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Laws Of Motion

Question
CBSEENPH11020304
Wired Faculty App

A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge (θ – θ0) and t is

Solution

A.

According to Newton's law of cooling.
dθ over dt space proportional to left parenthesis straight theta minus straight theta subscript straight o right parenthesis rightwards double arrow fraction numerator begin display style dθ end style over denominator begin display style dt end style end fraction space equals space minus straight k left parenthesis straight theta minus straight theta subscript straight o right parenthesis integral fraction numerator begin display style dθ end style over denominator begin display style straight theta minus straight theta subscript straight o end style end fraction space equals space integral negative kdt space rightwards double arrow space In space left parenthesis straight theta minus straight theta subscript straight o right parenthesis space equals space minus kt plus straight c
Hence the plot of ln(θ – θ0) vs t should be a straight line with negative slope.

Some More Questions From Laws of Motion Chapter

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