Question
A 146 g mass of aluminum (Al) is formed into a right circular cylinder shaped so that its diameter equals its height. (a) Find the resistance between the top and bottom faces of the cylinder at 20 °C. Use 2700 kg/m^3 as the density of Al and 2.82 × 10^8 · m as its resistivity. Answer in units of . (b) Find the resistance between opposite faces if the same mass of aluminum is formed into
a cube. Answer in units of .
Explanation / Answer
SOLUTION: Given the mass of the aluminium,
m=146x10-3 kg Density of the aluminium is,
=2.70x103 kg/m3 The specific resistance of the aluminium is,
=2.82 x10-8 /m Now the volume of the block is
V=m/ = 5.407 x 10-5 m3 (a) Given the diameter is equal to its height,
d= h We know the formula for the volume of the cylinder in terms of diameter is,
V=r2h =
(d/2)2d =
d3/4 From this,
d= (4V/)1/3 =(4 x 5.407 x 10-5m3 / 3.14)1/3 =0.0423 m We know the formula for the resistance is ,
R=L/A =
d/ (d2/4) =
4/d =4(2.82*10-8/m) /(3.14)(0.0423 m) =
8.49 x 10-7 ________________________________________________________ ________________________________________________________ (b) In this case we have to calculate the resistance between the opposite faces of the aluminium cube. we know the formula for volume of the cube is,
V=L3 L=V1/3 =(5.407 x 10-5m3)1/3 From this the length of the cube is,
L= 0.039 m Now the resistance between the opposite faces of the cube is,
R=L/A =L/L2 =/L =(2.82*10-8/m)/(0.039 m) =
7.23 x10-7 L=V1/3 =(5.407 x 10-5m3)1/3 From this the length of the cube is,
L= 0.039 m From this the length of the cube is,
L= 0.039 m Now the resistance between the opposite faces of the cube is,
R=L/A =L/L2 =/L =(2.82*10-8/m)/(0.039 m) =
7.23 x10-7 =/L =(2.82*10-8/m)/(0.039 m) =
7.23 x10-7