A)
B)
C)
D)
Correct Answer: A
Solution :
From 0 to 1 hour, \[N'={{N}_{0}}{{e}^{t}}\] From 1 hour onwards\[\frac{dN}{dt}=-5{{N}^{2}}\] So at t = 1 hour, \[N'=e{{N}_{0}}\] \[\frac{dN}{dt}=-5{{N}^{2}}\] \[\int\limits_{e{{N}_{0}}}^{N}{{{N}^{-2}}}dN=-5\int\limits_{1}^{t}{dt}\] \[\frac{1}{N}-\frac{1}{e{{N}_{0}}}=5(t-1)\] \[\frac{{{N}_{0}}}{N}-\frac{1}{e}=5{{N}_{0}}(t-1)\] \[\frac{{{N}_{0}}}{N}=5{{N}_{0}}(t-1)+\frac{1}{e}\] \[\frac{{{N}_{0}}}{N}=5{{N}_{0}}t+\left( \frac{1}{e}-5{{N}_{0}} \right)\] which is following \[y=mx+C\]You need to login to perform this action.
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