Question

A wire of 25 ohm is stretched to along get by 20% of its initial length its resistance will increase by

Answer 1

Answer:

$\boxed{\mathfrak{Percentage \ increase \ in \ resistance = 44 \%}}$

Explanation:

Initial resistance = $\rm R_1$

Initial area = $\rm A_1$

Initial length = $\rm L_1$

Final resistance = $\rm R_2$

Final area = $\rm A_2$

Final length = $\rm L_2$

By volume conservation,

$\rm A_1L_1 = A_2L_2$

Given that,

$\rm L_2 = L_1 + 30\%L_1 \\ \rm L_2 = 1.2L_1$

So,

$\rm \implies A_1 \cancel{L_1} = A_2 \times 1.2 \cancel{L_1} \\ \\ \rm \implies A_2 = \dfrac{1}{1.2} A_1$

We know that, Resistance:

$\bf R = \dfrac{\rho L}{A}$

So,

$\rm \implies \dfrac{R_1}{R_2} = \dfrac{\dfrac{\cancel{\rho} L_1}{A_1}}{\dfrac{\cancel{\rho} L_2}{A_2}} \\ \\ \rm \implies \dfrac{R_1}{R_2} = \dfrac{\dfrac{ L_1}{A_1}}{\dfrac{ L_2}{A_2}} \\ \\ \rm \implies \dfrac{R_1}{R_2} = \dfrac{\dfrac{ \cancel{L_1}}{ \cancel{A_1}}}{\dfrac{1.2 \cancel{L_1}}{ \dfrac{ \cancel{A_1}}{1.2} }} \\ \\ \rm \implies \dfrac{R_1}{R_2} = \dfrac{1}{1.2 \times 1.2} \\ \\ \rm \implies \dfrac{R_1}{R_2} = \dfrac{1}{1.44} \\ \\ \rm \implies R_2 = 1.44R_1 \\ \\ \rm \implies R_2 = R_1 + 44\%R_1$

$\therefore$ Percentage increase in resistance = 44%