Two wires $$A$$ and $$B$$ are made up of the same material and have the same mass. Wire $$A$$ has radius of $$2.0$$ mm and wire $$B$$ has radius of $$4.0$$ mm. The resistance of wire $$B$$ is $$2\Omega$$. The resistance of wire $$A$$ is _____ $$\Omega$$.

Given same material:

• Same resistivity ρ\rhoρ
• Same mass means same volume (since density same)

So

$$A_Al_A=A_Bl_B$$

where area

$$A=\pi r^2$$

Since

$$r_A=2mm,r_B​=4mm$$

$$\frac{A_A}{A_B}=\frac{2^2}{4^2}=\frac{1}{4}$$

Thus

$$A_Al_A=A_Bl_B$$

gives

$$\frac{1}{4}l_A=l_B$$

$$l_A=4l_B$$

Resistance:

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

So

$$\frac{R_A}{R_B}=\frac{l_A}{l_B}\cdot\frac{A_B}{A_A}$$

$$4\cdot4=16$$

Given

$$R_B=2Ω$$

therefore

$$R_A=16(2)=32Ω$$