In 10 days, 1/4th of the substance disappears and 3/4th of the substance is left.
It is given that at any given point of time the amount of substance disappearing is directly proportional to the amount of substance present at that moment.
In another 10 days 1/4th of the substance left will disappear, therefore, 3/16 of the substance will disappear.
∴ Total amount of substance disappeared in 20 days = 1/4 + 3/16 = 7/16 < 8/16 = 1/2
∴ Number of days taken for half the substance to disappear is more than 20 days.
∴We can eliminate options 1 and 3.
Amount of substance left after 20 days = 3/4 − 3/16 = 9/16
In another 10 days, the amount of substance that will disappear = 1/4 × Amount of substance left after 20 days
= 9/64
∴ Total amount of substance that has disappeared in 30 days = 7/16 + 9/64 = 37/64 > 32/64 = 1/2
∴ It will take less than 30 days for half of the substance to disappear (or for half of the substance to be left).
∴ We can eliminate option 4.
Hence, option 2.
Alternatively,
Let A be the amount of substance left at any time t.
Since the rate of decay of the radioactive substance is proportional to the amount of radioactive substance left.
We have,
dA/dt = −kA, where k is constant.
∴ dA/A = −kdt
Integrating both the sides we get,
ln(A) = –kt + c
Let the initial amount be A_{0}._{ }Then at t = 0,
c = ln(A_{0})
∴ln(A) = –kt + ln(A_{0})
∴ln(A/A_{0}) = –kt
∴ A = A_{0}e^{−kt} …(i)
By the condition given in the question, we have that,
At t = 10 days, we have, A = 0.75A_{0}, substituting the given values in (i), we get,
e^{−10k }= 0.75^{} Taking the natural logarithm of both sides, we get,
–10k = ln(0.75) = –0.29
∴ k = 0.029 …(ii)
For A = 0.5K, using (i) and (ii), we get
e^{−kt} = 0.5
Taking the natural logarithm of both sides, we get,
–0.029t = ln(0.5) = –0.693
∴ t ≈ 693/29 ≈ 23.8 days
Hence, option 2.
