87584
The rate of growth of bacteria is proportional to the bacteria present. If it is found that the number doubles in 3 hours, then the number of times the bacteria are increased in \(\mathbf{6}\) hours is
1 4 times the original
2 5 times the original
3 8 times the original
4 6 times the original
Explanation:
(A) : Let, ' \(N\) ' be the number of bacteria at any time \(t\). According to condition - \(\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}\) \(\frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{kN}\) \(\frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{kdt}\) Integrating on both sides, we get - \(\log \mathrm{N}=\mathrm{kt}+\mathrm{c}\) Initially when \(\mathrm{t}=0\) and \(\mathrm{N}=\mathrm{N}_{0}\) \(\log \mathrm{N}_{0}=\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log \mathrm{N}_{0}\) Then equation becomes \(\log \mathrm{N}=\mathrm{kt}+\log \mathrm{N}_{0}\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\mathrm{kt}\) When \(\mathrm{t}=3\) and \(\mathrm{N}=2 \mathrm{~N}_{0}\) \(\log \frac{2 \mathrm{~N}_{0}}{\mathrm{~N}_{0}}=\mathrm{k} \times 3\) \(\mathrm{k}=\frac{1}{3} \log 2\) Putting value of \(k\) then equation become \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{t} \frac{1}{3} \log 2\) When \(\mathrm{t}=6\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=6 \times \frac{1}{3} \log 2\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=\log (2)^{2}\) \(\frac{\mathrm{N}}{\mathrm{N}_{0}}=4\) \(\mathrm{N}=4 \mathrm{~N}_{0}\)
MHT CET-2020
Differential Equation
87585
A metal has half life period of 10 days. A sample originally has a mass of \(1000 \mathrm{mg}\). then the mass remaining after 50 days is
1 \(\frac{125}{4} \mathrm{mg}\)
2 \(\frac{225}{4} \mathrm{mg}\)
3 \(\frac{225}{8} \mathrm{mg}\)
4 \(\frac{125}{8} \mathrm{mg}\)
Explanation:
(A) : Let, \(\mathrm{N}\) be the mass of metal left at some instant of time therefore the half life period - \(\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}\) Now, be the initial mass of atom Given, \(\mathrm{T}=10 \text { days }\) \(\mathrm{t}=50 \text { day }\) \(\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}}=\frac{50}{10}=5 \text { days }\) Now, \(\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}\) \(\mathrm{N}=1000\left(\frac{1}{2}\right)^{5}=\frac{1000}{32}\) \(\mathrm{N}=\frac{125}{4} \mathrm{mg}\)
MHT CET-2020
Differential Equation
87586
Bismuth has half life of 5 days. If sample originally has a mass of \(800 \mathrm{mg}\). then the mass remaining after 30 days will be
1 \(12.5 \mathrm{mg}\)
2 \(10.5 \mathrm{mg}\)
3 \(12 \mathrm{mg}\)
4 \(10 \mathrm{mg}\)
Explanation:
(A) : Let ' \(x\) ' be the mass of Bismuth at time \(t\) \(\text { Then, } \frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{x}\) \(\frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{kx}\) \(\text { Now separate variable we get } \quad[\because \mathrm{k} \text { is constant }]\) \(\frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{kdt}\) On integrating both side we get \(\int \frac{\mathrm{dx}}{\mathrm{x}}=-\int \mathrm{kdt}\) \(\log \mathrm{x}=-\mathrm{kt}+\mathrm{c}\) For initially \(\mathrm{t}=0\) then \(\mathrm{x}=800 \mathrm{mg}\) \(\log (800)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log (800)\) Putting value of \(c\) then we get \(\log \mathrm{x}=-\mathrm{kt}+\log 800\) \(\log \mathrm{x}-\log 800=-\mathrm{kt}\) \(\log \left(\frac{\mathrm{x}}{800}\right)=-\mathrm{kt}\) \(\mathrm{x}\) \(\mathrm{x}=80 \mathrm{e}^{-\mathrm{kt}}\) \(\mathrm{x}=80 \mathrm{e}^{-\mathrm{kt}}\) Given, \(\mathrm{t}=5 \text { days and } \mathrm{x}=\frac{800}{2}\) \(\mathrm{x}=400\) Putting the value of \(x\) in equation (ii) we get \(400=800 \mathrm{e}^{-\mathrm{k} \times 5}\) \(\frac{1}{2}=\mathrm{e}^{-\mathrm{k} 5}\) \(\log \left(\frac{1}{2}\right)=-5 \mathrm{k}\) \(\mathrm{k}=-\frac{1}{5} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=\mathrm{t} \times \frac{1}{5} \log \left(\frac{1}{2}\right)\) For, \(t=30\) days \(\log \left(\frac{x}{800}\right)=30 \times \frac{1}{5} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=6 \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=\log \left(\frac{1}{2}\right)^{6}\) \(\frac{\mathrm{x}}{800}=\left(\frac{1}{2}\right)^{6}\) \(\mathrm{x}=800 \times \frac{1}{2^{6}}=800 \times \frac{1}{8 \times 8}\) \(\mathrm{x}=12.5^{\mathrm{mg}}\)
MHT CET-2020
Differential Equation
87587
A body is heated to \(110^{\circ} \mathrm{C}\) and placed in air at \(10^{\circ} \mathrm{C}\). After 1 hours its temperature is \(60^{\circ} \mathrm{C}\). The additional time required for it to cool to \(30^{\circ} \mathrm{C}\) is
1 \(\left(\frac{\log 2}{\log 5}\right)\) hours
2 \(\left(\frac{\log 2}{\log 5}+1\right)\) hours
3 \(\left(\frac{\log 5}{\log 2}\right)\) hours
4 \(\left(\frac{\log 5}{\log 2}-1\right)\) hours
Explanation:
(D) : Let \(\theta\) be the temperature of the body at any time \(\mathrm{t}-\) Temperature of air is \(10^{\circ} \mathrm{C}\) is \(\theta_{0}=10\) According to Newton's law of cooling - \(\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto \theta-\theta_{0}\) \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-10)\) \(\therefore \frac{1}{\theta-10} \mathrm{~d} \theta=-\mathrm{kdt}\) Integrating on both sides, we get - \(\int \frac{1}{\theta-10} \mathrm{~d} \theta=-\int \mathrm{kdt}\) \(\log (\theta-10)=-\mathrm{kt}+\mathrm{c}\) When, \(\mathrm{t}=0\) and \(\theta=110^{\circ} \mathrm{C}\) \(\therefore \log _{\mathrm{c}=\log 100}(110-10)=-\mathrm{k} \times 0+\mathrm{c}\) \(\log (\theta-10)=-\mathrm{kt}+\log 100\) \(\log (\theta-10)-\log (100)=-\mathrm{kt}\) When, \(\theta=60^{\circ} \mathrm{C}\) and \(\mathrm{t}=1\) \(\log \left(\frac{60-10}{100}\right)=-\mathrm{k} \times 1\) \(\mathrm{k}=-\log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\theta-10}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) Find \(t\) when \(\theta=30^{\circ} \mathrm{C}\) \(\log \left(\frac{30-10}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{20}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{1}{5}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\mathrm{t}=\frac{\log \left(\frac{1}{5}\right)}{(1)}\) \(\log \left(\frac{1}{2}\right)\) \(\mathrm{t}=\frac{\log 5}{\log 2}\) Now, additional time required for body to become cool at \(30^{\circ} \mathrm{C}\) \(\therefore \mathrm{t}=\left(\frac{\log 5}{\log 2}-1\right)\)
MHT CET-2020
Differential Equation
87588
The equation of a curve passing through \(\left(2, \frac{7}{2}\right)\) and having gradient \(1-\frac{1}{\mathrm{x}^{2}}\) is
1 \(x y=x^{2}+x+1\)
2 \(y=x^{2}+x+1\)
3 \(x y=x+1\)
4 none of these
Explanation:
(B) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}=1-\frac{1}{\mathrm{x}^{2}}\) On integrating we get- \(\mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}+\mathrm{c}\) This passes through \(\left(2, \frac{7}{2}\right)\) \(\frac{7}{2}=2+\frac{1}{2}+c \Rightarrow c=1\) Thus, the equation of the curve is \(y=x+\frac{1}{x}+1\) \(x y=x^{2}+x+1\)
87584
The rate of growth of bacteria is proportional to the bacteria present. If it is found that the number doubles in 3 hours, then the number of times the bacteria are increased in \(\mathbf{6}\) hours is
1 4 times the original
2 5 times the original
3 8 times the original
4 6 times the original
Explanation:
(A) : Let, ' \(N\) ' be the number of bacteria at any time \(t\). According to condition - \(\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}\) \(\frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{kN}\) \(\frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{kdt}\) Integrating on both sides, we get - \(\log \mathrm{N}=\mathrm{kt}+\mathrm{c}\) Initially when \(\mathrm{t}=0\) and \(\mathrm{N}=\mathrm{N}_{0}\) \(\log \mathrm{N}_{0}=\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log \mathrm{N}_{0}\) Then equation becomes \(\log \mathrm{N}=\mathrm{kt}+\log \mathrm{N}_{0}\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\mathrm{kt}\) When \(\mathrm{t}=3\) and \(\mathrm{N}=2 \mathrm{~N}_{0}\) \(\log \frac{2 \mathrm{~N}_{0}}{\mathrm{~N}_{0}}=\mathrm{k} \times 3\) \(\mathrm{k}=\frac{1}{3} \log 2\) Putting value of \(k\) then equation become \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{t} \frac{1}{3} \log 2\) When \(\mathrm{t}=6\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=6 \times \frac{1}{3} \log 2\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=\log (2)^{2}\) \(\frac{\mathrm{N}}{\mathrm{N}_{0}}=4\) \(\mathrm{N}=4 \mathrm{~N}_{0}\)
MHT CET-2020
Differential Equation
87585
A metal has half life period of 10 days. A sample originally has a mass of \(1000 \mathrm{mg}\). then the mass remaining after 50 days is
1 \(\frac{125}{4} \mathrm{mg}\)
2 \(\frac{225}{4} \mathrm{mg}\)
3 \(\frac{225}{8} \mathrm{mg}\)
4 \(\frac{125}{8} \mathrm{mg}\)
Explanation:
(A) : Let, \(\mathrm{N}\) be the mass of metal left at some instant of time therefore the half life period - \(\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}\) Now, be the initial mass of atom Given, \(\mathrm{T}=10 \text { days }\) \(\mathrm{t}=50 \text { day }\) \(\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}}=\frac{50}{10}=5 \text { days }\) Now, \(\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}\) \(\mathrm{N}=1000\left(\frac{1}{2}\right)^{5}=\frac{1000}{32}\) \(\mathrm{N}=\frac{125}{4} \mathrm{mg}\)
MHT CET-2020
Differential Equation
87586
Bismuth has half life of 5 days. If sample originally has a mass of \(800 \mathrm{mg}\). then the mass remaining after 30 days will be
1 \(12.5 \mathrm{mg}\)
2 \(10.5 \mathrm{mg}\)
3 \(12 \mathrm{mg}\)
4 \(10 \mathrm{mg}\)
Explanation:
(A) : Let ' \(x\) ' be the mass of Bismuth at time \(t\) \(\text { Then, } \frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{x}\) \(\frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{kx}\) \(\text { Now separate variable we get } \quad[\because \mathrm{k} \text { is constant }]\) \(\frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{kdt}\) On integrating both side we get \(\int \frac{\mathrm{dx}}{\mathrm{x}}=-\int \mathrm{kdt}\) \(\log \mathrm{x}=-\mathrm{kt}+\mathrm{c}\) For initially \(\mathrm{t}=0\) then \(\mathrm{x}=800 \mathrm{mg}\) \(\log (800)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log (800)\) Putting value of \(c\) then we get \(\log \mathrm{x}=-\mathrm{kt}+\log 800\) \(\log \mathrm{x}-\log 800=-\mathrm{kt}\) \(\log \left(\frac{\mathrm{x}}{800}\right)=-\mathrm{kt}\) \(\mathrm{x}\) \(\mathrm{x}=80 \mathrm{e}^{-\mathrm{kt}}\) \(\mathrm{x}=80 \mathrm{e}^{-\mathrm{kt}}\) Given, \(\mathrm{t}=5 \text { days and } \mathrm{x}=\frac{800}{2}\) \(\mathrm{x}=400\) Putting the value of \(x\) in equation (ii) we get \(400=800 \mathrm{e}^{-\mathrm{k} \times 5}\) \(\frac{1}{2}=\mathrm{e}^{-\mathrm{k} 5}\) \(\log \left(\frac{1}{2}\right)=-5 \mathrm{k}\) \(\mathrm{k}=-\frac{1}{5} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=\mathrm{t} \times \frac{1}{5} \log \left(\frac{1}{2}\right)\) For, \(t=30\) days \(\log \left(\frac{x}{800}\right)=30 \times \frac{1}{5} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=6 \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=\log \left(\frac{1}{2}\right)^{6}\) \(\frac{\mathrm{x}}{800}=\left(\frac{1}{2}\right)^{6}\) \(\mathrm{x}=800 \times \frac{1}{2^{6}}=800 \times \frac{1}{8 \times 8}\) \(\mathrm{x}=12.5^{\mathrm{mg}}\)
MHT CET-2020
Differential Equation
87587
A body is heated to \(110^{\circ} \mathrm{C}\) and placed in air at \(10^{\circ} \mathrm{C}\). After 1 hours its temperature is \(60^{\circ} \mathrm{C}\). The additional time required for it to cool to \(30^{\circ} \mathrm{C}\) is
1 \(\left(\frac{\log 2}{\log 5}\right)\) hours
2 \(\left(\frac{\log 2}{\log 5}+1\right)\) hours
3 \(\left(\frac{\log 5}{\log 2}\right)\) hours
4 \(\left(\frac{\log 5}{\log 2}-1\right)\) hours
Explanation:
(D) : Let \(\theta\) be the temperature of the body at any time \(\mathrm{t}-\) Temperature of air is \(10^{\circ} \mathrm{C}\) is \(\theta_{0}=10\) According to Newton's law of cooling - \(\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto \theta-\theta_{0}\) \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-10)\) \(\therefore \frac{1}{\theta-10} \mathrm{~d} \theta=-\mathrm{kdt}\) Integrating on both sides, we get - \(\int \frac{1}{\theta-10} \mathrm{~d} \theta=-\int \mathrm{kdt}\) \(\log (\theta-10)=-\mathrm{kt}+\mathrm{c}\) When, \(\mathrm{t}=0\) and \(\theta=110^{\circ} \mathrm{C}\) \(\therefore \log _{\mathrm{c}=\log 100}(110-10)=-\mathrm{k} \times 0+\mathrm{c}\) \(\log (\theta-10)=-\mathrm{kt}+\log 100\) \(\log (\theta-10)-\log (100)=-\mathrm{kt}\) When, \(\theta=60^{\circ} \mathrm{C}\) and \(\mathrm{t}=1\) \(\log \left(\frac{60-10}{100}\right)=-\mathrm{k} \times 1\) \(\mathrm{k}=-\log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\theta-10}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) Find \(t\) when \(\theta=30^{\circ} \mathrm{C}\) \(\log \left(\frac{30-10}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{20}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{1}{5}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\mathrm{t}=\frac{\log \left(\frac{1}{5}\right)}{(1)}\) \(\log \left(\frac{1}{2}\right)\) \(\mathrm{t}=\frac{\log 5}{\log 2}\) Now, additional time required for body to become cool at \(30^{\circ} \mathrm{C}\) \(\therefore \mathrm{t}=\left(\frac{\log 5}{\log 2}-1\right)\)
MHT CET-2020
Differential Equation
87588
The equation of a curve passing through \(\left(2, \frac{7}{2}\right)\) and having gradient \(1-\frac{1}{\mathrm{x}^{2}}\) is
1 \(x y=x^{2}+x+1\)
2 \(y=x^{2}+x+1\)
3 \(x y=x+1\)
4 none of these
Explanation:
(B) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}=1-\frac{1}{\mathrm{x}^{2}}\) On integrating we get- \(\mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}+\mathrm{c}\) This passes through \(\left(2, \frac{7}{2}\right)\) \(\frac{7}{2}=2+\frac{1}{2}+c \Rightarrow c=1\) Thus, the equation of the curve is \(y=x+\frac{1}{x}+1\) \(x y=x^{2}+x+1\)
87584
The rate of growth of bacteria is proportional to the bacteria present. If it is found that the number doubles in 3 hours, then the number of times the bacteria are increased in \(\mathbf{6}\) hours is
1 4 times the original
2 5 times the original
3 8 times the original
4 6 times the original
Explanation:
(A) : Let, ' \(N\) ' be the number of bacteria at any time \(t\). According to condition - \(\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}\) \(\frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{kN}\) \(\frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{kdt}\) Integrating on both sides, we get - \(\log \mathrm{N}=\mathrm{kt}+\mathrm{c}\) Initially when \(\mathrm{t}=0\) and \(\mathrm{N}=\mathrm{N}_{0}\) \(\log \mathrm{N}_{0}=\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log \mathrm{N}_{0}\) Then equation becomes \(\log \mathrm{N}=\mathrm{kt}+\log \mathrm{N}_{0}\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\mathrm{kt}\) When \(\mathrm{t}=3\) and \(\mathrm{N}=2 \mathrm{~N}_{0}\) \(\log \frac{2 \mathrm{~N}_{0}}{\mathrm{~N}_{0}}=\mathrm{k} \times 3\) \(\mathrm{k}=\frac{1}{3} \log 2\) Putting value of \(k\) then equation become \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{t} \frac{1}{3} \log 2\) When \(\mathrm{t}=6\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=6 \times \frac{1}{3} \log 2\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=\log (2)^{2}\) \(\frac{\mathrm{N}}{\mathrm{N}_{0}}=4\) \(\mathrm{N}=4 \mathrm{~N}_{0}\)
MHT CET-2020
Differential Equation
87585
A metal has half life period of 10 days. A sample originally has a mass of \(1000 \mathrm{mg}\). then the mass remaining after 50 days is
1 \(\frac{125}{4} \mathrm{mg}\)
2 \(\frac{225}{4} \mathrm{mg}\)
3 \(\frac{225}{8} \mathrm{mg}\)
4 \(\frac{125}{8} \mathrm{mg}\)
Explanation:
(A) : Let, \(\mathrm{N}\) be the mass of metal left at some instant of time therefore the half life period - \(\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}\) Now, be the initial mass of atom Given, \(\mathrm{T}=10 \text { days }\) \(\mathrm{t}=50 \text { day }\) \(\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}}=\frac{50}{10}=5 \text { days }\) Now, \(\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}\) \(\mathrm{N}=1000\left(\frac{1}{2}\right)^{5}=\frac{1000}{32}\) \(\mathrm{N}=\frac{125}{4} \mathrm{mg}\)
MHT CET-2020
Differential Equation
87586
Bismuth has half life of 5 days. If sample originally has a mass of \(800 \mathrm{mg}\). then the mass remaining after 30 days will be
1 \(12.5 \mathrm{mg}\)
2 \(10.5 \mathrm{mg}\)
3 \(12 \mathrm{mg}\)
4 \(10 \mathrm{mg}\)
Explanation:
(A) : Let ' \(x\) ' be the mass of Bismuth at time \(t\) \(\text { Then, } \frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{x}\) \(\frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{kx}\) \(\text { Now separate variable we get } \quad[\because \mathrm{k} \text { is constant }]\) \(\frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{kdt}\) On integrating both side we get \(\int \frac{\mathrm{dx}}{\mathrm{x}}=-\int \mathrm{kdt}\) \(\log \mathrm{x}=-\mathrm{kt}+\mathrm{c}\) For initially \(\mathrm{t}=0\) then \(\mathrm{x}=800 \mathrm{mg}\) \(\log (800)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log (800)\) Putting value of \(c\) then we get \(\log \mathrm{x}=-\mathrm{kt}+\log 800\) \(\log \mathrm{x}-\log 800=-\mathrm{kt}\) \(\log \left(\frac{\mathrm{x}}{800}\right)=-\mathrm{kt}\) \(\mathrm{x}\) \(\mathrm{x}=80 \mathrm{e}^{-\mathrm{kt}}\) \(\mathrm{x}=80 \mathrm{e}^{-\mathrm{kt}}\) Given, \(\mathrm{t}=5 \text { days and } \mathrm{x}=\frac{800}{2}\) \(\mathrm{x}=400\) Putting the value of \(x\) in equation (ii) we get \(400=800 \mathrm{e}^{-\mathrm{k} \times 5}\) \(\frac{1}{2}=\mathrm{e}^{-\mathrm{k} 5}\) \(\log \left(\frac{1}{2}\right)=-5 \mathrm{k}\) \(\mathrm{k}=-\frac{1}{5} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=\mathrm{t} \times \frac{1}{5} \log \left(\frac{1}{2}\right)\) For, \(t=30\) days \(\log \left(\frac{x}{800}\right)=30 \times \frac{1}{5} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=6 \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=\log \left(\frac{1}{2}\right)^{6}\) \(\frac{\mathrm{x}}{800}=\left(\frac{1}{2}\right)^{6}\) \(\mathrm{x}=800 \times \frac{1}{2^{6}}=800 \times \frac{1}{8 \times 8}\) \(\mathrm{x}=12.5^{\mathrm{mg}}\)
MHT CET-2020
Differential Equation
87587
A body is heated to \(110^{\circ} \mathrm{C}\) and placed in air at \(10^{\circ} \mathrm{C}\). After 1 hours its temperature is \(60^{\circ} \mathrm{C}\). The additional time required for it to cool to \(30^{\circ} \mathrm{C}\) is
1 \(\left(\frac{\log 2}{\log 5}\right)\) hours
2 \(\left(\frac{\log 2}{\log 5}+1\right)\) hours
3 \(\left(\frac{\log 5}{\log 2}\right)\) hours
4 \(\left(\frac{\log 5}{\log 2}-1\right)\) hours
Explanation:
(D) : Let \(\theta\) be the temperature of the body at any time \(\mathrm{t}-\) Temperature of air is \(10^{\circ} \mathrm{C}\) is \(\theta_{0}=10\) According to Newton's law of cooling - \(\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto \theta-\theta_{0}\) \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-10)\) \(\therefore \frac{1}{\theta-10} \mathrm{~d} \theta=-\mathrm{kdt}\) Integrating on both sides, we get - \(\int \frac{1}{\theta-10} \mathrm{~d} \theta=-\int \mathrm{kdt}\) \(\log (\theta-10)=-\mathrm{kt}+\mathrm{c}\) When, \(\mathrm{t}=0\) and \(\theta=110^{\circ} \mathrm{C}\) \(\therefore \log _{\mathrm{c}=\log 100}(110-10)=-\mathrm{k} \times 0+\mathrm{c}\) \(\log (\theta-10)=-\mathrm{kt}+\log 100\) \(\log (\theta-10)-\log (100)=-\mathrm{kt}\) When, \(\theta=60^{\circ} \mathrm{C}\) and \(\mathrm{t}=1\) \(\log \left(\frac{60-10}{100}\right)=-\mathrm{k} \times 1\) \(\mathrm{k}=-\log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\theta-10}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) Find \(t\) when \(\theta=30^{\circ} \mathrm{C}\) \(\log \left(\frac{30-10}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{20}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{1}{5}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\mathrm{t}=\frac{\log \left(\frac{1}{5}\right)}{(1)}\) \(\log \left(\frac{1}{2}\right)\) \(\mathrm{t}=\frac{\log 5}{\log 2}\) Now, additional time required for body to become cool at \(30^{\circ} \mathrm{C}\) \(\therefore \mathrm{t}=\left(\frac{\log 5}{\log 2}-1\right)\)
MHT CET-2020
Differential Equation
87588
The equation of a curve passing through \(\left(2, \frac{7}{2}\right)\) and having gradient \(1-\frac{1}{\mathrm{x}^{2}}\) is
1 \(x y=x^{2}+x+1\)
2 \(y=x^{2}+x+1\)
3 \(x y=x+1\)
4 none of these
Explanation:
(B) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}=1-\frac{1}{\mathrm{x}^{2}}\) On integrating we get- \(\mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}+\mathrm{c}\) This passes through \(\left(2, \frac{7}{2}\right)\) \(\frac{7}{2}=2+\frac{1}{2}+c \Rightarrow c=1\) Thus, the equation of the curve is \(y=x+\frac{1}{x}+1\) \(x y=x^{2}+x+1\)
87584
The rate of growth of bacteria is proportional to the bacteria present. If it is found that the number doubles in 3 hours, then the number of times the bacteria are increased in \(\mathbf{6}\) hours is
1 4 times the original
2 5 times the original
3 8 times the original
4 6 times the original
Explanation:
(A) : Let, ' \(N\) ' be the number of bacteria at any time \(t\). According to condition - \(\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}\) \(\frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{kN}\) \(\frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{kdt}\) Integrating on both sides, we get - \(\log \mathrm{N}=\mathrm{kt}+\mathrm{c}\) Initially when \(\mathrm{t}=0\) and \(\mathrm{N}=\mathrm{N}_{0}\) \(\log \mathrm{N}_{0}=\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log \mathrm{N}_{0}\) Then equation becomes \(\log \mathrm{N}=\mathrm{kt}+\log \mathrm{N}_{0}\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\mathrm{kt}\) When \(\mathrm{t}=3\) and \(\mathrm{N}=2 \mathrm{~N}_{0}\) \(\log \frac{2 \mathrm{~N}_{0}}{\mathrm{~N}_{0}}=\mathrm{k} \times 3\) \(\mathrm{k}=\frac{1}{3} \log 2\) Putting value of \(k\) then equation become \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{t} \frac{1}{3} \log 2\) When \(\mathrm{t}=6\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=6 \times \frac{1}{3} \log 2\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=\log (2)^{2}\) \(\frac{\mathrm{N}}{\mathrm{N}_{0}}=4\) \(\mathrm{N}=4 \mathrm{~N}_{0}\)
MHT CET-2020
Differential Equation
87585
A metal has half life period of 10 days. A sample originally has a mass of \(1000 \mathrm{mg}\). then the mass remaining after 50 days is
1 \(\frac{125}{4} \mathrm{mg}\)
2 \(\frac{225}{4} \mathrm{mg}\)
3 \(\frac{225}{8} \mathrm{mg}\)
4 \(\frac{125}{8} \mathrm{mg}\)
Explanation:
(A) : Let, \(\mathrm{N}\) be the mass of metal left at some instant of time therefore the half life period - \(\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}\) Now, be the initial mass of atom Given, \(\mathrm{T}=10 \text { days }\) \(\mathrm{t}=50 \text { day }\) \(\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}}=\frac{50}{10}=5 \text { days }\) Now, \(\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}\) \(\mathrm{N}=1000\left(\frac{1}{2}\right)^{5}=\frac{1000}{32}\) \(\mathrm{N}=\frac{125}{4} \mathrm{mg}\)
MHT CET-2020
Differential Equation
87586
Bismuth has half life of 5 days. If sample originally has a mass of \(800 \mathrm{mg}\). then the mass remaining after 30 days will be
1 \(12.5 \mathrm{mg}\)
2 \(10.5 \mathrm{mg}\)
3 \(12 \mathrm{mg}\)
4 \(10 \mathrm{mg}\)
Explanation:
(A) : Let ' \(x\) ' be the mass of Bismuth at time \(t\) \(\text { Then, } \frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{x}\) \(\frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{kx}\) \(\text { Now separate variable we get } \quad[\because \mathrm{k} \text { is constant }]\) \(\frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{kdt}\) On integrating both side we get \(\int \frac{\mathrm{dx}}{\mathrm{x}}=-\int \mathrm{kdt}\) \(\log \mathrm{x}=-\mathrm{kt}+\mathrm{c}\) For initially \(\mathrm{t}=0\) then \(\mathrm{x}=800 \mathrm{mg}\) \(\log (800)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log (800)\) Putting value of \(c\) then we get \(\log \mathrm{x}=-\mathrm{kt}+\log 800\) \(\log \mathrm{x}-\log 800=-\mathrm{kt}\) \(\log \left(\frac{\mathrm{x}}{800}\right)=-\mathrm{kt}\) \(\mathrm{x}\) \(\mathrm{x}=80 \mathrm{e}^{-\mathrm{kt}}\) \(\mathrm{x}=80 \mathrm{e}^{-\mathrm{kt}}\) Given, \(\mathrm{t}=5 \text { days and } \mathrm{x}=\frac{800}{2}\) \(\mathrm{x}=400\) Putting the value of \(x\) in equation (ii) we get \(400=800 \mathrm{e}^{-\mathrm{k} \times 5}\) \(\frac{1}{2}=\mathrm{e}^{-\mathrm{k} 5}\) \(\log \left(\frac{1}{2}\right)=-5 \mathrm{k}\) \(\mathrm{k}=-\frac{1}{5} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=\mathrm{t} \times \frac{1}{5} \log \left(\frac{1}{2}\right)\) For, \(t=30\) days \(\log \left(\frac{x}{800}\right)=30 \times \frac{1}{5} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=6 \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=\log \left(\frac{1}{2}\right)^{6}\) \(\frac{\mathrm{x}}{800}=\left(\frac{1}{2}\right)^{6}\) \(\mathrm{x}=800 \times \frac{1}{2^{6}}=800 \times \frac{1}{8 \times 8}\) \(\mathrm{x}=12.5^{\mathrm{mg}}\)
MHT CET-2020
Differential Equation
87587
A body is heated to \(110^{\circ} \mathrm{C}\) and placed in air at \(10^{\circ} \mathrm{C}\). After 1 hours its temperature is \(60^{\circ} \mathrm{C}\). The additional time required for it to cool to \(30^{\circ} \mathrm{C}\) is
1 \(\left(\frac{\log 2}{\log 5}\right)\) hours
2 \(\left(\frac{\log 2}{\log 5}+1\right)\) hours
3 \(\left(\frac{\log 5}{\log 2}\right)\) hours
4 \(\left(\frac{\log 5}{\log 2}-1\right)\) hours
Explanation:
(D) : Let \(\theta\) be the temperature of the body at any time \(\mathrm{t}-\) Temperature of air is \(10^{\circ} \mathrm{C}\) is \(\theta_{0}=10\) According to Newton's law of cooling - \(\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto \theta-\theta_{0}\) \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-10)\) \(\therefore \frac{1}{\theta-10} \mathrm{~d} \theta=-\mathrm{kdt}\) Integrating on both sides, we get - \(\int \frac{1}{\theta-10} \mathrm{~d} \theta=-\int \mathrm{kdt}\) \(\log (\theta-10)=-\mathrm{kt}+\mathrm{c}\) When, \(\mathrm{t}=0\) and \(\theta=110^{\circ} \mathrm{C}\) \(\therefore \log _{\mathrm{c}=\log 100}(110-10)=-\mathrm{k} \times 0+\mathrm{c}\) \(\log (\theta-10)=-\mathrm{kt}+\log 100\) \(\log (\theta-10)-\log (100)=-\mathrm{kt}\) When, \(\theta=60^{\circ} \mathrm{C}\) and \(\mathrm{t}=1\) \(\log \left(\frac{60-10}{100}\right)=-\mathrm{k} \times 1\) \(\mathrm{k}=-\log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\theta-10}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) Find \(t\) when \(\theta=30^{\circ} \mathrm{C}\) \(\log \left(\frac{30-10}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{20}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{1}{5}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\mathrm{t}=\frac{\log \left(\frac{1}{5}\right)}{(1)}\) \(\log \left(\frac{1}{2}\right)\) \(\mathrm{t}=\frac{\log 5}{\log 2}\) Now, additional time required for body to become cool at \(30^{\circ} \mathrm{C}\) \(\therefore \mathrm{t}=\left(\frac{\log 5}{\log 2}-1\right)\)
MHT CET-2020
Differential Equation
87588
The equation of a curve passing through \(\left(2, \frac{7}{2}\right)\) and having gradient \(1-\frac{1}{\mathrm{x}^{2}}\) is
1 \(x y=x^{2}+x+1\)
2 \(y=x^{2}+x+1\)
3 \(x y=x+1\)
4 none of these
Explanation:
(B) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}=1-\frac{1}{\mathrm{x}^{2}}\) On integrating we get- \(\mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}+\mathrm{c}\) This passes through \(\left(2, \frac{7}{2}\right)\) \(\frac{7}{2}=2+\frac{1}{2}+c \Rightarrow c=1\) Thus, the equation of the curve is \(y=x+\frac{1}{x}+1\) \(x y=x^{2}+x+1\)
87584
The rate of growth of bacteria is proportional to the bacteria present. If it is found that the number doubles in 3 hours, then the number of times the bacteria are increased in \(\mathbf{6}\) hours is
1 4 times the original
2 5 times the original
3 8 times the original
4 6 times the original
Explanation:
(A) : Let, ' \(N\) ' be the number of bacteria at any time \(t\). According to condition - \(\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}\) \(\frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{kN}\) \(\frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{kdt}\) Integrating on both sides, we get - \(\log \mathrm{N}=\mathrm{kt}+\mathrm{c}\) Initially when \(\mathrm{t}=0\) and \(\mathrm{N}=\mathrm{N}_{0}\) \(\log \mathrm{N}_{0}=\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log \mathrm{N}_{0}\) Then equation becomes \(\log \mathrm{N}=\mathrm{kt}+\log \mathrm{N}_{0}\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\mathrm{kt}\) When \(\mathrm{t}=3\) and \(\mathrm{N}=2 \mathrm{~N}_{0}\) \(\log \frac{2 \mathrm{~N}_{0}}{\mathrm{~N}_{0}}=\mathrm{k} \times 3\) \(\mathrm{k}=\frac{1}{3} \log 2\) Putting value of \(k\) then equation become \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{t} \frac{1}{3} \log 2\) When \(\mathrm{t}=6\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=6 \times \frac{1}{3} \log 2\) \(\log \frac{\mathrm{N}}{\mathrm{N}_{0}}=\log (2)^{2}\) \(\frac{\mathrm{N}}{\mathrm{N}_{0}}=4\) \(\mathrm{N}=4 \mathrm{~N}_{0}\)
MHT CET-2020
Differential Equation
87585
A metal has half life period of 10 days. A sample originally has a mass of \(1000 \mathrm{mg}\). then the mass remaining after 50 days is
1 \(\frac{125}{4} \mathrm{mg}\)
2 \(\frac{225}{4} \mathrm{mg}\)
3 \(\frac{225}{8} \mathrm{mg}\)
4 \(\frac{125}{8} \mathrm{mg}\)
Explanation:
(A) : Let, \(\mathrm{N}\) be the mass of metal left at some instant of time therefore the half life period - \(\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}\) Now, be the initial mass of atom Given, \(\mathrm{T}=10 \text { days }\) \(\mathrm{t}=50 \text { day }\) \(\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}}=\frac{50}{10}=5 \text { days }\) Now, \(\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}\) \(\mathrm{N}=1000\left(\frac{1}{2}\right)^{5}=\frac{1000}{32}\) \(\mathrm{N}=\frac{125}{4} \mathrm{mg}\)
MHT CET-2020
Differential Equation
87586
Bismuth has half life of 5 days. If sample originally has a mass of \(800 \mathrm{mg}\). then the mass remaining after 30 days will be
1 \(12.5 \mathrm{mg}\)
2 \(10.5 \mathrm{mg}\)
3 \(12 \mathrm{mg}\)
4 \(10 \mathrm{mg}\)
Explanation:
(A) : Let ' \(x\) ' be the mass of Bismuth at time \(t\) \(\text { Then, } \frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{x}\) \(\frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{kx}\) \(\text { Now separate variable we get } \quad[\because \mathrm{k} \text { is constant }]\) \(\frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{kdt}\) On integrating both side we get \(\int \frac{\mathrm{dx}}{\mathrm{x}}=-\int \mathrm{kdt}\) \(\log \mathrm{x}=-\mathrm{kt}+\mathrm{c}\) For initially \(\mathrm{t}=0\) then \(\mathrm{x}=800 \mathrm{mg}\) \(\log (800)=-\mathrm{k} \times 0+\mathrm{c}\) \(\mathrm{c}=\log (800)\) Putting value of \(c\) then we get \(\log \mathrm{x}=-\mathrm{kt}+\log 800\) \(\log \mathrm{x}-\log 800=-\mathrm{kt}\) \(\log \left(\frac{\mathrm{x}}{800}\right)=-\mathrm{kt}\) \(\mathrm{x}\) \(\mathrm{x}=80 \mathrm{e}^{-\mathrm{kt}}\) \(\mathrm{x}=80 \mathrm{e}^{-\mathrm{kt}}\) Given, \(\mathrm{t}=5 \text { days and } \mathrm{x}=\frac{800}{2}\) \(\mathrm{x}=400\) Putting the value of \(x\) in equation (ii) we get \(400=800 \mathrm{e}^{-\mathrm{k} \times 5}\) \(\frac{1}{2}=\mathrm{e}^{-\mathrm{k} 5}\) \(\log \left(\frac{1}{2}\right)=-5 \mathrm{k}\) \(\mathrm{k}=-\frac{1}{5} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=\mathrm{t} \times \frac{1}{5} \log \left(\frac{1}{2}\right)\) For, \(t=30\) days \(\log \left(\frac{x}{800}\right)=30 \times \frac{1}{5} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=6 \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\mathrm{x}}{800}\right)=\log \left(\frac{1}{2}\right)^{6}\) \(\frac{\mathrm{x}}{800}=\left(\frac{1}{2}\right)^{6}\) \(\mathrm{x}=800 \times \frac{1}{2^{6}}=800 \times \frac{1}{8 \times 8}\) \(\mathrm{x}=12.5^{\mathrm{mg}}\)
MHT CET-2020
Differential Equation
87587
A body is heated to \(110^{\circ} \mathrm{C}\) and placed in air at \(10^{\circ} \mathrm{C}\). After 1 hours its temperature is \(60^{\circ} \mathrm{C}\). The additional time required for it to cool to \(30^{\circ} \mathrm{C}\) is
1 \(\left(\frac{\log 2}{\log 5}\right)\) hours
2 \(\left(\frac{\log 2}{\log 5}+1\right)\) hours
3 \(\left(\frac{\log 5}{\log 2}\right)\) hours
4 \(\left(\frac{\log 5}{\log 2}-1\right)\) hours
Explanation:
(D) : Let \(\theta\) be the temperature of the body at any time \(\mathrm{t}-\) Temperature of air is \(10^{\circ} \mathrm{C}\) is \(\theta_{0}=10\) According to Newton's law of cooling - \(\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto \theta-\theta_{0}\) \(\frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}(\theta-10)\) \(\therefore \frac{1}{\theta-10} \mathrm{~d} \theta=-\mathrm{kdt}\) Integrating on both sides, we get - \(\int \frac{1}{\theta-10} \mathrm{~d} \theta=-\int \mathrm{kdt}\) \(\log (\theta-10)=-\mathrm{kt}+\mathrm{c}\) When, \(\mathrm{t}=0\) and \(\theta=110^{\circ} \mathrm{C}\) \(\therefore \log _{\mathrm{c}=\log 100}(110-10)=-\mathrm{k} \times 0+\mathrm{c}\) \(\log (\theta-10)=-\mathrm{kt}+\log 100\) \(\log (\theta-10)-\log (100)=-\mathrm{kt}\) When, \(\theta=60^{\circ} \mathrm{C}\) and \(\mathrm{t}=1\) \(\log \left(\frac{60-10}{100}\right)=-\mathrm{k} \times 1\) \(\mathrm{k}=-\log \left(\frac{1}{2}\right)\) \(\log \left(\frac{\theta-10}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) Find \(t\) when \(\theta=30^{\circ} \mathrm{C}\) \(\log \left(\frac{30-10}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{20}{100}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\log \left(\frac{1}{5}\right)=\mathrm{t} \log \left(\frac{1}{2}\right)\) \(\mathrm{t}=\frac{\log \left(\frac{1}{5}\right)}{(1)}\) \(\log \left(\frac{1}{2}\right)\) \(\mathrm{t}=\frac{\log 5}{\log 2}\) Now, additional time required for body to become cool at \(30^{\circ} \mathrm{C}\) \(\therefore \mathrm{t}=\left(\frac{\log 5}{\log 2}-1\right)\)
MHT CET-2020
Differential Equation
87588
The equation of a curve passing through \(\left(2, \frac{7}{2}\right)\) and having gradient \(1-\frac{1}{\mathrm{x}^{2}}\) is
1 \(x y=x^{2}+x+1\)
2 \(y=x^{2}+x+1\)
3 \(x y=x+1\)
4 none of these
Explanation:
(B) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}=1-\frac{1}{\mathrm{x}^{2}}\) On integrating we get- \(\mathrm{y}=\mathrm{x}+\frac{1}{\mathrm{x}}+\mathrm{c}\) This passes through \(\left(2, \frac{7}{2}\right)\) \(\frac{7}{2}=2+\frac{1}{2}+c \Rightarrow c=1\) Thus, the equation of the curve is \(y=x+\frac{1}{x}+1\) \(x y=x^{2}+x+1\)
