[Newton's Law of Cooling: The temperature of body changes at a rate, which is proportional to the difference in temperature between that of the surrounding medium and that of the body itself.]
Let 'T' be the temperature of the body at time t.
By Newton's Law of cooling,
\(\frac { dT }{ dt } =k\ \left( T-70 \right) \)
\(\Rightarrow \frac { dT }{ T-70 } k\quad dt.\)   
|Variables Separable 
Integrating, \(\int { \frac { dT }{ T-70 } } =k\int { 1.dt+log|c| } \)
\(\Rightarrow log|T-70|-logc|=kt\)
\(\Rightarrow log|\frac { T-70 }{ c } |=\quad kt\quad \Rightarrow \frac { T-70 }{ c } ={ e }^{ k\quad t }\)
\(\Rightarrow T=70+c { e }^{ k\ t }\)
When  \(t=0,T=94.6,\ 94.6=70+c { e }^{ 0 } \Rightarrow c=24.6\)
Putting in(1),  \(t=0,T=94.6,\quad 94.6=70+c{ e }^{ 0 } \Rightarrow c=24.6\)
When \(=60,T=93.4,\quad 93.4=70+24.6\quad { e }^{ 60kt }\) 
\(\Rightarrow 24.6 { e }^{ 60kt } =23.4 \Rightarrow { e }^{ 60kt }=\frac { 23.4 }{ 24.6 } =\frac { 117 }{ 123 } \)
\(\Rightarrow 60\ k=log\frac { 117 }{ 123 } \Rightarrow k=\frac { 1 }{ 60 } log\frac { 117 }{ 123 } <0\) ...(2)
Thus the time t, that has elapsed after death is given by:
\(98.6=70+(24.6){ e }^{ kt }\)
\(\Rightarrow \left( 24.6 \right) { e }^{ kt }=28.6\Rightarrow { e }^{ kt }=\frac { 28.6 }{ 24.6 } =\frac { 143 }{ 123 } \)
\(\Rightarrow \)  \(t=\frac { 1 }{ k } log\frac { 143 }{ 123 } \)
\(=-3.01\)
Hence, the estimated time of death  \(=11.30-3.01\)\(=8.30\) P.M. approx.

We have: \(m\frac { dv }{ dt } +V\frac { dm }{ dt } =0\)
\(\Rightarrow m\frac { dv }{ dt } +V\frac { dm }{ dt } =0\) ...(1)
Integrating,  \(\int { dv+V\int { \frac { dm }{ m } =C } } \)
\(\Rightarrow \) \(v+V\quad logm=C\)..(2)\(\left[ \because m>0 \right] \)
When \(v=0,m={ m }_{ 0 }.\)
\(\therefore\)   \(0+\ V\ log{ m }_{ 0 }=C\Rightarrow C=V\ log{ m }_{ 0 }\)
Putting In (2),  \(v+Vlogm=V\ log{ m }_{ 0 }\)
\(\Rightarrow\) \(v=V(log{ m }_{ 0 }-log\ m).\)
Hence,  \(v=V\log\left( \frac { { m }_{ 0 } }{ m } \right) \)
We should not encourage rocket technology because they can be used nuclear warheads to attack other countries.

Let 'x' be the bacteria at any time 't'
By the question,\(\frac { dx }{ dt } =kx\)
\(\Rightarrow \)  \(\frac { dx }{ x } =k\quad dt\) 
|Variables Separable
Integrating, \(log|x|=kt+c\) ...(1)
When  \(t=0,x={ x }_{ 0 }\)
\(\therefore\)   \(t=0,x={ x }_{ 0 }\) ...(2)
When\(t=2,x=2{ x }_{ 0 }\)
\(\therefore\)  \(t=2,x=2{ x }_{ 0 }\) ...(3)
Subtracting (2) from (3), \(log|2{ x }_{ 0 }|-log|{ x }_{ 0 }|=2k\)
\(\Rightarrow \)  \(log2=2k\Rightarrow k=\frac { 1 }{ 2 } log\quad 2\)
Putting in (1), \(log|x|=\frac { 1 }{ 2 } 2.t+log|{ x }_{ 0 }|\)
When \(x=5{ x }_{ 0 },log|5{ x }_{ 0 }|=2.t+log|{ x }_{ 0 }|\)
\(\Rightarrow \)  \(log5=\frac { 1 }{ 2 } log\quad 2.t\)
\(\Rightarrow \)   \(t=2\frac { log5 }{ log2 } \)
Hence, the number of bacteria will be five times after 2\(\frac { log5 }{ log2 } \) hours.

Let 'y' be the number of bacteria at any time t.
By the question,  \(\frac { dy }{ dt } =ky\)
\(\Rightarrow \)  \(\frac { dy }{ y } =k\quad dt\)             
|Variables Separable
Integrating,  \(\int { \frac { dy }{ y } } =k\int { 1. } dt+c\)      
\(\Rightarrow \)  \(log|y|=kt+c\)              
\(\Rightarrow \)  \(log\quad y\quad =kt+c\) ...(1) \(\left[ \because \quad y>0 \right] \)
Let   \(y={ y }_{ 0 }\left( =1,00,000 \right) \) when \(t=0\)
\(\therefore \)  \(log\quad { y }_{ 0 }=0+c\Rightarrow c=log\quad { y }_{ 0 }\)
Putting in(1),\(log\quad { y }_{ 0 }=0+c\Rightarrow c=log\quad { y }_{ 0 }\)
\(\Rightarrow \)  \(log\frac { y }{ { y }_{ 0 } } =kt\)...(2)
By the question, when \(log\frac { y }{ { y }_{ 0 } } =kt\)
\(\therefore \)   \(log\frac { \frac { 11{ y }_{ 0 } }{ 10 } }{ { y }_{ 0 } } =2k\Rightarrow k=\frac { 1 }{ 1 } log\frac { 11 }{ 10 } .\)
Putting in (2), \(log \frac { y }{ { y }_{ 0 } } =\frac { 1 }{ 2 } \left( log\frac { 11 }{ 10 } \right) t.\)
When \(y=2{ y }_{ 0 }\left( 2,00,000 \right) ,\)  
then  \(log\frac { 2{ y }_{ 0 } }{ { y }_{ 0 } } =\frac { 1 }{ 2 } log\frac { 11 }{ 10 } t\)
\(\Rightarrow \)  \(log2=\frac { 1 }{ 2 } log\left( \frac { 11 }{ 10 } t \right) \)
\(\Rightarrow \) \(t=\frac { 2log2 }{ log\frac { 11 }{ 10 } } \)
Hence, the bacteria count will reach \(2,00,000\) after \(\frac { 2log2 }{ log\frac { 11 }{ 10 } } \)hours.

Let 'x' be the number of bacteria present at time 't'.
By the question, \(\frac { dx }{ dt } =kx\)
\(\Rightarrow\)  \(\frac { dx }{ x } =k\quad dt\)                
|Variables Separable 
Integrating, \(log|x|=kt+c\)..(1)
When \(t=0,x={ x }_{ 0 }\) (say)
\(\therefore \)  \(log|x|=kt+log|{ x }_{ 0 }|\)
Putting in(1),   \(log|x|=kt+log|{ x }_{ 0 }|\) ...(2)
When \(t=5,x=3{ x }_{ 0 }.\)
\(\therefore \)  \(log|3{ x }_{ 0 }|\quad =5kt+log|{ x }_{ 0 }|\)
\(\Rightarrow\)  \(log3=5k\Rightarrow k=\frac { 1 }{ 5 } log3.\)
Putting in (2), \(log|x|=\frac { 1 }{ 5 } log\quad 3t+log|{ x }_{ 0 }|\)...(3)
(i) When \(t=10,log|x|=2log\quad 3+log|{ x }_{ 0 }|=log|9{ x }_{ 0 }|\)
\(\Rightarrow\)\(x=9{ x }_{ 0 }.\)
Hence, the bacteria will be 9 times after 10 hours.
(ii) When \(x=10{ x }_{ 0 }\)
\(\therefore\) From(3),\(log|10{ x }_{ 0 }|=\frac { 1 }{ 5 } log\quad 3.t+log|{ x }_{ 0 }|\)
\(\Rightarrow \)  \(log\quad 10=\frac { 1 }{ 5 } log\quad 3.t\)
\(\Rightarrow \)  \(t=5\frac { log10 }{ log3 } \)
Hence, the number of bacteria will be 10 times after \(5\frac { log10 }{ log3 } \) hours.