A cylindrical wire is stretched to increase its length by 10% Calculate the percentage increase in its resistance - Physics - Alternating Current

Question

A cylindrical wire is stretched to increase its length by 10%
Calculate the percentage increase in its resistance

Ruhhi Ralhan · Accepted Answer

Dear Student,

Given:

Length of the wire initially = l₁

Length of the wire after it increases by 10%, l₂ = l₁ + (10/100 )l₁ = l₁ + (1/10)l₁ = (11/10)l₁ Area of the wire initially : A₁

Area of the wire after it is stretched becomes : A₂

Resistance initially of the wire : R₁

Resistance after it is stretched : R₂

Now, the most important thing to be kept in mind while solving this problem is that volume of the wire remains constant.

Hence, l₁A₁ = l₂A₂

$O r, A_{2} = \frac{l_{1} A_{1}}{l_{2}}$

We know that resistance( R) is given by the following equation:

R = ρ(l/A)

Where ‘ρ’ is the resistivity of the wire which will remain the same throughout as it is the property of the material of which the wire is made up.

Taking the ratio of the resistances:

$\frac{R_{1}}{R_{2}} = \frac{ρ \frac{l_{1}}{A_{1}}}{ρ \frac{l_{2}}{A_{2}}} = \frac{l_{1} A_{2}}{l_{2} A_{1}}$

Substituting the values of l₂ and A₂ in the above equation we get,

$\frac{R_{1}}{R_{2}} = \frac{l_{1} (\frac{l_{1} A_{1}}{l_{2}})}{A_{1} l_{2}} = \frac{l_{1}^{2}}{l_{2}^{2}} = \frac{100}{121} R_{2} = \frac{121}{100} R_{1} Percentage Change in resistance, \frac{Δ R}{R} \times 100 = \frac{R_{2} - R_{1}}{R_{1}} \times 100 = 21 %$

A cylindrical wire is stretched to increase its length by 10%. Calculate the percentage increase in its resistance.