Properties Of Solids And Liquids Question 455
Question: A body cools in a surrounding which is at a constant temperature of $ {\theta _{0}} $ . Assume that it obeys Newton’s law of cooling. Its temperature$ \theta $ is plotted against time t. Tangents are drawn to the curve at the points$ P( \theta ={\theta _{2}} ) $ and$ Q( \theta ={\theta _{1}} ) $ . These tangents meet the time axis at angle of$ {\phi _{2}} $ and $ {\phi _{1}} $ , as shown, then
Options:
A) $ \frac{\tan {\phi _{2}}}{\tan {\phi _{1}}}=\frac{{\theta _{1}}-{\theta _{0}}}{{\theta _{2}}-{\theta _{0}}} $
B) $ \frac{\tan {\phi _{2}}}{\tan {\phi _{1}}}=\frac{{\theta _{2}}-{\theta _{0}}}{{\theta _{1}}-{\theta _{0}}} $
C) $ \frac{\tan {\phi _{1}}}{\tan {\phi _{2}}}=\frac{{\theta _{1}}}{{\theta _{2}}} $
D) $ \frac{\tan {\phi _{1}}}{\tan {\phi _{2}}}=\frac{{\theta _{2}}}{{\theta _{1}}} $
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Answer:
Correct Answer: B
Solution:
[b] For 0-t plot, rate of cooling $ =\frac{dQ}{dt}= $ slope of the curve. $ AT P,\frac{dQ}{dt}=| tan( 180{}^\circ -{\phi _{2}} ) | $
$ =tan{\phi _{2}}=k( {\theta _{2}}-{\theta _{1}} ) $ where k= constant. At Q, $ \frac{dQ}{dt} =| tan(180{}^\circ -{\varphi _{1}}) |= tan{\varphi _{1}}, = k({\theta _{1}}-{\theta _{0}}) $
$ \therefore \frac{\tan {\phi _{2}}}{\tan {\phi _{1}}}=\frac{{\theta _{2}}-{\theta _{0}}}{{\theta _{1}}-{\theta _{0}}} $
BETA
