Differential Equations Test

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Differential Equations Test - 11

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Question 1
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:

A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6°F. He took the temperature again after one hour; the temperature was lower than the first observation. It was 93.4°F. The room in which the cat was put is always at 70°F. The normal temperature of the cat is taken as 98.6°F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: dT/dt ∞ (T−70), where 70°F is the room temperature and T is the temperature of the object at time t.

Substituting the two different observations of T and t made, in the solution of the differential equation dT/dt = k(T - 70) where k is a constant of proportion, time of death is calculated.

Q. The solution of the differential equation dT/dt = k(T - 70) is given by,

A
A. log |T – 70| = kt + C

B
B. log |T – 70| = log |kt|+ C

C
C. T – 70 = kt + C

D
D. T – 70 = kt C
Question 2
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:

A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6°F. He took the temperature again after one hour; the temperature was lower than the first observation. It was 93.4°F. The room in which the cat was put is always at 70°F. The normal temperature of the cat is taken as 98.6°F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: dT/dt ∞ (T−70), where 70°F is the room temperature and T is the temperature of the object at time t.

Substituting the two different observations of T and t made, in the solution of the differential equation dT/dt = k(T - 70) where k is a constant of proportion, time of death is calculated.

Q. Which method of solving a differential equation helped in calculation of the time of death?

A
A. Variable separable method

B
B. Solving Homogeneous differential equation

C
C. Solving Linear differential equation

D
D. All of the above
Question 3
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:

A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6°F. He took the temperature again after one hour; the temperature was lower than the first observation. It was 93.4°F. The room in which the cat was put is always at 70°F. The normal temperature of the cat is taken as 98.6°F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: dT/dt ∞ (T−70), where 70°F is the room temperature and T is the temperature of the object at time t.

Substituting the two different observations of T and t made, in the solution of the differential equation dT/dt = k(T - 70) where k is a constant of proportion, time of death is calculated.

Q. What will be the degree of the above given differential equation.

A
A. 2

B
B. 1

C
C. 0

D
D. 3
Question 4
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:

A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6°F. He took the temperature again after one hour; the temperature was lower than the first observation. It was 93.4°F. The room in which the cat was put is always at 70°F. The normal temperature of the cat is taken as 98.6°F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: dT/dt ∞ (T−70), where 70°F is the room temperature and T is the temperature of the object at time t.

Substituting the two different observations of T and t made, in the solution of the differential equation dT/dt = k(T - 70) where k is a constant of proportion, time of death is calculated.

Q. If the temperature was measured 2 hours after 11.30 pm, what will be the change in time of death?

A
A. No change

B
B. Death time increased

C
C. Death time decreased

D
D. Death time always constant
Question 5
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:

A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6°F. He took the temperature again after one hour; the temperature was lower than the first observation. It was 93.4°F. The room in which the cat was put is always at 70°F. The normal temperature of the cat is taken as 98.6°F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation: dT/dt ∞ (T−70), where 70°F is the room temperature and T is the temperature of the object at time t.

Substituting the two different observations of T and t made, in the solution of the differential equation dT/dt = k(T - 70) where k is a constant of proportion, time of death is calculated.

Q. If t = 0 when T is 72, then the value of C is

A
A. -2

B
B. 0

C
C. 2

D
D. log 2
Question 6
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours.

Q. If N₀is the initial count of bacteria, after 10 hours the count is _______.

A
A. 1/5 log3

B
B. 3log N₀

C
C. 9N₀

D
D. 2N₀

Solution

Given when t = 0, N = N₀ .

From (i), log|N₀ | = C

∴ (i) ⇒ log|N| = Kt + log|N₀ |

⇒ ….(ii)

Given when t = 5, N = 3N₀ .

From (ii), log|3| = 5K

⇒ K = 1/5 log3

∴ The particular solution is

When t = 10,

N/N₀ = 9

⇒ N = 9N₀
Question 7
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours.

Q. If ‘N’ is the number of bacteria, the corresponding differential equation is _______.

A
A. dN/dt = Kt

B
B. dN/dt = KN

C
C. dK/dt = N

D
D. dK/dN = t

Solution

Given that N is the number of bacteria.

⇒ dN/dt = KN
Question 8
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours.

Q.

A
A. log|x| + C

B
B. log|Kx| + C

C
C.

D
D.

Solution
Question 9
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours.

Q. The general solution is _______.

A
A. log|N| = Kt + C

B
B. log|Nt| = K + C

C
C. log|N| = t

D
D. log|Kt| = N + C

Solution

log|N| = Kt + C ...(i)
Question 10
1 / -0.25

Direction: Read the following text and answer the following questions on the basis of the same:

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours.

Q. The bacteria becomes 10 times in _______ hours.

A
A. 5log 7

B
B. 5log10/log3

C
C. 5log/3

D
D. log(10⁵ /3)

Solution

Given N = 10N₀ ,