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Then,

$dtdN =kN$ where $k$ is the proportionality constant.

$NdN =kdt$

$∫_{N_{0}}NdN =∫_{0}kdt$

$ln(N_{0}N )=kt$

$N=N_{0}e_{kt}$ ...(i)

If $N_{0}=10_{5}$ and $t=2hours$ and

$N=(1+10010 )×10_{5}$

$=1.1×10_{5}$

Hence

$lnN_{0}N =kt$

$ln(10_{5}1.1×10_{5} )=2k$

$ln(1.1)=2k$

$k=21 ln(1.1)hours_{−1}$

$=ln(1.1 )hours_{−1}$

Now

$ln(N_{0}N )=kt$

N is now $2×10_{5}$

Hence

$N_{0}N $

$=10_{5}2×10_{5} $

$=2$

Thus

$ln(N_{0}N )=kt$ implies

$ln2=ln1.1 t$

$ln2=2ln(1.1) t$

$2ln2=ln(1.1)t$

$t=ln(1.1)ln4 $

$=14.54$ hours

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