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Thermodynamics

# A copper disc at $\;0^{0}C\;$ has a diameter of 50 cm and a hole is cut in the center with a diameter 5 cm . The diameter of the hole at $\;100^{0}C\;$ will be $\;(\alpha = 0.00016/^{0}C)\;$

$(a)\;5.08\;cm\qquad(b)\;50.8\;cm\qquad(c)\;0.508\;cm\qquad(d)\;100.08\;cm$

Answer : $\;5.08\;cm$
Explanation :
$d_{h} = d_{h} \;at\;0^{0}C\;(1+\alpha_{cu} \bigtriangleup t)$
$= 5\; cm \;(1+16 \times 10^{-5} \times 100)$
$= 5 \;(1+ 16 \times 10^{-3})$ cm
$= 5.08c\;cm$