NCERT Solutions for Exercise 9.5 Class 12 Maths Chapter 9- Differential Equations

# NCERT Solutions for Exercise 9.5 Class 12 Maths Chapter 9- Differential Equations

Edited By Ramraj Saini | Updated on Dec 04, 2023 08:26 AM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5

NCERT Solutions for Exercise 9.5 Class 12 Maths Chapter 9 Differential Equations are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 9.5 Class 12 Maths chapter 9 comes under the topic of homogeneous differential equations. Before going to the sample questions and exercise 9.5 Class 12 Maths, the NCERT Book explains what is a homogeneous equation is and how to identify it. Then a few examples are given and proceed to NCERT solutions for Class 12 Maths chapter 9 exercise 9.5. Questions based on homogeneous differential equations are given in the Class 12 Maths chapter 9 exercise 9.5. Questions to find both particular and general solutions of homogeneous differential equations are present in the Class 12th Maths chapter 6 exercise 9.5.

12th class Maths exercise 9.5 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Access NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5

Differential Equations Class 12 Chapter 9 Exercise: 9.5

The given diffrential eq can be written as
$\frac{dy}{dx}=\frac{x^{2}+y^{2}}{x^{2}+xy}$
Let $F(x,y)=\frac{x^{2}+y^{2}}{x^{2}+xy}$
Now, $F(\lambda x,\lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{(\lambda x)^{2}+(\lambda x)(\lambda y)}$
$=\frac{x^{2}+y^{2}}{x^{2}+xy} = \lambda ^{0}F(x,y)$ Hence, it is a homogeneous equation.

To solve it put y = vx
Diff
erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\v +x\frac{dv}{dx} = \frac{x^{2}+(vx)^{2}}{x^{2}+x(vx)}\\ v +x\frac{dv}{dx} = \frac{1+v^{2}}{1+v}$

$x\frac{dv}{dx} = \frac{(1+v^{2})-v(1+v)}{1+v} = \frac{1-v}{1+v}$

$( \frac{1+v}{1-v})dv = \frac{dx}{x}$

$( \frac{2}{1-v}-1)dv = \frac{dx}{x}$
Integrating on both side, we get;
$\\-2\log(1-v)-v=\log x -\log k\\ v= -2\log (1-v)-\log x+\log k\\ v= \log\frac{k}{x(1-v)^{2}}\\$
Again substitute the value $y = \frac{v}{x}$ ,we get;

$\\\frac{y}{x}= \log\frac{kx}{(x-y)^{2}}\\ \frac{kx}{(x-y)^{2}}=e^{y/x}\\ (x-y)^{2}=kxe^{-y/x}$
This is the required solution of given diff. equation

the above differential eq can be written as,

$\frac{dy}{dx} = F(x,y)=\frac{x+y}{x}$ ............................(i)

Now, $F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x} = \lambda ^{0}F(x,y)$
Thus the given differential eq is a homogeneous equaion
Now, to solve substitute y = vx
Diff erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx}= \frac{x+vx}{x} = 1+v$
$\\x\frac{dv}{dx}= 1\\ dv = \frac{dx}{x}$
Integrating on both sides, we get; (and substitute the value of $v =\frac{y}{x}$ )

$\\v =\log x+C\\ \frac{y}{x}=\log x+C\\ y = x\log x +Cx$
this is the required solution

$(x-y)dy - (x+y)dx = 0$

The given differential eq can be written as;

$\frac{dy}{dx}=\frac{x+y}{x-y} = F(x,y)(let\ say)$ ....................................(i)

$F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}= \lambda ^{0}F(x,y)$
Hence it is a homogeneous equation.

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\v+x\frac{dv}{dx}= \frac{1+v}{1-v}\\ x\frac{dv}{dx} = \frac{1+v}{1-v}-v =\frac{1+v^{2}}{1-v}$

$\frac{1-v}{1+v^{2}}dv = (\frac{1}{1+v^{2}}-\frac{v}{1-v^{2}})dv=\frac{dx}{x}$
Integrating on both sides, we get;

$\tan^{-1}v-1/2 \log(1+v^{2})=\log x+C$
again substitute the value of $v=y/x$
$\\\tan^{-1}(y/x)-1/2 \log(1+(y/x)^{2})=\log x+C\\ \tan^{-1}(y/x)-1/2 [\log(x^{2}+y^{2})-\log x^{2}]=\log x+C\\ tan^{-1}(y/x) = 1/2[\log (x^{2}+y^{2})]+C$

This is the required solution.

$(x^2 - y^2)dx + 2xydy = 0$

we can write it as;

$\frac{dy}{dx}= -\frac{(x^{2}-y^{2})}{2xy} = F(x,y)\ (let\ say)$ ...................................(i)

$F(\lambda x,\lambda y) = \frac{(\lambda x)^{2}-(\lambda y)^{2}}{2(\lambda x)(\lambda y)} = \lambda ^{0}.F(x,y)$
Hence it is a homogeneous equation

Now, to solve substitute y = vx
Diff erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx} = \frac{ x^{2}-(vx)^{2}}{2x(vx)} =\frac{v^{2}-1}{2v}$
$\\x\frac{dv}{dx} =\frac{v^{2}+1}{2v}\\ \frac{2v}{1+v^{2}}dv=\frac{dx}{x}$
integrating on both sides, we get

$\log (1+v^{2})= -\log x +\log C = \log C/x$
$\\= 1+v^{2} = C/x\\ = x^2+y^{2}=Cx$ .............[ $v =y/x$ ]
This is the required solution.

$x^2\frac{dy}{dx} = x^2 - 2y^2 +xy$

$\frac{dy}{dx}= \frac{x^{2}-2y^{2}+xy}{x^{2}} = F(x,y)\ (let\ say)$

$F(\lambda x,\lambda y)= \frac{(\lambda x)^{2}-2(\lambda y)^{2}+(\lambda .\lambda )xy}{(\lambda x)^{2}} = \lambda ^{0}.F(x,y)$ ............(i)
Hence it is a homogeneous eq

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\v+x\frac{dv}{dx}= 1-2v^{2}+v\\ x\frac{dv}{dx} = 1-2v^{2}\\ \frac{dv}{1-2v^{2}}=\frac{dx}{x}$

$1/2[\frac{dv}{(1/\sqrt{2})^{2}-v^{2}}] = \frac{dx}{x}$

On integrating both sides, we get;

$\frac{1}{2\sqrt{2}}\log (\frac{1/\sqrt{2}+v}{1/\sqrt{2}-v}) = \log x +C$
after substituting the value of $v= y/x$

$\frac{1}{2\sqrt{2}}\log (\frac{x+\sqrt{2}y}{x-\sqrt{2}y}) = \log \left | x \right | +C$

This is the required solution

$xdy - ydx = \sqrt{x^2 + y^2}dx$

$\frac{dy}{dx}=\frac{y+\sqrt{x^{2}+y^{2}}}{x} = F(x,y)$ .................................(i)

$F(\mu x,\mu y)=\frac{\mu y+\sqrt{(\mu x)^{2}+(\mu y)^{2}}}{\mu x} =\mu^{0}.F(x,y)$
henxe it is a homogeneous equation

Now, to solve substitute y = vx

Diff erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx}= v+\sqrt{1+v^{2}}=\sqrt{1+v^{2}}$

$=\frac{dv}{\sqrt{1+v^{2}}} =\frac{dx}{x}$

On integrating both sides,

$\Rightarrow \log \left | v+\sqrt{1+v^{2}} \right | = \log \left | x \right |+\log C$
Substitute the value of v=y/x , we get

$\\\Rightarrow \log \left | \frac{y+\sqrt{x^{2}+y^{2}}}{x} \right | = \log \left | Cx \right |\\ y+\sqrt{x^{2}+y^{2}} = Cx^{2}$

Required solution

Question:7 Solve.

$\left\{x\cos\left(\frac{y}{x} \right ) + y\sin\left(\frac{y}{x} \right ) \right \}ydx = \left\{y\sin\left(\frac{y}{x} \right ) - x\cos\left(\frac{y}{x} \right ) \right \}xdy$

$\frac{dy}{dx} =\frac{x \cos(y/x)+y\sin(y/x)}{y\sin(y/x)-x\cos(y/x)}.\frac{y}{x} = F(x,y)$ ......................(i)
By looking at the equation we can directly say that it is a homogenous equation.

Now, to solve substitute y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx} =\frac{v \cos v+v^{2}\sin v}{v\sin v-\cos v}\\ =x\frac{dv}{dx} = \frac{2v\cos v}{v\sin v-\cos v}\\ =(\tan v-1/v)dv = \frac{2dx}{x}$

integrating on both sides, we get

$\\=\log(\frac{\sec v}{v})= \log (Cx^{2})\\=\sec v/v =Cx^{2}$
substitute the value of v= y/x , we get

$\\\sec(y/x) =Cxy \\ xy \cos (y/x) = k$

Required solution

Question:8 Solve.

$x\frac{dy}{dx} - y + x\sin\left(\frac{y}{x}\right ) = 0$

$\frac{dy}{dx}=\frac{y-x \sin(y/x)}{x} = F(x,y)$ ...............................(i)

$F(\mu x, \mu y)=\frac{\mu y-\mu x \sin(\mu y/\mu x)}{\mu x} = \mu^{0}.F(x,y)$
it is a homogeneous equation

Now, to solve substitute y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$v+x\frac{dv}{dx}= v- \sin v = -\sin v$
$\Rightarrow -\frac{dv}{\sin v} = -(cosec\ v)dv=\frac{dx}{x}$

On integrating both sides we get;

$\\\Rightarrow \log \left | cosec\ v-\cot v \right |=-\log x+ \log C\\ \Rightarrow cosec (y/x) - \cot (y/x) = C/x$

$= x[1-\cos (y/x)] = C \sin (y/x)$ Required solution

Question:9 Solve.

$ydx + x\log\left(\frac{y}{x} \right ) -2xdy = 0$

$\frac{dy}{dx}= \frac{y}{2x-x \log(y/x)} = F(x,y)$ ..................(i)

$\frac{\mu y}{2\mu x-\mu x \log(\mu y/\mu x)} = F(\mu x,\mu y) = \mu^{0}.F(x,y)$

hence it is a homogeneous eq

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx}= \frac{v}{2-\log v}\\ =x\frac{dv}{dx} = \frac{v\log v-v}{2-\log v}\\ =[\frac{1}{v(\log v-1)}-\frac{1}{v}]dv=\frac{dx}{x}$
integrating on both sides, we get; ( substituting v =y/x)

$\\\Rightarrow \log[\log(y/x)-1]-\log(y/x)=\log(Cx)\\\Rightarrow \frac{x}{y}[\log(y/x)-1]=Cx\\ \Rightarrow \log (y/x)-1=Cy$

This is the required solution of the given differential eq

Question:10 Solve.

$\left(1 + e^{\frac{x}{y}} \right )dx + e^\frac{x}{y}\left(1-\frac{x}{y}\right )dy = 0$

$\frac{dx}{dy}=\frac{-e^{x/y}(1-x/y)}{1+e^{x/y}} = F(x,y)$ .......................................(i)

$= F(\mu x,\mu y)=\frac{-e^{\mu x/\mu y}(1-\mu x/\mu y)}{1+e^{\mu x/\mu y}} =\mu^{0}.F(x,y)$
Hence it is a homogeneous equation.

Now, to solve substitute x = yv

Diff erentiating on both sides wrt $x$
$\frac{dx}{dy}= v +y\frac{dv}{dy}$

Substitute this value in equation (i)

$\dpi{100} \\=v+y\frac{dv}{dy} = \frac{-e^{v}(1-v)}{1+e^{v}} \\ =y\frac{dv}{dy} = -\frac{v+e^{v}}{1+e^{v}}\\ =\frac{1+e^{v}}{v+e^{v}}dv=-\frac{dy}{y}$

Integrating on both sides, we get;

$\dpi{100} \log(v+e^{v})=-\log y+ \log c =\log (c/y)\\ =[\frac{x}{y}+e^{x/y}]= \frac{c}{y}\\\Rightarrow x+ye^{x/y}=c$
This is the required solution of the diff equation.

Question:11 Solve for particular solution.

$(x + y)dy + (x -y)dx = 0;\ y =1\ when \ x =1$

$\frac{dy}{dx}=\frac{-(x-y)}{x+y} =F(x,y)$ ..........................(i)

We can clearly say that it is a homogeneous equation.

Now, to solve substitute y = vx

Diff erentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\v+x\frac{dv}{dx}=\frac{v-1}{v+1}\\ \Rightarrow x\frac{dv}{dx} = -\frac{(1+v^{2})}{1+v}$

$\frac{1+v}{1+v^{2}}dv = [\frac{v}{1+v^{2}}+\frac{1}{1+v^{2}}]dv=-\frac{dx}{x}$

On integrating both sides

$\\=\frac{1}{2}[\log (1+v^{2})]+\tan^{-1}v = -\log x +k\\ =\log(1+v^{2})+2\tan^{-1}v=-2\log x +2k\\ =\log[(1+(y/x)^{2}).x^{2}]+2\tan^{-1}(y/x)=2k\\ =\log(x^{2}+y^{2})+2\tan^{-1}(y/x) = 2k$ ......................(ii)

Now, y=1 and x= 1

$\\=\log 2 +2\tan^{-1}1=2k\\ =\pi/2+\log 2 = 2k\\$

After substituting the value of 2k in eq. (ii)

$\log(x^{2}+y^2)+2\tan^{-1}(y/x)=\pi/2+\log 2$

This is the required solution.

Question:12 Solve for particular solution.

$x^2dy + (xy + y^2)dx = 0; y =1\ \textup{when}\ x = 1$

$\frac{dy}{dx}= \frac{-(xy+y^{2})}{x^{2}} = F(x,y)$ ...............................(i)

$F(\mu x, \mu y)=\frac{-\mu^{2}(xy+(\mu y)^{2})}{(\mu x)^{2}} =\mu ^{0}. F(x,y)$
Hence it is a homogeneous equation

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i), we get

$\\=v+\frac{xdv}{dx}= -v- v^{2}\\ =\frac{xdv}{dx}=-v(v+2)\\ =\frac{dv}{v+2}=-\frac{dx}{x}\\ =1/2[\frac{1}{v}-\frac{1}{v+2}]dv=-\frac{dx}{x}$

Integrating on both sides, we get;

$\\=\frac{1}{2}[\log v -\log(v+2)]= -\log x+\log C\\ =\frac{v}{v+2}=(C/x)^{2}$

replace the value of v=y/x

$\frac{x^{2}y}{y+2x}=C^{2}$ .............................(ii)

Now y =1 and x = 1

$C = 1/\sqrt{3}$
therefore,

$\frac{x^{2}y}{y+2x}=1/3$

Required solution

Question:13 Solve for particular solution.

$\left [x\sin^2\left(\frac{y}{x} \right ) - y \right ]dx + xdy = 0;\ y =\frac{\pi}{4}\ when \ x = 1$

$\frac{dy}{dx}=\frac{-[x\sin^{2}(y/x)-y]}{x} = F(x,y)$ ..................(i)

$F(\mu x,\mu y)=\frac{-[\mu x\sin^{2}(\mu y/\mu x)-\mu y]}{\mu x}=\mu ^{0}.F(x,y)$

Hence it is a homogeneous eq

Now, to solve substitute y = vx

Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

on integrating both sides, we get;

$\\-\cot v =\log\left | x \right | -C\\ =\cot v = \log\left | x \right |+\log C$

On substituting v =y/x

$=\cot (y/x) = \log\left | Cx \right |$ ............................(ii)

Now, $y = \pi/4\ @ x=1$

$\\\cot (\pi/4) = \log C \\ =C=e^{1}$

put this value of C in eq (ii)

$\cot (y/x)=\log\left | ex \right |$

Required solution.

Question:14 Solve for particular solution.

$\frac{dy}{dx} - \frac{y}{x} + \textup{cosec}\left (\frac{y}{x} \right ) = 0;\ y = 0 \ \textup{when}\ x = 1$

$\frac{dy}{dx} = \frac{y}{x} -cosec(y/x) =F(x,y)$ ....................................(i)

the above eq is homogeneous. So,
Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx}=v- cosec\ v\\ =x\frac{dv}{dx} = -cosec\ v\\ =-\frac{dv}{cosec\ v}= \frac{dx}{x}\\ =-\sin v dv = \frac{dx}{x}$

on integrating both sides, we get;

$\\=cos\ v = \log x +\log C =\log Cx\\ =\cos(y/x)= \log Cx$ .................................(ii)

now y = 0 and x =1 , we get

$C =e^{1}$

put the value of C in eq 2

$\cos(y/x)=\log \left | ex \right |$

Question:15 Solve for particular solution.

$2xy + y^2 - 2x^2\frac{dy}{dx} = 0 ;\ y = 2\ \textup{when}\ x = 1$

The above eq can be written as;

$\frac{dy}{dx}=\frac{2xy+y^{2}}{2x^{2}} = F(x,y)$
By looking, we can say that it is a homogeneous equation.

Now, to solve substitute y = vx
Differentiating on both sides wrt $x$
$\frac{dy}{dx}= v +x\frac{dv}{dx}$

Substitute this value in equation (i)

$\\=v+x\frac{dv}{dx}= \frac{2v+v^{2}}{2}\\ =x\frac{dv}{dx} = v^{2}/2\\ = \frac{2dv}{v^{2}}=\frac{dx}{x}$

integrating on both sides, we get;

$\\=-2/v=\log \left | x \right |+C\\ =-\frac{2x}{y}=\log \left | x \right |+C$ .............................(ii)

Now, y = 2 and x =1, we get

C =-1
put this value in equation(ii)

$\\=-\frac{2x}{y}=\log \left | x \right |-1\\ \Rightarrow y = \frac{2x}{1- \log x}$

(A) $y = vx$

(B) $v = yx$

(C) $x = vy$

(D) $x =v$

$\frac{dx}{dy}= h\left(\frac{x}{y} \right )$
for solving this type of equation put x/y = v
x = vy

option C is correct

(A) $(4x + 6x +5)dy - (3y + 2x +4)dx = 0$

(B) $(xy)dx - (x^3 + y^3)dy = 0$

(C) $(x^3 +2y^2)dx + 2xydy =0$

(D) $y^2dx + (x^2 -xy -y^2)dy = 0$

Option D is the right answer.

$y^2dx + (x^2 -xy -y^2)dy = 0$
$\frac{dy}{dx}=\frac{y^{2}}{x^{2}-xy-y^{2}} = F(x,y)$
we can take out lambda as a common factor and it can be cancelled out

## More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5

To practice the concepts given in topic 9.5.2, 4 example questions are given prior to the exercise 9.5 Class 12 Maths. There are 17 questions in the Class 12 Maths chapter 9 exercise 9.5 and 2 questions of this have 4 choices. As mentioned in the first para certain questions in the Class 12th Maths chapter 6 exercise 9.5 are to find the general solutions and some questions are to find the particular solutions of homogeneous differential equations.

## Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.5

• Utilising the exercise 9.5 Class 12 Maths students can master the concepts of identifying the homogeneous equations and finding the solutions to these equations.
• Not only for CBSE board exams but also for various state board exams and Engineering Entrance Exams, the NCERT solutions for Class 12 Maths chapter 9 exercise 9.5 can be used.
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## Key Features Of NCERT Solutions for Exercise 9.5 Class 12 Maths Chapter 9

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 9.5 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 9.5, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 9.5 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 9.5 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 9.5 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 9.5 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

## Subject Wise NCERT Exemplar Solutions

1. What number of multiple-choice questions are present in the exercise 9.5 Class 12 Maths?

Two objective type questions with 4 choices are given in the NCERT solutions for Class 12 Maths chapter 9 exercise 9.5.

2. What is the first step in solving the problems given in the Class 12 Maths chapter 9 exercise 9.5?

The first step is to verify the given differential equation is homogeneous or not.

3. How many examples are solved in topic 9.5.2?

4 examples are given.

4. Whether the concept of integration is required to solve the problems of Class 12th Maths chapter 6 exercise 9.5?

Yes, the concepts of basic integrals studied in the NCERT Class 12 Maths chapter 7 is required to solve exercise 9.5

5. What is the topic discussed before the homogeneous differrential equation?

The differential equations of  variable separable form.

6. The number of exercises coming under 9.5 are……….

Three exercises are coming under the topic .5

7. What is the topic discussed after exercise 9.5?

The idea of a linear differential equation is discussed after exercise 9.5 Class 12 Maths.

8. How many questions are present in exercise 9.5?

Five

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Hope you are doing fine

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Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9