Thermometry Calorimetry And Thermal Expansion Question 197
A cylindrical steel alloy is inserted into a circular hole of diameter 2.60 m in a brass plate. When the plug and the plate are at a temperature $ 20{}^\circ C $ , the diameter of the plug is 0.010 mm smaller than that of the hole. The temperature at which the plug will just fit in is (given and $ {\alpha_{steel}}=11\times 10^{-6}/{}^\circ C $ and $ {\alpha_{brass}} =19\times 10^{-6}/{}^\circ C $ )
Options:
A) $ -,20{}^\circ C $
B) $ -,48{}^\circ C $
C) $ 10{}^\circ C $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ 2.60+2.60\times {\alpha_{steel}}\Delta T= 2.60 {\alpha_{brass}}\Delta T $
BETA
