R' = 1.44R = 14.4ohm

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