Question

two copper wires of same length but the second has diameter twice that of the first if the resistance of the first 2 ohm, resistance of the second is​

Answer 1

1). R = p * l/A = p * l/(pi*d^2)

2). R1 = p * l/A1 = p*l/(pi*2d^2)

= p*l/4*(pi*d^2)

= 1/4 * p*l/A

= 1/4 * 2 = 1/2 ohm

Answer 2

The resistance of second wire is 0.5 ohms.

Explanation:

The resistance of a wire is given by :

$R=\rho \dfrac{l}{A}$

It is clear that resistance is inversely proportional to the area of cross section. So,

$\dfrac{R_1}{R_2}=\dfrac{A_2}{A_1}$

R₁, R₂ are resistances of two wires and A₁, A₂ are their area of cross sections.

According to given condition, two copper wires of same length but the second has diameter twice that of the first, d₂ = 2d₁

So,

$\dfrac{R_1}{R_2}=\dfrac{(d_2/2)^2}{(d_1/2)^2}\\\\\dfrac{R_1}{R_2}=\dfrac{d_2^2}{d_1^2}\\\dfrac{R_1}{R_2}=\dfrac{(2d_1)^2}{d_1^2}\\\\\dfrac{R_1}{R_2}=4\\\\R_2=\dfrac{R_1}{4}\\\\R_2=\dfrac{2}{4}=0.5\ \Omega$

So, the resistance of second wire is 0.5 ohms.

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