


Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 9.3
Ex 9.3, 2
Ex 9.3, 3
Ex 9.3, 4 Important
Ex 9.3, 5
Ex 9.3, 6
Ex 9.3, 7 Important
Ex 9.3, 8
Ex 9.3, 9 Important
Ex 9.3, 10 Important
Ex 9.3, 11 Important
Ex 9.3, 12
Ex 9.3, 13
Ex 9.3, 14
Ex 9.3, 15 Important
Ex 9.3, 16
Ex 9.3, 17 Important
Ex 9.3, 18
Ex 9.3, 19 Important
Ex 9.3, 20 Important
Ex 9.3, 21
Ex 9.3, 22 Important You are here
Ex 9.3, 23 (MCQ)
Last updated at May 29, 2023 by Teachoo
Ex 9.3, 22 In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours, In how many hours will the count reach 2,00,000 , if the rate of growth of bacteria is proportional to the number present? Let the number of bacteria at time t be y Given that rate of growth of bacteria is proportional to the number present 𝑑𝑦/𝑑𝑡 ∝ y 𝑑𝑦/𝑑𝑡 = ky 𝑑𝑦/𝑦 = kdt Integrating both sides ∫1▒〖𝑑𝑦/𝑦=𝑘〗 ∫1▒𝑑𝑡 log y = kt + C Now, according to question The bacteria count is 1,00,000. The number is increased by 10% in 2 hours, In how many hours will the count reach 2,00,000 Putting t = 0 and y = 1,00,000 in (1) log 1,00,000 = k × 0 + C C = log 1,00,000 …(1) Putting value of C in (1) log y = kt + C log y = kt + log 1,00,000 Now, Putting t = 2 and y = 1,00,000 in (2) log 1,10,000 = 2k + log 1,00,000 log 1,10,000 − log 1,00,000 = 2k log (1,10,000/1,00,000) = 2k 1/2 log (11/10) = k …(2) Putting value of k in (2) log y = kt + log 1,00,000 log y = 1/2 log (11/10) t + log 1,00,000 Now, If Bacterial = 2,00,000, we have to find t Putting y = 2,00,000 in (3) log 2,00,000 = 1/2 log (11/10) t + log (1,00,000) log 2,00,000 − log 1,00,000 = 1/2 log (11/10) t log (2,00,000/1,00,000) = 1/2 log (11/10) t log 2 = 1/2 log (11/10) t t = (𝟐 𝐥𝐨𝐠𝟐)/𝐥𝐨𝐠〖 (𝟏𝟏/𝟏𝟎)〗 …(3)