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Question: The rate at which ice-ball melts is proportional to the amount of ice in it. If half of it melts in ...

The rate at which ice-ball melts is proportional to the amount of ice in it. If half of it melts in 20 minutes, the amount of ice after 40 minutes compared to it original amount is
A.(18)th\left( {\dfrac{1}{8}} \right)th
B.(116)th\left( {\dfrac{1}{{16}}} \right)th
C.(14)th\left( {\dfrac{1}{4}} \right)th
D.(132)th\left( {\dfrac{1}{{32}}} \right)th

Explanation

Solution

Hint: Here, we will first rewrite the given conditions in the mathematical form, that is, dxdtx\dfrac{{dx}}{{dt}} \propto x, where xx is the amount of ice and tt is time. Then integrate the given equation using separate variables methods. Apply this, and then use the given conditions to find the required value.

Complete step-by-step answer:
We are given that the rate at which ice-ball melts is proportional to the amount of ice in it.
Let us assume that the amount of ice is xx.
We will now rewrite the given statement in mathematical form, that is, dxdtx\dfrac{{dx}}{{dt}} \propto x, where the amount of ice is x and t is time.
Separating the variables and in the above form, we get
dxx=kdt\dfrac{{dx}}{x} = kdt.
, where is any constantxx
Integrating the above equation on each of the sides, we get

dxx=kdt logex=kt  \Rightarrow \int {\dfrac{{dx}}{x}} = \int {kdt} \\\ \Rightarrow {\log _e}x = kt \\\

Applying exponential on both sides in above equation, we get

elogex=ekt x=ekt ......(1)  \Rightarrow {e^{{{\log }_e}x}} = {e^{kt}} \\\ \Rightarrow x = {e^{kt}}{\text{ ......}}\left( 1 \right) \\\

Replacing 12\dfrac{1}{2} for xx and 20 minutes for tt in the above equation, we get

12=e20k (12)120=ek ek=(12)120  \Rightarrow \dfrac{1}{2} = {e^{20k}} \\\ \Rightarrow {\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{{20}}}} = {e^k} \\\ \Rightarrow {e^k} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{{20}}}} \\\

We will now replace 40 for xx in the equation (1)\left( 1 \right) to find the amount of ice after 40 minutes, we get

x=e40k x=(ek)40  \Rightarrow x = {e^{40k}} \\\ \Rightarrow x = \left( {{e^k}} \right)40 \\\

Substituting the value of ek{e^k} in the above equation, we get

x=[(12)120]40 x=(12)2 x=14  \Rightarrow x = {\left[ {{{\left( {\dfrac{1}{2}} \right)}^{\dfrac{1}{{20}}}}} \right]^{40}} \\\ \Rightarrow x = {\left( {\dfrac{1}{2}} \right)^2} \\\ \Rightarrow x = \dfrac{1}{4} \\\

Therefore, we have found that the amount of ice left is (14)th\left( {\dfrac{1}{4}} \right)th of the original amount.

Hence, the option is C will be correct.

Note: In this question, we will first assume the amount of ice by any variable. Students should write the given information into the mathematical form for better understanding. We should know that the equation for proportional, dxdtx\dfrac{{dx}}{{dt}} \propto x, where xx is the amount of ice and tt is time, can be written in an equal form by multiplying the left hand side by some constant kk, dxx=kdt\dfrac{{dx}}{x} = kdt.