NEET Test Series from KOTA - 10 Papers In MS WORD
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Differential Equation
87580
Bactria increases at the rate proportional to the number of bacteria present. If the original number \(\mathbf{N}\) doubles in \(\mathbf{4}\) hours, then the number of bacteria will be \(4 \mathbf{N}\) in
1 2 hours
2 8 hours
3 4 hours
4 6 hours
Explanation:
(B) : Given, \(\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}\) Integrating on both sides we get - \(\int_{\mathrm{N}}^{\mathrm{N}_{\mathrm{I}}} \frac{\mathrm{dN}}{\mathrm{N}}=\int_{0}^{\mathrm{t}} \mathrm{kdt}\) \([\log \mathrm{N}]_{\mathrm{N}}^{\mathrm{N}_{1}}=\mathrm{k}[\mathrm{t}]_{0}^{\mathrm{t}}\) \(\log \mathrm{N}_{1}-\log \mathrm{N}=\mathrm{kt}\) \(\log \frac{\mathrm{N}_{1}}{\mathrm{~N}}=\mathrm{kt}+\mathrm{c}\) \(\frac{\mathrm{N}_{1}}{\mathrm{~N}}=\mathrm{e}^{\mathrm{kt}}+\mathrm{c}\) \(\mathrm{N}_{1}=2 \mathrm{~N}\) When \(\mathrm{t}=4\) \(2 \mathrm{~N}=\mathrm{N}^{\mathrm{k}} \mathrm{e}^{\mathrm{k} .4}\) \(2=\mathrm{e}^{4}\) \((2)^{1 / 4}=e^{\mathrm{k}}\) \(\mathrm{N}_{1}=\mathrm{N} \mathrm{e}^{\mathrm{kt}}\) \(\mathrm{N}_{1}=\mathrm{N}\left(2^{\frac{1}{4} \mathrm{t}}\right)\) Now \(4 \mathrm{~N}=\mathrm{N}(2)^{1 / 4}\) \(2^{\frac{\mathrm{t}}{4}}=4\) \(\frac{\mathrm{t}}{4}=2\) \(\mathrm{t}=8\) Hour
MHT CET-2020
Differential Equation
87581
If the population grows at the rate of \(8 \%\) per year, then the time taken for the population to be doubled is (given \(\log 2=\mathbf{0 . 6 9 1 2}\) )
1 10.27 years
2 8.64 years
3 4.3 years
4 6.8 years
Explanation:
(B) : Let ' \(\mathrm{P}\) ' be the population at any time \(\mathrm{t}\) we have given. The population grows at the rate of \(8 \%\) per year Then \(\frac{\mathrm{dP}}{\mathrm{dt}}=8 \% \mathrm{P}\) \(\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{8}{100} \mathrm{P}\) On integrating both sides we get - \(\int \frac{\mathrm{dP}}{\mathrm{P}}=\int \frac{8}{100} \mathrm{dt}\) \(\ln \mathrm{P}=\frac{8}{100} \mathrm{t}+\mathrm{c}\) When \(\mathrm{t}=0 \quad\) then \(\mathrm{P}=\mathrm{P}_{0}\) \(\ln \mathrm{P}_{0}=\mathrm{c}\) \(\text { We find when } \mathrm{P}=2 \mathrm{P}_{0}\) \(\ln 2 \mathrm{P}_{0}=\frac{8}{100} \mathrm{t}+\log \mathrm{P}_{\mathrm{o}}\) \(\ln \left(\frac{2 \mathrm{P}_{0}}{\mathrm{P}_{0}}\right)=\frac{8}{100} \mathrm{t}\) \(\ln 2=\frac{8}{100} \mathrm{t}\) \(\mathrm{t}=0.6912 \times \frac{100}{8}=8.64 \text { years }\)
MHT CET-2020
Differential Equation
87582
Water at \(100^{\circ} \mathrm{C}\) cools in 15 minutes to \(75^{\circ} \mathrm{C}\) at a room temperature of \(25^{\circ} \mathrm{C}\). Then the temperature of the water after 30 minutes is
(C) : Let ' \(\mathrm{T}\) ' be the temperature at any time ' \(\mathrm{t}\) ' for given condition \(\frac{\mathrm{dT}}{\mathrm{dt}} \propto(\mathrm{T}-25)\) \(\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-25)\) \(\frac{\mathrm{dT}}{\mathrm{T}-25}=-\mathrm{kdt}\) Integrating on both sides, we get - \(\log (\mathrm{T}-25)=-\mathrm{kt}+\mathrm{c}\) When \(\mathrm{t}=0\) and \(\mathrm{T}=100^{\circ} \mathrm{C}\) \(\log (100-25)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log 75\) Then equation become \(\log (\mathrm{T}-25)=-\mathrm{kt}+\log 75\) Again, when \(\mathrm{t}=15\) and \(\mathrm{T}=75^{\circ} \mathrm{C}\) \(\log \left(\frac{\mathrm{T}-25}{75}\right)=-\mathrm{kt}\) \(\mathrm{k} \times 15=\log \left(\frac{75}{75-25}\right)\) \(\mathrm{k}=\frac{1}{15} \log \left(\frac{3}{2}\right)\) Putting the value of \(\mathrm{k}\) in equation (ii), we get - \(\frac{1}{15} \log \left(\frac{3}{2}\right) \mathrm{t}=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) When \(\mathrm{t}=30\) then find \(\mathrm{T}\) \(30 \times \frac{1}{15} \log \left(\frac{3}{2}\right)=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) \(2 \log \left(\frac{3}{2}\right)=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) \(\log \frac{9}{4}=\log \frac{75}{\mathrm{~T}-25}\) \(\frac{9}{4}=\frac{75}{\mathrm{~T}-25}\) \(9 \mathrm{~T}-225=300\) \(9 \mathrm{~T}=525\) \(\mathrm{T}=\frac{525}{9}=\left(\frac{175}{3}\right)^{\circ} \mathrm{C}\)
MHT CET-2020
Differential Equation
87583
A body cools according to Newton's law from \(1000 \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) in 20 minutes. If the temperature of the surrounding is \(20^{\circ} \mathrm{C}\), then the temperature of the body after one hour is
1 \(40^{\circ} \mathrm{C}\)
2 \(15^{\circ} \mathrm{C}\)
3 \(30^{\circ} \mathrm{C}\)
4 \(20^{\circ} \mathrm{C}\)
Explanation:
(C) : Let ' \(\theta\) ' be the temperature of body at time ' \(t\) '. The temperature of surrounding is \(20^{\circ} \mathrm{C}\). According to Newton's law of cooling - \(\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto \theta-20\) \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-20)\) Integrating on both sides, we get - \(\int \frac{\mathrm{d} \theta}{\theta-20}=\int-\mathrm{kdt}\) \(\log (\theta-20)=-\mathrm{kt}+\mathrm{c}\) Given, when \(\mathrm{t}=0\) and \(\theta=100\) \(\log (100-20)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log 80\) Thus, equation become \(\log \left(\frac{\theta-20}{80}\right)=-\mathrm{kt}\) Now, When \(\mathrm{t}=20\) then \(\theta=60\) \(\log \left(\frac{60-20}{80}\right)=-\mathrm{k} \times 20\) \(\mathrm{k}=-\frac{1}{20} \log \left(\frac{1}{2}\right)\) Putting value of \(\mathrm{k}\) the equation becomes \(\log \left(\frac{\theta-20}{80}\right)=\frac{\mathrm{t}}{20} \log \left(\frac{1}{2}\right)\) We find temperature after 1 hour \(\mathrm{t}=60\) \(\log \left(\frac{\theta-20}{80}\right)=\frac{60}{20} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\theta-20}{80}\right)=\log \left(\frac{1}{2}\right)^{3}\) \(\frac{\theta-20}{80}=\frac{1}{8}\) \(\theta-20=10\) \(\theta=30^{\circ} \mathrm{C}\)
87580
Bactria increases at the rate proportional to the number of bacteria present. If the original number \(\mathbf{N}\) doubles in \(\mathbf{4}\) hours, then the number of bacteria will be \(4 \mathbf{N}\) in
1 2 hours
2 8 hours
3 4 hours
4 6 hours
Explanation:
(B) : Given, \(\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}\) Integrating on both sides we get - \(\int_{\mathrm{N}}^{\mathrm{N}_{\mathrm{I}}} \frac{\mathrm{dN}}{\mathrm{N}}=\int_{0}^{\mathrm{t}} \mathrm{kdt}\) \([\log \mathrm{N}]_{\mathrm{N}}^{\mathrm{N}_{1}}=\mathrm{k}[\mathrm{t}]_{0}^{\mathrm{t}}\) \(\log \mathrm{N}_{1}-\log \mathrm{N}=\mathrm{kt}\) \(\log \frac{\mathrm{N}_{1}}{\mathrm{~N}}=\mathrm{kt}+\mathrm{c}\) \(\frac{\mathrm{N}_{1}}{\mathrm{~N}}=\mathrm{e}^{\mathrm{kt}}+\mathrm{c}\) \(\mathrm{N}_{1}=2 \mathrm{~N}\) When \(\mathrm{t}=4\) \(2 \mathrm{~N}=\mathrm{N}^{\mathrm{k}} \mathrm{e}^{\mathrm{k} .4}\) \(2=\mathrm{e}^{4}\) \((2)^{1 / 4}=e^{\mathrm{k}}\) \(\mathrm{N}_{1}=\mathrm{N} \mathrm{e}^{\mathrm{kt}}\) \(\mathrm{N}_{1}=\mathrm{N}\left(2^{\frac{1}{4} \mathrm{t}}\right)\) Now \(4 \mathrm{~N}=\mathrm{N}(2)^{1 / 4}\) \(2^{\frac{\mathrm{t}}{4}}=4\) \(\frac{\mathrm{t}}{4}=2\) \(\mathrm{t}=8\) Hour
MHT CET-2020
Differential Equation
87581
If the population grows at the rate of \(8 \%\) per year, then the time taken for the population to be doubled is (given \(\log 2=\mathbf{0 . 6 9 1 2}\) )
1 10.27 years
2 8.64 years
3 4.3 years
4 6.8 years
Explanation:
(B) : Let ' \(\mathrm{P}\) ' be the population at any time \(\mathrm{t}\) we have given. The population grows at the rate of \(8 \%\) per year Then \(\frac{\mathrm{dP}}{\mathrm{dt}}=8 \% \mathrm{P}\) \(\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{8}{100} \mathrm{P}\) On integrating both sides we get - \(\int \frac{\mathrm{dP}}{\mathrm{P}}=\int \frac{8}{100} \mathrm{dt}\) \(\ln \mathrm{P}=\frac{8}{100} \mathrm{t}+\mathrm{c}\) When \(\mathrm{t}=0 \quad\) then \(\mathrm{P}=\mathrm{P}_{0}\) \(\ln \mathrm{P}_{0}=\mathrm{c}\) \(\text { We find when } \mathrm{P}=2 \mathrm{P}_{0}\) \(\ln 2 \mathrm{P}_{0}=\frac{8}{100} \mathrm{t}+\log \mathrm{P}_{\mathrm{o}}\) \(\ln \left(\frac{2 \mathrm{P}_{0}}{\mathrm{P}_{0}}\right)=\frac{8}{100} \mathrm{t}\) \(\ln 2=\frac{8}{100} \mathrm{t}\) \(\mathrm{t}=0.6912 \times \frac{100}{8}=8.64 \text { years }\)
MHT CET-2020
Differential Equation
87582
Water at \(100^{\circ} \mathrm{C}\) cools in 15 minutes to \(75^{\circ} \mathrm{C}\) at a room temperature of \(25^{\circ} \mathrm{C}\). Then the temperature of the water after 30 minutes is
(C) : Let ' \(\mathrm{T}\) ' be the temperature at any time ' \(\mathrm{t}\) ' for given condition \(\frac{\mathrm{dT}}{\mathrm{dt}} \propto(\mathrm{T}-25)\) \(\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-25)\) \(\frac{\mathrm{dT}}{\mathrm{T}-25}=-\mathrm{kdt}\) Integrating on both sides, we get - \(\log (\mathrm{T}-25)=-\mathrm{kt}+\mathrm{c}\) When \(\mathrm{t}=0\) and \(\mathrm{T}=100^{\circ} \mathrm{C}\) \(\log (100-25)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log 75\) Then equation become \(\log (\mathrm{T}-25)=-\mathrm{kt}+\log 75\) Again, when \(\mathrm{t}=15\) and \(\mathrm{T}=75^{\circ} \mathrm{C}\) \(\log \left(\frac{\mathrm{T}-25}{75}\right)=-\mathrm{kt}\) \(\mathrm{k} \times 15=\log \left(\frac{75}{75-25}\right)\) \(\mathrm{k}=\frac{1}{15} \log \left(\frac{3}{2}\right)\) Putting the value of \(\mathrm{k}\) in equation (ii), we get - \(\frac{1}{15} \log \left(\frac{3}{2}\right) \mathrm{t}=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) When \(\mathrm{t}=30\) then find \(\mathrm{T}\) \(30 \times \frac{1}{15} \log \left(\frac{3}{2}\right)=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) \(2 \log \left(\frac{3}{2}\right)=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) \(\log \frac{9}{4}=\log \frac{75}{\mathrm{~T}-25}\) \(\frac{9}{4}=\frac{75}{\mathrm{~T}-25}\) \(9 \mathrm{~T}-225=300\) \(9 \mathrm{~T}=525\) \(\mathrm{T}=\frac{525}{9}=\left(\frac{175}{3}\right)^{\circ} \mathrm{C}\)
MHT CET-2020
Differential Equation
87583
A body cools according to Newton's law from \(1000 \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) in 20 minutes. If the temperature of the surrounding is \(20^{\circ} \mathrm{C}\), then the temperature of the body after one hour is
1 \(40^{\circ} \mathrm{C}\)
2 \(15^{\circ} \mathrm{C}\)
3 \(30^{\circ} \mathrm{C}\)
4 \(20^{\circ} \mathrm{C}\)
Explanation:
(C) : Let ' \(\theta\) ' be the temperature of body at time ' \(t\) '. The temperature of surrounding is \(20^{\circ} \mathrm{C}\). According to Newton's law of cooling - \(\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto \theta-20\) \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-20)\) Integrating on both sides, we get - \(\int \frac{\mathrm{d} \theta}{\theta-20}=\int-\mathrm{kdt}\) \(\log (\theta-20)=-\mathrm{kt}+\mathrm{c}\) Given, when \(\mathrm{t}=0\) and \(\theta=100\) \(\log (100-20)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log 80\) Thus, equation become \(\log \left(\frac{\theta-20}{80}\right)=-\mathrm{kt}\) Now, When \(\mathrm{t}=20\) then \(\theta=60\) \(\log \left(\frac{60-20}{80}\right)=-\mathrm{k} \times 20\) \(\mathrm{k}=-\frac{1}{20} \log \left(\frac{1}{2}\right)\) Putting value of \(\mathrm{k}\) the equation becomes \(\log \left(\frac{\theta-20}{80}\right)=\frac{\mathrm{t}}{20} \log \left(\frac{1}{2}\right)\) We find temperature after 1 hour \(\mathrm{t}=60\) \(\log \left(\frac{\theta-20}{80}\right)=\frac{60}{20} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\theta-20}{80}\right)=\log \left(\frac{1}{2}\right)^{3}\) \(\frac{\theta-20}{80}=\frac{1}{8}\) \(\theta-20=10\) \(\theta=30^{\circ} \mathrm{C}\)
87580
Bactria increases at the rate proportional to the number of bacteria present. If the original number \(\mathbf{N}\) doubles in \(\mathbf{4}\) hours, then the number of bacteria will be \(4 \mathbf{N}\) in
1 2 hours
2 8 hours
3 4 hours
4 6 hours
Explanation:
(B) : Given, \(\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}\) Integrating on both sides we get - \(\int_{\mathrm{N}}^{\mathrm{N}_{\mathrm{I}}} \frac{\mathrm{dN}}{\mathrm{N}}=\int_{0}^{\mathrm{t}} \mathrm{kdt}\) \([\log \mathrm{N}]_{\mathrm{N}}^{\mathrm{N}_{1}}=\mathrm{k}[\mathrm{t}]_{0}^{\mathrm{t}}\) \(\log \mathrm{N}_{1}-\log \mathrm{N}=\mathrm{kt}\) \(\log \frac{\mathrm{N}_{1}}{\mathrm{~N}}=\mathrm{kt}+\mathrm{c}\) \(\frac{\mathrm{N}_{1}}{\mathrm{~N}}=\mathrm{e}^{\mathrm{kt}}+\mathrm{c}\) \(\mathrm{N}_{1}=2 \mathrm{~N}\) When \(\mathrm{t}=4\) \(2 \mathrm{~N}=\mathrm{N}^{\mathrm{k}} \mathrm{e}^{\mathrm{k} .4}\) \(2=\mathrm{e}^{4}\) \((2)^{1 / 4}=e^{\mathrm{k}}\) \(\mathrm{N}_{1}=\mathrm{N} \mathrm{e}^{\mathrm{kt}}\) \(\mathrm{N}_{1}=\mathrm{N}\left(2^{\frac{1}{4} \mathrm{t}}\right)\) Now \(4 \mathrm{~N}=\mathrm{N}(2)^{1 / 4}\) \(2^{\frac{\mathrm{t}}{4}}=4\) \(\frac{\mathrm{t}}{4}=2\) \(\mathrm{t}=8\) Hour
MHT CET-2020
Differential Equation
87581
If the population grows at the rate of \(8 \%\) per year, then the time taken for the population to be doubled is (given \(\log 2=\mathbf{0 . 6 9 1 2}\) )
1 10.27 years
2 8.64 years
3 4.3 years
4 6.8 years
Explanation:
(B) : Let ' \(\mathrm{P}\) ' be the population at any time \(\mathrm{t}\) we have given. The population grows at the rate of \(8 \%\) per year Then \(\frac{\mathrm{dP}}{\mathrm{dt}}=8 \% \mathrm{P}\) \(\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{8}{100} \mathrm{P}\) On integrating both sides we get - \(\int \frac{\mathrm{dP}}{\mathrm{P}}=\int \frac{8}{100} \mathrm{dt}\) \(\ln \mathrm{P}=\frac{8}{100} \mathrm{t}+\mathrm{c}\) When \(\mathrm{t}=0 \quad\) then \(\mathrm{P}=\mathrm{P}_{0}\) \(\ln \mathrm{P}_{0}=\mathrm{c}\) \(\text { We find when } \mathrm{P}=2 \mathrm{P}_{0}\) \(\ln 2 \mathrm{P}_{0}=\frac{8}{100} \mathrm{t}+\log \mathrm{P}_{\mathrm{o}}\) \(\ln \left(\frac{2 \mathrm{P}_{0}}{\mathrm{P}_{0}}\right)=\frac{8}{100} \mathrm{t}\) \(\ln 2=\frac{8}{100} \mathrm{t}\) \(\mathrm{t}=0.6912 \times \frac{100}{8}=8.64 \text { years }\)
MHT CET-2020
Differential Equation
87582
Water at \(100^{\circ} \mathrm{C}\) cools in 15 minutes to \(75^{\circ} \mathrm{C}\) at a room temperature of \(25^{\circ} \mathrm{C}\). Then the temperature of the water after 30 minutes is
(C) : Let ' \(\mathrm{T}\) ' be the temperature at any time ' \(\mathrm{t}\) ' for given condition \(\frac{\mathrm{dT}}{\mathrm{dt}} \propto(\mathrm{T}-25)\) \(\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-25)\) \(\frac{\mathrm{dT}}{\mathrm{T}-25}=-\mathrm{kdt}\) Integrating on both sides, we get - \(\log (\mathrm{T}-25)=-\mathrm{kt}+\mathrm{c}\) When \(\mathrm{t}=0\) and \(\mathrm{T}=100^{\circ} \mathrm{C}\) \(\log (100-25)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log 75\) Then equation become \(\log (\mathrm{T}-25)=-\mathrm{kt}+\log 75\) Again, when \(\mathrm{t}=15\) and \(\mathrm{T}=75^{\circ} \mathrm{C}\) \(\log \left(\frac{\mathrm{T}-25}{75}\right)=-\mathrm{kt}\) \(\mathrm{k} \times 15=\log \left(\frac{75}{75-25}\right)\) \(\mathrm{k}=\frac{1}{15} \log \left(\frac{3}{2}\right)\) Putting the value of \(\mathrm{k}\) in equation (ii), we get - \(\frac{1}{15} \log \left(\frac{3}{2}\right) \mathrm{t}=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) When \(\mathrm{t}=30\) then find \(\mathrm{T}\) \(30 \times \frac{1}{15} \log \left(\frac{3}{2}\right)=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) \(2 \log \left(\frac{3}{2}\right)=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) \(\log \frac{9}{4}=\log \frac{75}{\mathrm{~T}-25}\) \(\frac{9}{4}=\frac{75}{\mathrm{~T}-25}\) \(9 \mathrm{~T}-225=300\) \(9 \mathrm{~T}=525\) \(\mathrm{T}=\frac{525}{9}=\left(\frac{175}{3}\right)^{\circ} \mathrm{C}\)
MHT CET-2020
Differential Equation
87583
A body cools according to Newton's law from \(1000 \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) in 20 minutes. If the temperature of the surrounding is \(20^{\circ} \mathrm{C}\), then the temperature of the body after one hour is
1 \(40^{\circ} \mathrm{C}\)
2 \(15^{\circ} \mathrm{C}\)
3 \(30^{\circ} \mathrm{C}\)
4 \(20^{\circ} \mathrm{C}\)
Explanation:
(C) : Let ' \(\theta\) ' be the temperature of body at time ' \(t\) '. The temperature of surrounding is \(20^{\circ} \mathrm{C}\). According to Newton's law of cooling - \(\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto \theta-20\) \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-20)\) Integrating on both sides, we get - \(\int \frac{\mathrm{d} \theta}{\theta-20}=\int-\mathrm{kdt}\) \(\log (\theta-20)=-\mathrm{kt}+\mathrm{c}\) Given, when \(\mathrm{t}=0\) and \(\theta=100\) \(\log (100-20)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log 80\) Thus, equation become \(\log \left(\frac{\theta-20}{80}\right)=-\mathrm{kt}\) Now, When \(\mathrm{t}=20\) then \(\theta=60\) \(\log \left(\frac{60-20}{80}\right)=-\mathrm{k} \times 20\) \(\mathrm{k}=-\frac{1}{20} \log \left(\frac{1}{2}\right)\) Putting value of \(\mathrm{k}\) the equation becomes \(\log \left(\frac{\theta-20}{80}\right)=\frac{\mathrm{t}}{20} \log \left(\frac{1}{2}\right)\) We find temperature after 1 hour \(\mathrm{t}=60\) \(\log \left(\frac{\theta-20}{80}\right)=\frac{60}{20} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\theta-20}{80}\right)=\log \left(\frac{1}{2}\right)^{3}\) \(\frac{\theta-20}{80}=\frac{1}{8}\) \(\theta-20=10\) \(\theta=30^{\circ} \mathrm{C}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Differential Equation
87580
Bactria increases at the rate proportional to the number of bacteria present. If the original number \(\mathbf{N}\) doubles in \(\mathbf{4}\) hours, then the number of bacteria will be \(4 \mathbf{N}\) in
1 2 hours
2 8 hours
3 4 hours
4 6 hours
Explanation:
(B) : Given, \(\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}\) Integrating on both sides we get - \(\int_{\mathrm{N}}^{\mathrm{N}_{\mathrm{I}}} \frac{\mathrm{dN}}{\mathrm{N}}=\int_{0}^{\mathrm{t}} \mathrm{kdt}\) \([\log \mathrm{N}]_{\mathrm{N}}^{\mathrm{N}_{1}}=\mathrm{k}[\mathrm{t}]_{0}^{\mathrm{t}}\) \(\log \mathrm{N}_{1}-\log \mathrm{N}=\mathrm{kt}\) \(\log \frac{\mathrm{N}_{1}}{\mathrm{~N}}=\mathrm{kt}+\mathrm{c}\) \(\frac{\mathrm{N}_{1}}{\mathrm{~N}}=\mathrm{e}^{\mathrm{kt}}+\mathrm{c}\) \(\mathrm{N}_{1}=2 \mathrm{~N}\) When \(\mathrm{t}=4\) \(2 \mathrm{~N}=\mathrm{N}^{\mathrm{k}} \mathrm{e}^{\mathrm{k} .4}\) \(2=\mathrm{e}^{4}\) \((2)^{1 / 4}=e^{\mathrm{k}}\) \(\mathrm{N}_{1}=\mathrm{N} \mathrm{e}^{\mathrm{kt}}\) \(\mathrm{N}_{1}=\mathrm{N}\left(2^{\frac{1}{4} \mathrm{t}}\right)\) Now \(4 \mathrm{~N}=\mathrm{N}(2)^{1 / 4}\) \(2^{\frac{\mathrm{t}}{4}}=4\) \(\frac{\mathrm{t}}{4}=2\) \(\mathrm{t}=8\) Hour
MHT CET-2020
Differential Equation
87581
If the population grows at the rate of \(8 \%\) per year, then the time taken for the population to be doubled is (given \(\log 2=\mathbf{0 . 6 9 1 2}\) )
1 10.27 years
2 8.64 years
3 4.3 years
4 6.8 years
Explanation:
(B) : Let ' \(\mathrm{P}\) ' be the population at any time \(\mathrm{t}\) we have given. The population grows at the rate of \(8 \%\) per year Then \(\frac{\mathrm{dP}}{\mathrm{dt}}=8 \% \mathrm{P}\) \(\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{8}{100} \mathrm{P}\) On integrating both sides we get - \(\int \frac{\mathrm{dP}}{\mathrm{P}}=\int \frac{8}{100} \mathrm{dt}\) \(\ln \mathrm{P}=\frac{8}{100} \mathrm{t}+\mathrm{c}\) When \(\mathrm{t}=0 \quad\) then \(\mathrm{P}=\mathrm{P}_{0}\) \(\ln \mathrm{P}_{0}=\mathrm{c}\) \(\text { We find when } \mathrm{P}=2 \mathrm{P}_{0}\) \(\ln 2 \mathrm{P}_{0}=\frac{8}{100} \mathrm{t}+\log \mathrm{P}_{\mathrm{o}}\) \(\ln \left(\frac{2 \mathrm{P}_{0}}{\mathrm{P}_{0}}\right)=\frac{8}{100} \mathrm{t}\) \(\ln 2=\frac{8}{100} \mathrm{t}\) \(\mathrm{t}=0.6912 \times \frac{100}{8}=8.64 \text { years }\)
MHT CET-2020
Differential Equation
87582
Water at \(100^{\circ} \mathrm{C}\) cools in 15 minutes to \(75^{\circ} \mathrm{C}\) at a room temperature of \(25^{\circ} \mathrm{C}\). Then the temperature of the water after 30 minutes is
(C) : Let ' \(\mathrm{T}\) ' be the temperature at any time ' \(\mathrm{t}\) ' for given condition \(\frac{\mathrm{dT}}{\mathrm{dt}} \propto(\mathrm{T}-25)\) \(\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-25)\) \(\frac{\mathrm{dT}}{\mathrm{T}-25}=-\mathrm{kdt}\) Integrating on both sides, we get - \(\log (\mathrm{T}-25)=-\mathrm{kt}+\mathrm{c}\) When \(\mathrm{t}=0\) and \(\mathrm{T}=100^{\circ} \mathrm{C}\) \(\log (100-25)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log 75\) Then equation become \(\log (\mathrm{T}-25)=-\mathrm{kt}+\log 75\) Again, when \(\mathrm{t}=15\) and \(\mathrm{T}=75^{\circ} \mathrm{C}\) \(\log \left(\frac{\mathrm{T}-25}{75}\right)=-\mathrm{kt}\) \(\mathrm{k} \times 15=\log \left(\frac{75}{75-25}\right)\) \(\mathrm{k}=\frac{1}{15} \log \left(\frac{3}{2}\right)\) Putting the value of \(\mathrm{k}\) in equation (ii), we get - \(\frac{1}{15} \log \left(\frac{3}{2}\right) \mathrm{t}=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) When \(\mathrm{t}=30\) then find \(\mathrm{T}\) \(30 \times \frac{1}{15} \log \left(\frac{3}{2}\right)=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) \(2 \log \left(\frac{3}{2}\right)=\log \left(\frac{75}{\mathrm{~T}-25}\right)\) \(\log \frac{9}{4}=\log \frac{75}{\mathrm{~T}-25}\) \(\frac{9}{4}=\frac{75}{\mathrm{~T}-25}\) \(9 \mathrm{~T}-225=300\) \(9 \mathrm{~T}=525\) \(\mathrm{T}=\frac{525}{9}=\left(\frac{175}{3}\right)^{\circ} \mathrm{C}\)
MHT CET-2020
Differential Equation
87583
A body cools according to Newton's law from \(1000 \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) in 20 minutes. If the temperature of the surrounding is \(20^{\circ} \mathrm{C}\), then the temperature of the body after one hour is
1 \(40^{\circ} \mathrm{C}\)
2 \(15^{\circ} \mathrm{C}\)
3 \(30^{\circ} \mathrm{C}\)
4 \(20^{\circ} \mathrm{C}\)
Explanation:
(C) : Let ' \(\theta\) ' be the temperature of body at time ' \(t\) '. The temperature of surrounding is \(20^{\circ} \mathrm{C}\). According to Newton's law of cooling - \(\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto \theta-20\) \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-20)\) Integrating on both sides, we get - \(\int \frac{\mathrm{d} \theta}{\theta-20}=\int-\mathrm{kdt}\) \(\log (\theta-20)=-\mathrm{kt}+\mathrm{c}\) Given, when \(\mathrm{t}=0\) and \(\theta=100\) \(\log (100-20)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log 80\) Thus, equation become \(\log \left(\frac{\theta-20}{80}\right)=-\mathrm{kt}\) Now, When \(\mathrm{t}=20\) then \(\theta=60\) \(\log \left(\frac{60-20}{80}\right)=-\mathrm{k} \times 20\) \(\mathrm{k}=-\frac{1}{20} \log \left(\frac{1}{2}\right)\) Putting value of \(\mathrm{k}\) the equation becomes \(\log \left(\frac{\theta-20}{80}\right)=\frac{\mathrm{t}}{20} \log \left(\frac{1}{2}\right)\) We find temperature after 1 hour \(\mathrm{t}=60\) \(\log \left(\frac{\theta-20}{80}\right)=\frac{60}{20} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\theta-20}{80}\right)=\log \left(\frac{1}{2}\right)^{3}\) \(\frac{\theta-20}{80}=\frac{1}{8}\) \(\theta-20=10\) \(\theta=30^{\circ} \mathrm{C}\)