First slide

Thermal expansion

Question

An iron ball of diameter 6cm and is 0.01 mm too large to pass through a hole in a brass plate when the ball and plate are at a temperature of 200C. The temperature at which (both for ball and plate) the ball just pass through the hole is (α iron = 12 x 10^-6/⁰C; αbrass = 18 x 10^-6/⁰C)

Moderate

A

68⁰C
B

48⁰C
C

28⁰C
D

40⁰C

Solution

$\large \Delta l=l\alpha\Delta t$
$\large (l_b\alpha_b-l_{Fe}\alpha_{Fe})(t_2-t_1)=\Delta l$
$\large [5.99\times 18\times10^6-6\times 12\times 10^{-6}](t_2-20)=-0.001$
$\large 36(t-20)=1000\Rightarrow t=48^0C$
$\large \Rightarrow 35.82(t-20)=1000$
$\large \Rightarrow t=47.91^0C\simeq 48^0C$

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