Question

1 kg piece of copper is drawn into a wire of cross-sectional area 1 mm

²

, and then the same wire is remoulded again into a wire of cross-sectional area 2 mm

²

. Find the ratio of the resistance of the first wire to the second wire.

• A. 2:1
• B. 4:1
• C. 8:1
• D. 16:1

Answer 1

Answer: (B)

Step 1: Given that:

Mass(m) of copper = $1kg$

The cross-sectional area(A) of copper wire = $1m{m}^2$ = $1\times {10}^{-6}{m}^2$

The area of cross-section(A') of the same wire after remoulding = $2m{m}^2$ = $2\times {10}^{-6}{m}^2$

Step 2: Formula and Concept Used:

resistance of a wire is given as; $R=\frac{\text{ρ}l}{A}$

Where; ρ is the resistivity of the material of the wire, A is the area of cross-section and l is the length of the wire.

The volume of a solid is the product of length and cross-sectional area. $V=AL$

Since volume is the same in both cases, the length of the second wire will be half of the length of 1st wire.

Step 3: Calculation of the ratio of the resistance of the first wire to the second wire:

Let the length of 1st wire = L
Area of 1st wire = A =$1m{m}^2$

$R=\frac{\text{ρ}\times L}{1\times {10}^{-6}{m}^2}$ ........(1)

Now,
Length of the second wire($L^{\prime }$) = $\frac{L}{2}$

Now, if the resistance of the second wire is R' then;

$R^{\prime }=\text{ρ}\frac{\frac{L}{2}}{2\times {10}^{-6}{m}^2}$

$R^{\prime }=\text{ρ}\frac{L}{4\times {10}^{-6}{m}^2}$...........(2)

from equation 1) divided by equation 2), we get;

$\frac{R}{R^{\prime }}=\frac{\frac{\text{ρ}\times L}{1\times {10}^{-6}{m}^2}}{\frac{\text{ρ}\times L}{4\times {10}^{-6}{m}^2}}$

$\frac{R}{R^{\prime }}=\frac{4}{1}$

$R:R^{\prime }=4:1$

Thus,

The ratio of the resistance of the first and second wire is $4:1$