5 Marks Questions

Question 15 Marks

The population of a village increases at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25,000 in the year 2004, what was its population in 2009 ?

Answer

Let the population at time t be x.
Given, rate of population increase is proportional to number of inhabitants.
i.e., $\frac{d x}{d t} \propto x$
$\Rightarrow \quad \frac{d x}{d t}=k x$
where k is constant of proportionality $\frac{d x}{x}=k d t$
$\Rightarrow \quad \int \frac{d x}{x}=k \int d t$
$\Rightarrow \quad \log |x|=k t+C$ ....(i)
In 1999, t = 0 and population = 20000
$\therefore \quad \log 20,000= C$
Put value of C in eq (i), we get
log |x| = kt + log 20000
$\Rightarrow \log |x|-\log 20,000=k t$
$\Rightarrow \quad \log \left(\frac{x}{20,000}\right)=k t$
In 2004, t = 5 x = 25000
$\therefore \quad \log \left(\frac{25000}{20000}\right)=k \times 5$
$\Rightarrow \quad k=\frac{1}{5} \log \left(\frac{5}{4}\right)$
$\therefore$ From eq (ii) becomes
$\log \left|\frac{x}{20,000}\right|=\left(\frac{1}{5} \log \frac{5}{4}\right) t$
In 2009, t = 10
$\therefore \quad \log \left|\frac{x}{20,000}\right|=\left(\frac{1}{5} \log \frac{5}{4}\right) \times 10$
$\Rightarrow \quad \log \left|\frac{x}{20,000}\right|=2 \log \left(\frac{5}{4}\right)$
$=\log \left(\frac{5}{4}\right)^2$
$=\log \left(\frac{25}{16}\right)$
$\frac{x}{20,000}=\frac{25}{16}$
$\Rightarrow \quad x=\frac{25 \times 20,000}{16}$
= 31,250
Hence, in 2009 the population was 31,250.

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Question 25 Marks

It is given that radium decomposes at a rate proportional to the amount present. If P% of the original amount of radium disappears in I years, what percentage of it will remain after 2l years?

Answer

Let $A_0$ be the original amount of radium and A be the amount of radium at any time t. Then, the rate of decomposition of radium is $\frac{d A}{d t}$.
Given,$\frac{d A}{d t} \propto A$
$\Rightarrow \quad \frac{d A}{d t}=-k A$, where $k$ is positive constant
$\Rightarrow \quad \frac{d A}{A}=-k d t$
$\Rightarrow \quad \int \frac{d A}{A}=-k \int d t$
$\Rightarrow \log |A|=-k t+C$ .....(i)
At $t=0$, we have $A=A_0$, therefore from eq (i), we get
$\log \left|A_0\right|=0+C \Rightarrow C=\log A_0$
$\therefore E q$ (i) becomes,
$\log A=-k t+\log A_0$
$\Rightarrow \quad \log \left|\frac{A}{A_0}\right|=-k t$ .....(ii)
Given that $p \%$ of the original amount of radium disintegrates in I years.
$\therefore$ Amount of radium present at $t=l$ is
$\left(A_0-\frac{p}{100} \times A_0\right)=\left(1-\frac{p}{100}\right) A_0$
Put $A=A_0\left(1-\frac{p}{100}\right)$ and $t=l$ in eq. (ii), we get
$\log \left(1-\frac{p}{100}\right)=-k l$
$\Rightarrow \quad k=\frac{-1}{l} \log \left(1-\frac{p}{100}\right)$
Substitute value of k in eq (ii), we get
$\log \left(\frac{A}{A_0}\right)=\frac{t}{l} \log \left(1-\frac{p}{100}\right)$ ....(iii)
Let A be the amount of radium available after 2l years
Put t = 2l in eq (iii), we get
$\log \left|\frac{A}{A_0}\right|=2 \log \left|1-\frac{p}{100}\right|$
$\Rightarrow \quad\left(\frac{A}{A_0}\right)=\left(1-\frac{p}{100}\right)^2$
Multiplying both sides by 100
$\Rightarrow \quad \frac{A}{A_0} \times 100=\left(1-\frac{p}{100}\right)^2 \times 100$
$\Rightarrow \quad \frac{A}{A_0} \times 100=\left(10-\frac{p}{10}\right)^2$
Hence, required percentage is $\left(10-\frac{p}{10}\right)^2$

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Question 35 Marks

Suppose the growth of a population is proportional to the number present. If the population of a colony doubles in 25 days, in how many days will the population become triple ?

Answer

Let $P_0$ be the initial population and P be the population at any time t.
Given, $\frac{d P}{d t} \propto P$
$\Rightarrow \quad \frac{d P}{d t}=k P, k$ is constant
$\Rightarrow \quad \frac{d P}{P}=k d t$
$\Rightarrow \quad \int \frac{d P}{P}=k \int d t$
$\Rightarrow \quad \log |P|=k t+ C$ ....(i)
At t = 0 we have $P=P_0$
$\therefore$ From eq (i) $\log \mid P_{0 \mid}=0+C$
$\Rightarrow \quad C=\log \left|P_{0 \mid}\right|$
Therefore, eq (i) becomes $\log P=k t+\log P_0$
$\Rightarrow \quad \log \left|\frac{P}{P_0}\right|=k t$
Also, given that $t P=2 P_0$ when $t=25$ days
$\therefore$ From eq (ii), we have
$\log \left|\frac{2 P_0}{P_0}\right|=25 k$
$\Rightarrow \quad k=\frac{1}{25} \log 2$
Put value of k in eq (ii), we get
$\log \left|\frac{P}{P_0}\right|=\left(\frac{1}{25} \log 2\right) t$ ....(iii)
Put value of k in eq (ii), we get
$\log \left|\frac{P}{P_0}\right|=\left(\frac{1}{25} \log 2\right) t$ ....(iii)
Suppose the population is tripled in $t_1$ days i.e.,
$P=3 P_0$ when $t=t_1$
Put $P=3 P_0$ and $t=t_1$ in eq (ii), we get
$\log \left|\frac{3 P_0}{P_0}\right|=\left(\frac{1}{25} \log 2\right) t_1$
$\Rightarrow \quad t_1=25\left(\frac{\log 3}{\log 2}\right)$ days
Hence, population is tripled in $25\left(\frac{\log 3}{\log 2}\right)$ days.

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Question 45 Marks

A radio active substance has a half life of h days. Find a formula for its mass m in terms of t the time, if the initial mass is $m_0$. What is the initial decay rate ?

Answer

According to question,
The equation of decay is $\frac{d m}{d t}=-k m$ ...(i)
$\therefore \quad \frac{d m}{m}=-k d t$
Integrating both sides, we get
$\int \frac{d m}{m}=-k \int d t$
$\therefore \quad \log m=-k t+C$ ...(ii)
Initially, $t=0, m=m_0 \therefore \log m_0=C$
and $\log m=-k t+\log m_0$
Again when $t=h$ (half life), $m=\frac{m_0}{2}$
[ $\because$ half life means - half mass decays in $h$ days]
$\Rightarrow \quad \log \left|\frac{m_0}{2}\right|=-k h+\log \left|m_0\right|$
$\Rightarrow \quad k h=\log \left|m_0\right|-\log \left|\frac{m_0}{2}\right|$
$=\log \left|\frac{m_0}{m_0 / 2}\right|=\log 2$
$\Rightarrow \quad k=\frac{\log 2}{h}$
From eq. (ii),
$\log m=\frac{-\log 2}{h} \cdot t+\log m_0$ $\left[\therefore C=\log m_0\right]$
$\therefore \quad \log |m|-\log \left|m_0\right|=\frac{-t}{h} \log 2$
$\Rightarrow \quad \log \left|\frac{m}{m_0}\right|=\log 2^{-\left(\frac{t}{h}\right)}$
$\Rightarrow \quad \frac{m}{m_0}=2^{-\left(\frac{t}{h}\right)}$
$m=m_0 .2^{-\left(\frac{t}{h}\right)}$....(ii)
Which is the formula for its mass $m$ in terms of $t$. The initial decay rate is given by $\frac{d m}{d t}$ when $m=$
$m_0$ and $k=\frac{\log 2}{h}$
Hence, $\frac{d m}{d t}=-k m=\frac{-\log 2}{h} \times m_0$

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Question 55 Marks

Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally in 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the rain drop at any time.

Answer

Let r(t) denotes the radius of the raindrop after t minutes.
Since, given the radius is decreasing as t increases, the rate of change of r must be negative.
Let V be the volume of the raindrop and S be its surface area
i.e., $V=\frac{4}{3} \pi r^3$ and $S=4 \pi r^2$
Also given, $\frac{d v}{d t} \propto S$
$\Rightarrow \quad \frac{d v}{d t}=-k S$
$[\therefore$ as the rate of change of radiusb with time is negative]
$\therefore \quad \frac{4}{3} \pi 3 r^2 \frac{d r}{d t}=-k 4 \pi r^2\left[\because \frac{d v}{d t}=\frac{4}{3} \pi\left(3 r^2\right)\right]$
or $\frac{d r}{d t}=-k$
Integrating (i), we get
$\int \frac{d r}{d t}=-\int k$ .....(i)
$\Rightarrow \quad \int d r=-\int k d t$
$\Rightarrow \quad r=-k t+C$ ....(ii)
Let at time t = 0 radius is $r_0$
$\therefore$ From $( ii ), r_0=0+C=C$
Hence $r=-k t+r_0$ is the required expression.

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Question 65 Marks

Solve the differential equation: $(x-y-2) d x-(2 x-2 y-3) d y=0$

Answer

Given, $(x-y-2) d x-(2 x-2 y-3) d y=0$
$\Rightarrow \quad(x-y-2) d x-[2(x-y)-3] d y=0$
$\Rightarrow \quad \frac{x-y-2}{2(x-y)-3}=\frac{d y}{d x}$
Let x - y - 2 = t
$\Rightarrow \quad 1-\frac{d y}{d x}=\frac{d t}{d x}$
$\Rightarrow \quad \frac{d y}{d x}=1-\frac{d t}{d x}$
$\frac{t}{2(t+2)-3}=1-\frac{d t}{d x}$
$\binom{\because x-y-2=t}{x-y=t+2}$
$\Rightarrow \quad \frac{t}{2 t+4-3}=1-\frac{d t}{d x}$
$\Rightarrow \quad \frac{t}{2 t+1}=1-\frac{d t}{d x}$
$\Rightarrow \quad \frac{d t}{d x}=1-\frac{t}{2 t+1} \Rightarrow \frac{d t}{d x}=\frac{2 t+1-t}{2 t+1}$
$\Rightarrow \quad \frac{d t}{d x}=\frac{t+1}{2 t+1}$
$\Rightarrow \quad \frac{2 t+1}{t+1} d t=d x$
Integrating both sides
$\int \frac{2 t+1}{t+1} d t=\int d x$
$\begin{array}{l}\Rightarrow \quad \int \frac{2 t+2-2+1}{t+1} d t=\int d x \\ \Rightarrow \quad \int \frac{2(t+1)-1}{t+1} d t=x+c \\ \Rightarrow \quad \int 2 d t-\int \frac{1}{t+1} d t=x+c\end{array}$
$2 t-\log |t+1|=x+c$
$\begin{aligned} \Rightarrow & 2(x-y-2)-\log |x-y-2+1| =x+c \\ \Rightarrow & 2 x-2 y-4-\log |x-y-1| =x+c \\ \Rightarrow & x-2 y-4-\log |x-y-1| =c \\ \Rightarrow & x-2 y-\log |x-y-1| =C_1\left(C_1=c+4\right)\end{aligned}$
is the required solution.

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Question 75 Marks

Solve the following differential equation. $\sqrt{1+x^2+y^2+x^2 y^2}+x y \frac{d y}{d x}=0$

Answer

Given differential equation is
$\sqrt{1+x^2+y^2+x^2 y^2}+x y \frac{d y}{d x}=0$
or, $\sqrt{\left(1+x^2\right)+y^2\left(1+x^2\right)}=-x y \frac{d y}{d x}$
or, $\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=-x y \frac{d y}{d x}$
or, $\sqrt{1+x^2} \cdot \sqrt{1+y^2}=-x y \frac{d y}{d x}$
or, $\frac{y}{\sqrt{1+y^2}} d y=-\frac{\sqrt{1+x^2}}{x} d x$
On integrating both sides, we get
$\int \frac{y}{\sqrt{1+y^2}} d y=-\int \frac{\sqrt{1+x^2}}{x^2} x d x$
On putting $1+y^2=t$ and $1+x^2=u^2$
or 2ydy = dt and 2x dx = 2u du
$\Rightarrow y d y=\frac{d t}{2}$
and x dx = u du
$\therefore \quad \frac{1}{2} \int t^{-1 / 2} d t=-\int \frac{u}{u^2-1} u d u$
or, $\frac{1}{2} \int t^{-1 / 2} d t=-\int \frac{u^2}{u^2-1} d u$
or, $\frac{1}{2} \frac{t^{1 / 2}}{\frac{1}{2}}=-\int \frac{\left(u^2-1+1\right)}{u^2-1} d u$
or$t^{1 / 2}=-\int \frac{u^2-1}{u^2-1} d u-\int \frac{1}{u^2-1} d u$
or $\quad \sqrt{1+y^2}=-\int d u-\int \frac{1}{u^2-(1)^2} d u$
or, $\sqrt{1+y^2}=-u-\frac{1}{2} \log \left|\frac{u-1}{u+1}\right|+C$
$\left[\because \int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\right]$
$\therefore \quad \sqrt{1+y^2}=-\sqrt{1+x^2}-\frac{1}{2} \log \left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C$
which is the required solution.

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Question 85 Marks

Obtain the differential equation of all the circles of radius r.

Answer

$(x-a)^2+(y-b)^2=r^2$ ....(i)
or, $2(x-a)+2(y-b) \frac{d y}{d x}=0$ .....(ii)
or, $1+(y-b) \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0$ ....(iii)
$\therefore(y-b)=-\frac{\left(1+y_1^2\right)}{y_2}$
From (ii), $(x-a)=\frac{y_1\left(1+y_1^2\right)}{y_2}$
Putting these values in eqn. (i),$\frac{y_1^2\left(1+y_1^2\right)^2}{y_2^2}+\frac{\left(1+y_1^2\right)^2}{y_2^2}=r^2$
or, $\left[1+\left(\frac{d y}{d x}\right)^2\right]^3=r^2\left(\frac{d^2 y}{d x^2}\right)^2$

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Applied Maths 5 Marks Questions

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