(A) The resistance of the block of aluminum is given by:
[tex]R=\frac{\rho L}{A}[/tex]
where
[tex]\rho[/tex] is the aluminum resistivity
L is the length of the cylinder
A is the cross-sectional area of the cylinder
We already know the aluminum resistivity ([tex]\rho=2.65 \cdot 10^{-8} \Omega m[/tex]) and the length of the cylinder (L=20 m), so we must find the cross-sectional area A, which is given by the difference between the area of the larger cylinder and the area of the radial hole:
[tex]A=\pi (R^2 -r^2)[/tex]
where [tex]R=2.5 cm=0.025 m[/tex] and [tex]r=20 mm=0.02 m[/tex] (assuming that the 20 mm removed radially refers to the radius of the hole).
Therefore, the cross-sectional area is
[tex]A=\pi ((0.025 m)^2-(0.020 m)^2)=7.06 \cdot 10^{-4} m^2[/tex]
Substituting into the initial formula of the resistance, we find:
[tex]R=\frac{\rho L}{A}=\frac{(2.65 \cdot 10^{-8} \Omega m)(20 m)}{7.06 \cdot 10^{-4} m^2}=7.51 \cdot 10^{-4} \Omega[/tex]
(B) The resistivity of the tungsten is [tex]\rho=5.6 \cdot 10^{-8} \Omega m[/tex], so a cylinder of tungsten of the same size would have a resistance of
[tex]R=\frac{\rho L}{A}=\frac{(5.6 \cdot 10^{-8} \Omega m)(20 m)}{7.06 \cdot 10^{-4} m^2}= 1.59 \cdot 10^{-3} \Omega[/tex]
The behaviour of the resistance versus temperature is given by:
[tex]R=R_0 (1 + \alpha \Delta T)[/tex]
where [tex]\alpha[/tex] is a coefficient that for aluminum is equal to [tex]\alpha = 0.004308[/tex], R0 is the resistance of the piece of aluminum we found at point (A), and R is the resistance of the tungsten. Re-arranging the formula and substituting, we find
[tex]\Delta T = \frac{R/R_0 -1}{\alpha}=\frac{\frac{1.59 \cdot 10^{-3} \Omega}{7.51 \cdot 10^{-4} \Omega}-1}{0.004308}=259.3^{\circ}C[/tex]
So, the temperature must increase by 259.3 degrees.
(C) The power dissipated is given by:
[tex]P=I^2 R[/tex]
where I=30 A is the current. Substituting the numbers into the formula, we find
[tex]P=(30 A)^2 (7.51 \cdot 10^{-4} \Omega)=0.68 W[/tex]
