This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “Newton’s Law of Cooling and Escape Velocity”.

1. According to Newton’s law of cooling “The change of temperature of a body is proportional to the difference between the temperature of a body and that of the surrounding medium”. If t_{1}℃ is the initial temperature of the body and t_{2}℃ is the constant temperature of the medium, T℃ be the temperature of the body at any time t then find the expression for T℃ as a function of t_{1}℃, t_{2}℃ and time t.

a) T=t_{1}+(t_{2}) e^{-kt}

b) T=t_{2}+(t_{1}-t_{2}) e^{-kt}

c) T=t_{1}+(t_{1}-t_{2}) e^{kt}

d) T=t_{2}+(t_{1}) e^{kt}

View Answer

Explanation: According to the definition of Newton’s law of cooling \(\frac{dT}{dt} ∝ (T-t_2) \,or\, \frac{dT}{dt} = -k(T-t_2)\) ….k is a constant of proportionality and negative sign indicates the cooling of a body with increase of the time. since t

_{1}℃ initial temperature of the body at t=0 T=t

_{1}–> T(0) = t

_{1}℃. \(\frac{dT}{dt} = -k(T-t_2)\)…….at T(0) = t

_{1}℃, now solving DE the above equation is of variable separable form i.e \(\int \frac{dt}{T-t_2} = \int -kdt + c\) =log (T-t

_{2}) = -kt + c –> T-t

_{2}= pe

^{-kt}…where p=e

^{c}=constant, using initial condition i.e T(0)= t

_{1}we get t

_{1}-t

_{2}=p substituting back in equation we obtain T=t

_{2}+(t

_{1}-t

_{2}) e

^{-kt}.

2. A body in air at 25℃ cools from 100℃ to 75℃ in 1 minute. What is the temperature of the body at the end of 3 minutes? (Take log(1.5)=0.4)

a) 40℃

b) 47.5℃

c) 42.5℃

d) 50℃

View Answer

Explanation: By Newton’s law of cooling w.k.t T = t

_{2}+ (t

_{1}-t

_{2}) e

^{-kt}, given t

_{1}=100℃, t

_{2}=25℃

when t=1 –> T(1) = 25 + 75 e

^{-k}= 75℃ –> 50/75 = 2/3 = e

^{-k}

–> 3/2=e

^{k}taking log k=log(1.5)=0.4.

to find T when t=3 minute using the value of k we get

T = 25 + 75e

^{-0.4*3}= 47.5℃……e

^{-1.2}=0.3.

3. A bottle of mineral water at a room temperature of 72℉ is kept in a refrigerator where the temperature is 44℉.After half an hour water cooled to 61℉.What is the temperature of the body in another half an hour?(Take log \(\frac{28}{17}\) = 0.498, e^{-0.99}=0.37)

a) 18℉

b) 9.4℉

c) 54.4℉

d) 36.4℉

View Answer

Explanation: By Newton’s law of cooling w.k.t T=t

_{2}+(t

_{1}-t

_{2}) e

^{-kt}, given t

_{1}=72℉, t

_{2}=44℉

At t=half an hour = 30mts T=61℉, finding k using the given values i.e

61=44+28e

^{-k30}–> \(\frac{17}{28}\) = e

^{-k30}or \(\frac{28}{17}\) = e

^{k30}taking log, log \(\frac{28}{17}\) = 30k –> k=0.0166

to find T when t = 30mts + 30mts = 60mts

T = 44 + 28e

^{-(0.0166)30}= 54.4℉.

4. The radius of the moon is roughly 2000km. The acceleration of gravity at the surface of the moon is about \(\frac{g}{6}\), where g is the acceleration of gravity at the surface of the earth. What is the velocity of escape for the moon?(Take g=10ms^{-2})

a) 2.58 kms^{-1}

b) 4.58 kms^{-1}

c) 6.28 kms^{-1}

d) 12.28 kms^{-1}

View Answer

Explanation: Let R be radius of the earth and r be the variable distance from Newton’s law \(a = \frac{dv}{dt} = \frac{k}{r^2}\) when r=R a=-g due to retardation of a body -gR

^{2}=k

substituting the value of k back we get \(\frac{dv}{dt} = \frac{-gR^2}{r^2} = \frac{dr}{dt} \frac{dv}{dr} = v \frac{dv}{dr}\) solving DE for v

we get \(\int v \,dv = \int \frac{-gR^2}{r^2} dr + c \rightarrow v^2 = \frac{2gR^2}{r} + C\) to find c we use at r=R

v=v

_{e}thus we get \(v_e^2 – \frac{2gR^2}{R} = C\)… where 2c=C=constant substituting value of c we get \(v^2=\frac{2gR^2}{r} + v_e^2 – 2gR\)…..if r>>R \(\frac{2gR^2}{r} = 0\) and particle to get escape from earth v≥0 –> v

_{e}

^{2}– 2gR≥0 –> \(v_e=\sqrt{2gR}\) to find v

_{e}

from the moon g becomes \(\frac{g}{6}\), R=2000km=2×10

^{6}m, g=10ms

^{-2}

therefore \(v_e = \sqrt{2*\frac{10}{6}(2×10^6)} = 2.58kms^{-1}.\).

**Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.**

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