Question

Match list I with list II and choose the correct options.

List I	List II
(a) Pacific salmon fish	1. Verhulst-Pearl logistic growth
(b) $N_t=N_0{e}^{et}$	2. Breeds only once in lifetime
(c) Oyster	3. Exponential growth
(d) $dN/dt=rN\left(\frac{K-N}{K}\right)$	4. A large number of small sized offspring

• A. (a)- 4, (b)- 3, (c )- 1, (d)- 2
• B. (a)- 3, (b)- 4, (c)- 1, (d)- 2
• C. (a)- 3, (b)- 1, (c)- 4, (d - 2
• D. (a)- 2, (b)- 3, (c)- 4, (d)- 1

Answer 1

The correct option is D (a)- 2, (b)- 3, (c)- 4, (d)- 1

(a)- 2: Pacific salmon fish is called as a semelparous type of fish. These fishes live in the ocean water. They swim and then lay eggs in the fresh water. They spawn and then die. These type of animals spawn only once in their lifetime and die after spawning.

(b)- 3: The exponential function appears when original quantity grows exponentially over time. If a quantity N increases at a rate proportional to the amount present at time t, then the quantity can be written as N(t) = N${}_0$ekt, where N${}_0$ is the value of N at time t = 0 and N increases as t increases and k is called an exponential growth function.

(c)- 4: There are variations found in the production of offspring in different animals. Some of the animals produce small sized offsprings which are many in number. The examples of such animals are oysters and pelagic fishes.

(d)- 1: In a finite world, no population can grow exponentially for very long. Sooner or later every population must encounter either difficult environmental conditions or shortages of its requisites for reproduction. This is explained by the Verhulst-Pearl Logistic growth equation which is given in the form of dN/dt=rN(K−N/K), where N denotes population, K is the carrying capacity and r shows the rate of increase in the population.

So, the correct answer is option D.