A wire of resistance 160 is melted and drawn into a wire of one-f… — Solution | JEE Physics

MCQEasyJEE 2023Resistivity & Conductivity

A wire of resistance $160 \, \Omega$ is melted and drawn into a wire of one-fourth of its length. The new resistance of the wire will be:

Answer

Correct answer:A

Standard Method

Given: Initial resistance is $R_1 = 160 \, \Omega$ . The new wire has length $L_2 = \frac{L_1}{4}$ . Volume of the wire remains constant.

Find: The new resistance of the wire.

For constant volume,

A_1 L_1 = A_2 L_2

Since

L_2 = \frac{L_1}{4}

we get

A_2 = 4A_1

Now use the resistance formula,

R = \rho \frac{L}{A}

So for the new wire,

R_2 = \rho \frac{L_2}{A_2}

Substituting $L_2 = \frac{L_1}{4}$ and $A_2 = 4A_1$ ,

R_2 = \rho \frac{L_1/4}{4A_1} = \rho \frac{L_1}{16A_1} = \frac{1}{16}R_1

Hence,

R_2 = \frac{1}{16} \times 160 = 10 \, \Omega

Therefore, the new resistance is $10 \, \Omega$ and the correct option is A.

Using resistance as directly proportional only to length and forgetting that the cross-sectional area also changes. This is wrong because melting and redrawing keeps volume constant, so $A$ cannot remain unchanged. First use $A_1L_1 = A_2L_2$ , then apply $R = \rho \frac{L}{A}$ .
Taking $A_2 = \frac{A_1}{4}$ when the new length becomes $\frac{1}{4}$ of the original. This is wrong because for constant volume, reducing length increases area. The correct relation is $A_2 = 4A_1$ .
Calculating the new resistance as $\frac{R_1}{4}$ instead of $\frac{R_1}{16}$ . This misses the combined effect of shorter length and larger area. Both changes reduce resistance, so multiply the effects: $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$ .

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