Applications of Differential Equations

Find the differential equation form of $V=iR$ with respect to change in time. If $V=26$ volts, $i=2$ Amperes, then find $R$ in ohms.

Find the differential equation form of $V=iR$ with respect to change in time. If $V=12$ volts, $i=4$ Amperes, then find $R$ in ohms.

Find the differential equation form of $V=iR$ with respect to change in time. If $V=24$ volts, $i=4$ Amperes, then find $R$ in ohms.

Find the differential equation form of $V=iR$ with respect to change in time. If $V=24$ volts, $i=2$ Amperes, then find $R$ in ohms.

Find the differential equation form of $V=iR$ with respect to change in time. If $V=20$ volts, $i=2$ Amperes, then find $R$ in ohms.

A tank contains $40 \mathrm{l}$ of solution containing $2 \mathrm{g}$ of substance per liter. Salt water containing $3 \mathrm{g}$ of this substance per liter runs in at the rate of $4 \frac{l}{\min }$ and the well - stirred mixture runs out at the same rate. Find the amount of substance in the tank after $50$ minutes. [Use, $\frac{1}{e^5}=0.006737947$]

A tank contains $40 \mathrm{l}$ of solution containing $2 \mathrm{g}$ of substance per liter. Salt water containing $3 \mathrm{g}$ of this substance per liter runs in at the rate of $4 \frac{l}{\min }$ and the well - stirred mixture runs out at the same rate. Find the amount of substance in the tank after $30$ minutes. [Use, $\frac{1}{e^3}=0.0497870684$]

A tank contains $40 \mathrm{l}$ of solution containing $2 \mathrm{g}$ of substance per liter. Salt water containing $3 \mathrm{g}$ of this substance per liter runs in at the rate of $4 \frac{l}{\min }$ and the well - stirred mixture runs out at the same rate. Find the amount of substance in the tank after $20$ minutes. [Use, $\frac{1}{e^2}=0.135335283$]

A tank contains $40 \mathrm{l}$ of solution containing $2 \mathrm{g}$ of substance per liter. Salt water containing $3 \mathrm{g}$ of this substance per liter runs in at the rate of $4 \frac{l}{\min }$ and the well - stirred mixture runs out at the same rate. Find the amount of substance in the tank after $10$ minutes. [Use, $\frac{1}{e}=0.367879441$]

A tank contains $40 \mathrm{l}$ of solution containing $2 \mathrm{g}$ of substance per liter. Salt water containing $3 \mathrm{g}$ of this substance per liter runs in at the rate of $4 \frac{l}{\min }$ and the well - stirred mixture runs out at the same rate. Find the amount of substance in the tank after $15$ minutes. [Use, $e^{-1.5}=0.22313016$]

The population grows at the rate of $8\%$ per year. Find the time taken for the population to become double in years. Given: $\log 2=0.6912$

In a culture of yeast the active ferment doubles itself in $3$ hours. Assuming that the quantity increases at a rate proportional to itself, determine the number of times it multiplies itself in $15$ hours.

Find the half life of uranium, which disintegrates at a rate proportional to the amount present at any instant. Given that $m_1$ and $m_2$ grams of uranium are present at time $t_1$ and $t_2$ respectively

The bacteria culture grows at a rate proportional to its size. After $2$ hours, there are $600$ bacteria and after $8$ hours the count is $75,000$. Find the initial population and when will the population reach $2,00,000$ ?

The rate of growth of the population is proportional to the number present. If the population doubled in the last $25$ years and the present population is $1$ lac, when will the city have population $4,00,000$$?$

A body is heated to$110°C$ and placed in air at $10°C$. After $1$ hour its temperature is $60°C$. How much additional time is required for it to cool to $30°C$ in hours $?$

The rate of decay of certain substance is directly proportional to the amount present at that instant. Initially, there are $27$ gms of certain substance and three hours later it is found that $8$ gms are left. Find the amount left after one more hour.

The rate of growth of bacteria is proportional to the number present. If initially, there were $1000$ bacteria and the number doubles in $1$ hour, find the number of bacteria after $2\frac{1}{2}$ hours. [Take $\sqrt{2}=1.414$]

If the population grows at the rate of $5\%$ per year, then the time taken for the population to become double is (Given $\log 2=0·6912$)

The bacteria increases at the rate proportional to the number of bacteria present. If the original number $N_0$ doubles in $4$ hours, then the number of bacteria in $12$ hours will be