RD Sharma Solutions Class 12 Mathematics Chapter 21 CSBQ

Satyajeet KumarUpdated on 25 Jan 2022, 05:59 PM IST

NCERT solutions can be precious for all students who will show up for their board exams. The RD Sharma class 12th exercise CSBQ Solutions especially will be helpful for all high school students who have a maths paper on their boards. RD Sharma class 12 chapter 21 exercise CSBQ will amaze you with its various benefits and uses. Students and teachers in India consider this book the best study guide for all students who want to ace their maths papers.

RD Sharma class 12 solutions Differential Equations ex CSBQ is your one-stop answer for maths. This material contains two questions that are divided into subparts summing up to 10 questions. RD Sharma Solutions It includes case studies based on fundamental concepts and can be solved by students with a bit of basic knowledge.

RD Sharma Class 12 Solutions Chapter21 CSBQ Differential Equation- Other Exercise

KCC Institute of Legal and Higher Education, Greater Noida

Law Admissions 2026 Open | Avail Merit-Based Scholarships

Apply

Global Institute of Business Studies, Bengaluru

BBA/MBA/PGDM Early Bird Applications Deadline 06th Nov'26. Apply now using coupon code CAREER250 and get ₹250 off the application fee

Apply

Differential Equations Excercise: CSBQ

Differential Equations Exercise Case Study Based Question Question 1 (i)

Answer: Option(a) $\frac{d V}{d t}=-\lambda, \lambda>0$ is a constant
Hint: Volume of sphere $=\frac{4}{3} \pi r^{3}$ & then differentiate.
Given: Volume of a spherical balloon being deflated changes at a constant rate.
Solution: Volume of a balloon at a time, $t= V$.
Now as volume changes at constant rate, we can say
$\begin{aligned} &\frac{d}{d t}(V) \alpha 1 \\ &\frac{d}{d t}(V)=\lambda \end{aligned}$

Where $\lambda$ is some constant.

$\frac{d V}{d t}=\lambda$
But as the volume of balloon is deflated i. e., reduced so $\lambda$ will be represented by -ve value.
Thus,
$\frac{d V}{d t}=-\lambda, \lambda>0$ is a constant .

Differential Equations Exercise Case Study Based Question Question 1 (ii)

Answer: option(a) $4 \pi r^{2} \frac{d r}{d t}=-\lambda, \lambda>0$
Hint: Use formula of volume of $\text { sphere }=\frac{4}{3} \pi r^{3}$
Given: r is the radius of balloon.
Solution: $\frac{d V}{d t}=-\lambda$ as framed above
So, using formula ,$\mathrm{V}=\frac{4}{3} \pi r^{3}$
Where, r = radius of spherical balloon.
$\begin{aligned} &\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=-\lambda \\ &\frac{4}{3} \pi \frac{d}{d t}\left(r^{3}\right)=-\lambda \\ &\frac{4 \pi}{3} 3 r^{2} \frac{d r}{d t}=-\lambda \\ &4 \pi r^{2} \frac{d r}{d t}=-\lambda \end{aligned}$

Thus the radius at instant t is

$4 \pi r^{2} \frac{d r}{d t}=-\lambda, \lambda>0$

Differential Equations Exercise Case Study Based Question Question 1 (iii)

Answer: option(c)
Hint: Try to understand the given equation is whether of linear or non- linear or homogeneous.
Given: Differential equation in r and t
Solution: obtained differential equation $4 \pi r^{2} \frac{d r}{d t}=-\lambda$
The above formed equation is of the order 1st. So, it is a first order differential equation.
It is of non – linear type and variable can be separated.
“So it is first order non- linear differential equation in variable separable form”.
Note: Linear equation is in the form of $\frac{d y}{d x}+y=x \ldots$
While homogeneous equation is having same degree.

Differential Equations Exercise Case Study Based Question Question 1 (iv)

Answer: option(a) $\therefore r=\left(27-\frac{3 \lambda t}{4 \pi}\right)^{\frac{1}{3}}$
Hint: Solve the differential equation & find value of radius at time t.
Given: radius of balloon is 3 units, time is t and r is the radius.
Solution: $4 \pi r^{2} \frac{d r}{d t}=-\lambda$
$4 \pi r^{2} d r=-\lambda d t$
Integrating on both sides,
$\begin{aligned} &\int 4 \pi r^{2} d r=\int-\lambda d t \\ &4 \pi \int r^{2} d r=-\lambda \int d t \\ &4 \pi \frac{r^{3}}{3}=-\lambda t+c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i) \end{aligned}$

Now given radius is 3 and & initially i.e., time, $t= 0$

Integrating on both sides,
$\begin{aligned} 4 \pi \frac{(3)^{3}}{3} &=-\lambda(0)+c \\ 4 \pi \frac{9 \times 3}{3} &=0+c \\ 36 \pi &=c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (ii) \end{aligned}$

Put (ii) in (i)
$\begin{aligned} 4 \pi \frac{r^{3}}{3} &=-\lambda t+36 \pi \\ 4 \pi r^{3} &=-3 \lambda t+108 \pi \\ r^{3} &=\frac{-3 \lambda t+108 \pi}{4 \pi} \\ r^{3} &=\frac{108}{4}-\frac{3 \lambda t}{4 \pi} \\ \therefore \quad r &=\left(27-\frac{3 \lambda t}{4 \pi}\right)^{\frac{1}{3}} \end{aligned}$

Differential Equations Exercise Case Study Based Question Question 1 (v)

Answer: option(b) $r=(27+63 t)^{\frac{1}{3}}$
Hint: Here the case is given of inflate , so use formula $4 \pi r^{2} \frac{d r}{d t}=-\lambda$ and solve it.
Given: radius of balloon is 6 units.
Solution: as the case seems to be of radius increasing, after 3 seconds the differential equation gets modified as ,
$\begin{aligned} &4 \pi r^{2} \frac{d r}{d t}=-\lambda \\ &4 \pi r^{2} d r=-\lambda d t \end{aligned}$
Integrating both the sides.
$\begin{aligned} &\int 4 \pi r^{2} d r=\int \lambda d t \\ &4 \pi \int r^{2} d r=\lambda \int d t \\ &4 \pi \frac{r^{3}}{3}=\lambda t+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(i) \end{aligned}$
Now as stated before the radius was initially 3 units.
$\begin{aligned} 4 \pi \frac{r^{3}}{3} &=\lambda t+c \\ 4 \pi \frac{(3)^{3}}{3} &=\lambda(0)+c \\ 36 \pi &=c\; \; \; \; \; \; \; \; \; \; \; \; \; .......(ii) \end{aligned}$
Put (ii) in (i)
$\therefore 4 \pi \frac{r^{3}}{3}=\lambda t+36 \pi$
Now it is given that after 3 seconds radius is 6 times.
$\begin{gathered} 4 \pi \frac{(6)^{3}}{3}=\lambda(3)+c \\ 288 \pi=3 \lambda+c \\ 288 \pi-3 \lambda=c\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ......(iii) \end{gathered}$
Put (ii) in (iii)
$\begin{aligned} &288 \pi-3 \lambda=36 \pi \\ &288 \pi-36 \pi=3 \lambda \\ &\therefore \lambda=96 \pi-12 \pi \\ &\therefore \lambda=84 \pi \; \; \; \; \; \; \; \; \; \; \; \; ....(iv) \end{aligned}$
Now put (ii) & (iv) in (i)
$\\\frac{4}{3} \pi r^{3}=84 \pi t+36 \pi\\ \begin{gathered} 4 \pi r^{3}=252 \pi t+108 \pi \\ r^{3}=63 t+\frac{108 \pi}{4 \pi} \\ r=(63 t+27)^{\frac{1}{3}} \\ \therefore r=(27+63 t)^{\frac{1}{3}} \end{gathered}$

Differential Equations Exercise Case Study Based Question Question 2 (i)

Answer: option(b) $\frac{d P}{d t}=\frac{P R}{100}$
Hint: Solve the problem by forming differential equation.
Given: principal= p
Time= t
Rate of interest= r% per annum
Solution: We know,
$I=\frac{P R T}{100}$
Where, I = rate of interest.
Interest = change in principal w .r .t to time.
So we say,
$\begin{aligned} &\frac{d}{d t}(P)=\frac{P R}{100} \\ &\therefore \frac{d P}{d t}=\frac{P R}{100} \end{aligned}$

Differential Equations Exercise Case Study Based Question Question 2 (ii)

Answer: option(c) $\log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100}$
Hint: Solve the equation by integrating.
Given: $P_{o}$ is the initial principal
Solution: $\frac{d P}{d t}=\frac{P r}{100}$
$\frac{d P}{P}=\frac{r}{100} d t$
Integrating both sides,
$\int\left(\frac{1}{P}\right) d P=\int\left(\frac{r}{100}\right) d t$
We know, $\int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c$
$\log P=\frac{r t}{100}+c \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i)$
Now, initially Principal, $P= P_{o}$
$\text { i.e., at } \mathrm{t}=0, \mathrm{P}=\mathrm{P}_{0}$
put values in (i)
$\begin{aligned} &\log \mathrm{P}_{0}=\frac{r(0)}{100}+c \\ &\mathrm{c}=\log \mathrm{P}_{0} \end{aligned}$ ........(ii)
put (ii) in (i)
$\begin{aligned} &\therefore \log P=\frac{r t}{100}+\log P_{0} \\ &\therefore \log P-\log P_{0}=\frac{r t}{100} \\ &\text { Use, } \log x-\log y=\log \left(\frac{x}{y}\right) \\ &\therefore \log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100} \end{aligned}$

Differential Equations Exercise Case Study Based Question Question 2 (iii)

Answer: option(a) $t=20 \log _{\mathrm{e}} 2$
Hint: Just put values in obtained equation.
Given $r= 5%$,
Solution: If $P_{0}=100 \text { then } P=2\; P_{2}=200$
Put all the values in ,
$\begin{aligned} &\log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100} \\ &\log \left(\frac{200}{100}\right)=\frac{5 \times t}{100} \\ &\log 2=\frac{t}{20} \end{aligned}$
$\therefore t=20 \log _{\mathrm{e}} 2$

Differential Equations Exercise Case Study Based Question Question 2 (iv)

Answer: option(c) $r=10 \log _{\mathrm{e}} 2 \%$
Hint: Put all the values & obtain value of ‘r’
Given: Rs 100 doubles itself in 10 years.
Solution : $P_{0}=100 \text { then } P=200=2 P_{0}$
$t = 10$ years
put values in following equation,
$\begin{aligned} &\log \left(\frac{200}{100}\right)=\frac{r \times 10}{100} \\ &\log 2=\frac{\mathrm{r}}{10} \\ &\therefore r=10 \log _{\mathrm{e}} 2 \% \end{aligned}$

Differential Equations Exercise Case Study Based Question Question 2 (v)

Answer: (a) $P=1000 e^{0.5}$
Hint: Substitute values in given equation & get the solution.
Given:
$\begin{aligned} &P_{0}=1000 \\ &r=5 \% \\ &t=10 \text { years } \end{aligned}$
Solution: we use the given data accordingly,
$\begin{aligned} &\log \left(\frac{P}{P_{0}}\right)=\frac{r t}{100} \\ &\log \left(\frac{p}{1000}\right)=\frac{5 \times 10}{100} \\ &\log P-\log 1000=\frac{50}{100} \\ &\log \left(\frac{P}{1000}\right)=0.5 \end{aligned}$
$\begin{aligned} &\frac{P}{1000}=e^{0.5} \\ \\&\therefore P=1000 e^{0.5} \end{aligned}$

RD Sharma Solutions can be the perfect guide for students in self-practice. There are several reasons why their NCERT solutions are the absolute best among so many other options. Here is a list of benefits that come with using class 12 RD Sharma chapter 21 exercise CSBQ solution:-

The RD Sharma class 12 solutions chapter 21 ex CSBQ has answers to all questions of the NCERT textbooks. The class 12 RD Sharma chapter 21 exercise CSBQ solution pdf is updated frequently in accordance with the latest syllabus of CBSE.
RD Sharma class 12 solutions chapter 21 ex CSBQ material is prepared by experts and is exam-oriented. Students can take advantage of the solutions by referring to them as a guide for their textbooks. Moreover, this material is updated to the latest version, which means it contains answers from the newest version of the book. Students can rest assured that the material they are referring to provides the best information possible.
The main advantage of having solved material is that students can revise whenever they want. RD Sharma class 12th exercise CSBQ Is designed specifically for the convenience of students. As it contains questions and answers in the same place, students can refer to it whenever they have doubts. RD Sharma's class 12th exercise CSBQ comes in handy as it saves a lot of time in their preparation.
The RD Sharma class 12th exercise CSBQ Material is available on career360's website and is free to access.

RD Sharma Chapter wise Solutions