(B) Let $x$ be the assets at time $t$. Given $\frac{dx}{dt} = k \sqrt{x}$.Separating variables,we get $\frac{dx}{\sqrt{x}} = k dt$.Integrating both si

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(B) Let $x$ be the assets at time $t$. Given $\frac{dx}{dt} = k \sqrt{x}$.Separating variables,we get $\frac{dx}{\sqrt{x}} = k dt$.Integrating both sides,$\int x^{-1/2} dx = \int k dt$,which gives $2 \sqrt{x} = kt + C$.At $t = 0$,$x = 9$,so $2 \sqrt{9} = k(0) + C \Rightarrow C = 6$.Thus,$2 \sqrt{x} = kt + 6$.At $t = 2$,$x = 16$,so $2 \sqrt{16} = k(2) + 6 \Rightarrow 8 = 2k + 6 \Rightarrow 2k = 2 \Rightarrow k = 1$.The equation becomes $2 \sqrt{x} = t + 6$.After $5$ more years,$t = 2 + 5 = 7$.Substituting $t = 7$,$2 \sqrt{x} = 7 + 6 = 13$.$\sqrt{x} = 6.5$.$x = (6.5)^2 = 42.25$ crores.

Class 12 Mathematics Differential Equations Application of differential equations

Difficult

The assets of a person are increasing at a rate proportional to the square root of the assets at a given time. His assets increase from Rupees $9$ crores to Rupees $16$ crores in $2$ years,then his assets at the end of $5$ more years will be

MHT CET 2022 All MHT CET

A
$46.25$ crores
B
$42.25$ crores
C
$30.25$ crores
D
$56.25$ crores

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Differential Equations

Subtopic

Application of differential equations

Previous Year

MHT CET 2022 Paper All MHT CET years →

The solution of $\frac{d^2 y}{d x^2}=0$ represents

EasyAP EAMCET 2020

View Solution

$A$ spherical balloon expands at a rate proportional to its surface area. Initially, its radius is $2 \text{ cm}$ and $5 \text{ minutes}$ later it increases to $7 \text{ cm}$. What will be the surface area of the spherical balloon after $12 \text{ minutes}$ (in $\text{ cm}^2$)?

DifficultMHT CET 2026

View Solution

In a certain culture of bacteria,the rate of increase is proportional to the number of bacteria present at that instant. It is found that there are $10,000$ bacteria at the end of $3$ hours and $40,000$ bacteria at the end of $5$ hours. Find the number of bacteria present in the beginning.

DifficultMHT CET 2023

View Solution

The slope of the tangent to a curve $C : y = y(x)$ at any point $(x, y)$ on it is $\frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$. If $C$ passes through the points $(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}})$ and $(\alpha, \frac{1}{2}e^{2\alpha})$,then $e^{\alpha}$ is equal to:

JEE MAIN 2022

View Solution

Let $f:[0,2] \rightarrow R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$. Let $F(x)=\int_0^{x^2} f(\sqrt{t}) dt$ for $x \in [0,2]$. If $F'(x)=f'(x)$ for all $x \in (0,2)$,then $F(2)$ equals

DifficultIIT 2014

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The assets of a person are increasing at a rate proportional to the square root of the assets at a given time. His assets increase from Rupees $9$ crores to Rupees $16$ crores in $2$ years,then his assets at the end of $5$ more years will be

Similar Questions

The solution of $\frac{d^2 y}{d x^2}=0$ represents

$A$ spherical balloon expands at a rate proportional to its surface area. Initially, its radius is $2 \text{ cm}$ and $5 \text{ minutes}$ later it increases to $7 \text{ cm}$. What will be the surface area of the spherical balloon after $12 \text{ minutes}$ (in $\text{ cm}^2$)?

In a certain culture of bacteria,the rate of increase is proportional to the number of bacteria present at that instant. It is found that there are $10,000$ bacteria at the end of $3$ hours and $40,000$ bacteria at the end of $5$ hours. Find the number of bacteria present in the beginning.

The slope of the tangent to a curve $C : y = y(x)$ at any point $(x, y)$ on it is $\frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$. If $C$ passes through the points $(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}})$ and $(\alpha, \frac{1}{2}e^{2\alpha})$,then $e^{\alpha}$ is equal to:

Let $f:[0,2] \rightarrow R$ be a function which is continuous on $[0,2]$ and is differentiable on $(0,2)$ with $f(0)=1$. Let $F(x)=\int_0^{x^2} f(\sqrt{t}) dt$ for $x \in [0,2]$. If $F'(x)=f'(x)$ for all $x \in (0,2)$,then $F(2)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module