1.The population of India in the year 2000 was 1 billion and it increased expone
ID: 1766918 • Letter: 1
Question
1.The population of India in the year 2000 was 1 billion and it increased exponentially at a rate of 1.6% per year. If the growth rate is maintained, what will be the population in the year 2020? If the growth rate is decreased to 1.2% per year from 2020 onwards and is maintained at that level, what will be the population in the year 2050? Assuming the average human exhales 2.3 pounds of carbon dioxide on an average day, what is the total amount of Carbon released in the atmosphere annually by human exhalation (in billion tonnes) around the year 2020?
2.Calculate the suspended particulate concentration (in g/m3) in a sample collected through a hi-vol. sampler: Weight of clean filter = 5.00 g, Weight of the filter after exposure for 24 hours = 5.38 g, Average air flow = 2000 m3 in 24 hours.
Explanation / Answer
(1) The population of India in in 2000(A0) = 1billion
Rate of exponential increase/year(k) = 1.6% = 1.6/100 = 0.016
Time (t) = 20 years
The growth function is given by , A = A0 e(kt) = 1billion *(e(0.016*20))
Population of india in year 2020 = 1billion * 1.3770 = 1.377 billion.
If the growth rate is reduced to 1.2% from 2020, k= 1.2/100= 0.012
Time (t) = 2020-2050= 30years
Now , the new A0 (population in 2020) will be 1.377 billion
So the population of India in 2050, A = A0 (e(kt)) = 1.377 (e(0.012*30)) = 1.9736 billion.
Assuming the average human exales 2.3 pounds of CO2/ day i.e 0.00104326 tonnes
Then annual human exalation around the year 2020 = 1.377*0.00104326*365 = 0.5243 billion tonnes/year
(2) Weight of the clean filter = 5.00g
Weight of the filter after exposure for 24 hours = 5.38g
So, weight of the perticulates = 5.38-5 = 0.38g
Average air flow = 2000 m3
concentration in microgram/m3 = (0.38*1000000)/2000 = 190 microgram/m3
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