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JEE Main 2012Offline
Exam
JEE Main
Year
2012
Mode
Offline
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2012 · Offline · Physics · Thermal Properties of Matter

A liquid in a beaker has temperature θ(t) at time t and θ_0 is...

A liquid in a beaker has temperature θ(t)\theta(t) at time tandt and\theta_0istemperatureofsurroundings,thenaccordingtoNewton′slawofcoolingthecorrectgraphbetweenis temperature of surroundings, then according to Newton's law of cooling the correct graph between\log _e(θ−θ0)\left(\theta-\theta_0\right) and tt isA liquid in a beaker has temperature θ(t) at time t and θ_0 is temperature of surroundings, then according to Newton'sA liquid in a beaker has temperature θ(t) at time t and θ_0 is temperature of surroundings, then according to Newton'sA liquid in a beaker has temperature θ(t) at time t and θ_0 is temperature of surroundings, then according to Newton'sA liquid in a beaker has temperature θ(t) at time t and θ_0 is temperature of surroundings, then according to Newton's

Options

  1. AWhat is the most serious form of malaria caused by Plasmodium?
  2. BWhat is the most serious form of malaria caused by Plasmodium?
  3. CWhat is the most serious form of malaria caused by Plasmodium?
  4. DWhat is the most serious form of malaria caused by Plasmodium?

Correct answer

A. What is the most serious form of malaria caused by Plasmodium?

Step-by-step solution

According to Newtons law of cooling. dθdt∝−(θ−θ0)\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto-\left(\theta-\theta_0\right)\Rightarrow \quad \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}\left(\theta-\theta_0\right)∫dθθ−θ0=∫−kdt\int \frac{d \theta}{\theta-\theta_0}=\int-k d t ⇒ln⁡(θ−θ0)=−kt+c\Rightarrow \quad \ln \left(\theta-\theta_0\right)=-k t+c Hence the plot of ln⁡(θ−θ0)\ln \left(\theta-\theta_0\right) vs tt should be a straight line with negative slope.
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