Question

An iron ring is to fit snugly on a cylindrical iron rod. At 20 oC, the diameter of the rod is

6.453cm and the inside diameter of the ring is 6.420cm. To what temperature must the ring

be brought if its hole is be large enough so it will slip over the rod? The coefficient of linear

expansivity of iron is 12 x 10-6 oC-1.

Answer 1

Solution.

$d=6.453\sdot10^{-2}m;$

$d_0=6.420\sdot 10^{-2}m;$

$\alpha=12\sdot10^{-6} C^{-1};$

$L=L_0(1+\alpha T)\implies T=\dfrac{L-L_0}{L_0\alpha};$

$T=\dfrac{6.453-6.42}{6.42\sdot12\sdot10^{-6}}=428K;$

Answer: $T=428K.$