# Find the resistance between the top and bottom faces of the cylinder

1. Jun 22, 2009

### mba444

1. The problem statement, all variables and given/known data

A 361 g mass of aluminum (Al) is formed into
a right circular cylinder shaped so that its
diameter equals its height.
Find the resistance between the top and
bottom faces of the cylinder at 20◦C. Use
2700 kg/m3 as the density of Al and
2.82 × 10−8 OHM· m as its resistivity. Answer
in units of
OHM.

2. Relevant equations
R= raw *(L/A)
D=(m/v)
height = diameter .... therefore r will (h/2)
V=pi*r^2*h

3. The attempt at a solution

what i did is that i solved for the height by substituting it in the density problem getting D(given)= M(given)/(Pi*(h/2)^2*h) .. which after solving is h= (1.444/(2700*Pi))^(1/3)
after that i substitute it back into the resistance equation.

2. Jun 22, 2009

### tiny-tim

Hi mba444!

(have a pi: π and a rho: ρ and try using the X2 tag just above the Reply box )
Well, that looks ok so far …

what number did you get, and what did you get next?

3. Jun 23, 2009

### mba444

i got for h .. 0.0554m
after substituting into the R equation i got 1.169e-7

but my answer is wrong !!

4. Jun 23, 2009

### drizzle

I think you need to use the Callendar–Van Dusen equation [which is a relation between temperature and resistance]

5. Jun 23, 2009

### tiny-tim

how do you get 1.169e-7 ?

6. Jun 23, 2009

### mba444

ignore the previous value i think its a calculation mistake
i calculated it again i got 3.65586e-8
is it right ??

7. Jun 23, 2009

### tiny-tim

how did you get it?

8. Jun 23, 2009

### mba444

R= (2.82e-8)[(0.055)/(PI*(2*0.055)^2)]

9. Jun 23, 2009

### tiny-tim

(and what happened to that π I gave you?)

10. Jun 23, 2009

### mba444

ya since in the question they told us that the height equal the diameter

11. Jun 23, 2009

### mba444

i really dont understand .. my home work is due today .. and i really feel lost

12. Jun 23, 2009

### Redbelly98

Staff Emeritus
What values are you using for raw, L, and A?