The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t = 0. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after k/(log ₑ( 6/5 )) hours, then ( k/(log ₑ2) )² is equal to :

Solution

Maths sir · 4 steps

Use the 20% growth to find the growth constant, then ask how long it takes to double.

1. Separate and integrate. Growth proportional to population means dN/dt = λ N. Separating gives dN/N = λ dt, and with N = 1000 at t = 0 this integrates to N = 1000 e^(λ t).
2. Find the constant. After 2 hours, N = 1200, so e^(2λ) = 6/5. This gives λ = 1/2 logₑ(6/5).
3. Time to double. For N = 2000 we need e^(λ t) = 2, so t = (logₑ 2)/λ.

   t = (2 logₑ 2)/logₑ(6/5) = k/logₑ(6/5) ⇒ k = 2 logₑ 2

4. So. k/(logₑ 2) = 2, and squaring gives (k/(logₑ 2))² = 4. The answer is (d).

The idea to keep. Proportional growth always gives N = N₀ e^(λ t). Use one given data point to fix λ, then take logs to find any time you need.

Answer from the NTA final answer key. Solution by PrepWiser.