NCERT Solutions for Exercise 9.4 Class 12 Maths Chapter 9- Differential Equations

NCERT Solutions for Exercise 9.4 Class 12 Maths Chapter 9- Differential Equations

Edited By Ramraj Saini | Updated on Dec 04, 2023 08:20 AM IST | #CBSE Class 12th
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NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.4

NCERT Solutions for Exercise 9.4 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 9.4 Class 12 Maths chapter 9 gives an idea about how to solve differential equations of variable separable type. The given differential equations are converted to variable separable forms and are solved using integration. The NCERT solutions for Class 12 Maths chapter 9 exercise 9.4 presents such 23 questions. In Class 12 Maths chapter 9 exercise 9.4 certain cases are discussed where the differential equations are to be formed using the given data and then solve using the variable separable method. Two more methods were discussed in the coming exercises to solve the differential equations.

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  2. Access NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.4
  3. Differential Equations Class 12 Chapter 9 Exercise: 9.4
  4. More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.4
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  7. Key Features Of NCERT Solutions for Exercise 9.4 Class 12 Maths Chapter 9
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12th class Maths exercise 9.4 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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Differential Equations Class 12 Chapter 9 Exercise: 9.4

Question:1 Find the general solution: \frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}

Answer:

Given,

\frac{dy}{dx} = \frac{1-\cos x}{1 + \cos x}

\\ \implies\frac{dy}{dx} = \frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}} = tan^2\frac{x}{2} \\ \implies dy = (sec^2\frac{x}{2} - 1)dx

\\ \implies \int dy = \int sec^2\frac{x}{2}dx - \int dx \\ \implies y = 2tan^{-1}\frac{x}{2} - x + C


Question:2 Find the general solution: \frac{dy}{dx} = \sqrt{4-y^2}\ (-2 < y < 2)

Answer:

Given, in the question

\frac{dy}{dx} = \sqrt{4-y^2}

\\ \implies \frac{dy}{\sqrt{4-y^2}} = dx \\ \implies \int \frac{dy}{\sqrt{4-y^2}} = \int dx

\\ (\int \frac{dy}{\sqrt{a^2-y^2}} = sin^{-1}\frac{y}{a})\\

The required general solution:

\\ \implies sin^{-1}\frac{y}{2} = x + C

Question:3 Find the general solution: \frac{dy}{dx} + y = 1 (y\neq 1)

Answer:

Given, in the question

\frac{dy}{dx} + y = 1

\\ \implies \frac{dy}{dx} = 1- y \\ \implies \int\frac{dy}{1-y} = \int dx

(\int\frac{dx}{x} = lnx)

\\ \implies -log(1-y) = x + C\ \ (We\ can\ write\ C= log k) \\ \implies log k(1-y) = -x \\ \implies 1- y = \frac{1}{k}e^{-x} \\

The required general equation

\implies y = 1 -\frac{1}{k}e^{-x}

Question:4 Find the general solution: \sec^2 x \tan y dx + \sec^2 y \tan x dy = 0

Answer:

Given,

\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0

\\ \implies \frac{sec^2 y}{tan y}dy = -\frac{sec^2 x}{tan x}dx \\ \implies \int \frac{sec^2 y}{tan y}dy = - \int \frac{sec^2 x}{tan x}dx

Now, let tany = t and tanx = u

sec^2 y dy = dt\ and\ sec^2 x dx = du

\\ \implies \int \frac{dt}{t} = -\int \frac{du}{u} \\ \implies log t = -log u +logk \\ \implies t = \frac{1}{ku} \\ \implies tany = \frac{1}{ktanx}

Question:5 Find the general solution:

(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0

Answer:

Given, in the question

(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0

\\ \implies dy = \frac{(e^x - e^{-x})}{(e^x + e^{-x})}dx

Let,

\\ (e^x + e^{-x}) = t \\ \implies (e^x - e^{-x})dx = dt

\\ \implies \int dy = \int \frac{dt}{t} \\ \implies y = log t + C \\ \implies y = log(e^x + e^{-x}) + C

This is the general solution

Question:6 Find the general solution: \frac{dy}{dx} = (1+x^2)(1+y^2)

Answer:

Given, in the question

\frac{dy}{dx} = (1+x^2)(1+y^2)

\\ \implies \int \frac{dy}{(1+y^2)} = \int (1+x^2)dx

(\int \frac{dx}{(1+x^2)} =tan^{-1}x +c)

\\ \implies tan^{-1}y = x+\frac{x^3}{3} + C

Question:7 Find the general solution: y\log y dx - x dy = 0

Answer:

Given,

y\log y dx - x dy = 0

\\ \implies \frac{1}{ylog y}dy = \frac{1}{x}dx

let logy = t

=> 1/ydy = dt

\\ \implies \int \frac{dt}{t} = \int \frac{1}{x}dx \\ \implies \log t = \log x + \log k \\ \implies t = kx \\ \implies \log y = kx

This is the general solution

Question:8 Find the general solution: x^5\frac{dy}{dx} = - y^5

Answer:

Given, in the question

x^5\frac{dy}{dx} = - y^5

\\ \implies \int \frac{dy}{y^5} = - \int \frac{dx}{x^5} \\ \implies \frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C \\ \implies \frac{1}{y^4} + \frac{1}{x^4} = C

This is the required general equation.

Question:9 Find the general solution: \frac{dy}{dx} = \sin^{-1}x

Answer:

Given, in the question

\frac{dy}{dx} = \sin^{-1}x

\implies \int dy = \int \sin^{-1}xdx

Now,

\int (u.v)dx = u\int vdx - \int(\frac{du}{dx}.\int vdx)dx

Here, u = \sin^{-1}x and v = 1

\implies y = \sin^{-1}x .x - \int(\frac{1}{\sqrt{1-x^2}}.x)dx

\\ Let\ 1- x^2 = t \\ \implies -2xdx = dt \implies xdx = -dt/2

\\ \implies y = x\sin^{-1}x+ \int(\frac{dt}{2\sqrt{t}}) \\ \implies y = x\sin^{-1}x + \frac{1}{2}.2\sqrt{t} + C \\ \implies y = x\sin^{-1}x + \sqrt{1-x^2} + C

Question:10 Find the general solution e^x\tan y dx + (1-e^x)\sec^2 y dy = 0

Answer:

Given,

e^x\tan y dx + (1-e^x)\sec^2 y dy = 0

\\ \implies e^x\tan y dx = - (1-e^x)\sec^2 y dy \\ \implies \int \frac{\sec^2 y }{\tan y}dy = -\int \frac{e^x }{(1-e^x)}dx

\\ let\ tany = t \ and \ 1-e^x = u \\ \implies \sec^2 ydy = dt\ and \ -e^xdx = du

\\ \therefore \int \frac{dt }{t} = \int \frac{du }{u} \\ \implies \log t = \log u + \log k \\ \implies t = ku \\ \implies \tan y= k (1-e^x)

Question:11 Find a particular solution satisfying the given condition:

(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x; \ y = 1\ \textup{when}\ x = 0

Answer:

Given, in the question

(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x

\\ \implies \int dy = \int\frac{2x^2 + x}{(x^3 + x^2 + x + 1)}dx

(x^3 + x^2 + x + 1) = (x +1)(x^2+1)

Now,

1517900463071792

1517900463878580

1517900464704155

1517900465489721

Now comparing the coefficients

A + B = 2; B + C = 1; A + C = 0

Solving these:

1517900467056911

Putting the values of A,B,C:

1517900467842360

Therefore,

1517900468626376

1517900469359944

1517900470082760

1517900470844580

151790047163329

151790047240963

1517900473174127

1517900473936125

1517900474702185

1517900475486768

Now, y= 1 when x = 0

1517900477940526

c = 1

Putting the value of c, we get:

1517900478708257

Question:12 Find a particular solution satisfying the given condition:

x(x^2 -1)\frac{dy}{dx} =1;\ y = 0\ \textup{when} \ x = 2

Answer:

Given, in the question

x(x^2 -1)\frac{dy}{dx} =1

\\ \implies \int dy=\int \frac{dx}{x(x^2 -1)} \\ \implies \int dy=\int \frac{dx}{x(x -1)(x+1)}

Let,

1628485409903

1517900483388423

1517900484190962

Now comparing the values of A,B,C

A + B + C = 0; B-C = 0; A = -1

Solving these:

1517900484965476

Now putting the values of A,B,C

15179004857815100

1517900486545318

1517900487390967

1517900488152880

Given, y =0 when x =2

1517900489745379

1517900490509113

1517900492095338

Therefore,

\\ \implies y = \frac{1}{2}\log[\frac{4(x-1)(x+1)}{3x^2}]

\\ \implies y = \frac{1}{2}\log[\frac{4(x^2-1)}{3x^2}]

Question:13 Find a particular solution satisfying the given condition:

\cos\left(\frac{dy}{dx} \right ) = a\ (a\in R);\ y = 1\ \textup{when}\ x = 0

Answer:

Given,

\cos\left(\frac{dy}{dx} \right ) = a

\\ \implies \frac{dy}{dx} = \cos^{-1}a \\ \implies \int dy = \int\cos^{-1}a\ dx \\ \implies y = x\cos^{-1}a + c

Now, y =1 when x =0

1 = 0 + c

Therefore, c = 1

Putting the value of c:

\implies y = x\cos^{-1}a + 1

Question:14 Find a particular solution satisfying the given condition:

\frac{dy}{dx} = y\tan x; \ y =1\ \textup{when}\ x = 0

Answer:

Given,

\frac{dy}{dx} = y\tan x

\\ \implies \int \frac{dy}{y} = \int \tan x\ dx \\ \implies \log y = \log \sec x + \log k \\ \implies y = k\sec x

Now, y=1 when x =0

1 = ksec0

\implies k = 1

Putting the vlue of k:

y = sec x

Question:15 Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = e^x\sin x .

Answer:

We first find the general solution of the given differential equation

Given,

y' = e^x\sin x

\\ \implies \int dy = \int e^x\sin xdx

\\ Let I = \int e^x\sin xdx \\ \implies I = \sin x.e^x - \int(\cos x. e^x)dx \\ \implies I = e^x\sin x - [e^x\cos x - \int(-\sin x.e^x)dx] \\ \implies 2I = e^x(\sin x - \cos x) \\ \implies I = \frac{1}{2}e^x(\sin x - \cos x)

\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + c

Now, Since the curve passes through (0,0)

y = 0 when x =0

\\ \therefore 0 = \frac{1}{2}e^0(\sin 0 - \cos 0) + c \\ \implies c = \frac{1}{2}

Putting the value of c, we get:

\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + \frac{1}{2} \\ \implies 2y -1 = e^x(\sin x - \cos x)

Question:16 For the differential equation xy\frac{dy}{dx} = (x+2)(y+2) , find the solution curve passing through the point (1, –1).

Answer:

We first find the general solution of the given differential equation

Given,

xy\frac{dy}{dx} = (x+2)(y+2)

\\ \implies \int \frac{y}{y+2}dy = \int \frac{x+2}{x}dx \\ \implies \int \frac{(y+2)-2}{y+2}dy = \int (1 + \frac{2}{x})dx \\ \implies \int (1 - \frac{2}{y+2})dy = \int (1 + \frac{2}{x})dx \\ \implies y - 2\log (y+2) = x + 2\log x + C

Now, Since the curve passes through (1,-1)

y = -1 when x = 1

\\ \therefore -1 - 2\log (-1+2) = 1 + 2\log 1 + C \\ \implies -1 -0 = 1 + 0 +C \\ \implies C = -2

Putting the value of C:

\\ y - 2\log (y+2) = x + 2\log x + -2 \\ \implies y -x + 2 = 2\log x(y+2)

Question:17 Find the equation of a curve passing through the point ( 0 ,-2) given that at any point (x,y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Answer:

According to the question,

y\frac{dy}{dx} =x

\\ \implies \int ydy =\int xdx \\ \implies \frac{y^2}{2} = \frac{x^2}{2} + c

Now, Since the curve passes through (0,-2).

x =0 and y = -2

\\ \implies \frac{(-2)^2}{2} = \frac{0^2}{2} + c \\ \implies c = 2

Putting the value of c, we get

\\ \frac{y^2}{2} = \frac{x^2}{2} + 2 \\ \implies y^2 = x^2 + 4

Question:18 At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Answer:

Slope m of line joining (x,y) and (-4,-3) is \frac{y+3}{x+4}

According to the question,

\\ \frac{dy}{dx} = 2(\frac{y+3}{x+4}) \\ \implies \int \frac{dy}{y+3} = 2\int \frac{dx}{x+4} \\ \implies \log (y+3) = 2\log (x+4) + \log k \\ \implies (y+3) = k(x+4)^2

Now, Since the curve passes through (-2,1)

x = -2 , y =1

\\ \implies (1+3) = k(-2+4)^2 \\ \implies k =1

Putting the value of k, we get

\\ \implies y+3 = (x+4)^2

Question:19 The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Answer:

Volume of a sphere, V = \frac{4}{3}\pi r ^3

Given, Rate of change is constant.

\\ \therefore \frac{dV}{dt} = c \\ \implies \frac{d}{dt} (\frac{4}{3}\pi r ^3) = c \\ \implies \int d(\frac{4}{3}\pi r ^3) = c\int dt \\ \implies \frac{4}{3}\pi r ^3 = ct + k

Now, at t=0, r=3 and at t=3 , r =6

Putting these value:

\frac{4}{3}\pi (3) ^3 = c(0) + k \\ \implies k = 36\pi

Also,

\frac{4}{3}\pi (6) ^3 = c(3) + 36\pi \\ \implies 3c = 252\pi \\ \implies c = 84\pi

Putting the value of c and k:

\\ \frac{4}{3}\pi r ^3 = 84\pi t + 36\pi \\ \implies r ^3 = (21 t + 9)(3) = 62t + 27 \\ \implies r = \sqrt[3]{62t + 27}

Question:20 In a bank, principal increases continuously at the rate of r % per year. Find the value of r if Rs 100 double itself in 10 years (log e 2 = 0.6931).

Answer:

Let p be the principal amount and t be the time.

According to question,

\frac{dp}{dt} = (\frac{r}{100})p

\\ \implies \int\frac{dp}{p} = \int (\frac{r}{100})dt \\ \implies \log p = \frac{r}{100}t + C

\\ \implies p = e^{\frac{rt}{100} + C}

Now, at t =0 , p = 100

and at t =10, p = 200

Putting these values,

\\ \implies 100 = e^{\frac{r(0)}{100} + C} = e^C

Also,

, \\ \implies 200 = e^{\frac{r(10)}{100} + C} = e^{\frac{r}{10}}.e^C = e^{\frac{r}{10}}.100 \\ \implies e^{\frac{r}{10}} = 2 \\ \implies \frac{r}{10} = \ln 2 = 0.6931 \\ \implies r = 6.93

So value of r = 6.93%

Question:21 In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e 0.5 = 1.648).

Answer:

Let p be the principal amount and t be the time.

According to question,

\frac{dp}{dt} = (\frac{5}{100})p

\\ \implies \int\frac{dp}{p} = \int (\frac{1}{20})dt \\ \implies \log p = \frac{1}{20}t + C

\\ \implies p = e^{\frac{t}{20} + C}

Now, at t =0 , p = 1000

Putting these values,

\\ \implies 1000 = e^{\frac{(0)}{20} + C} = e^C

Also, At t=10

, \\ \implies p = e^{\frac{(10)}{20} + C} = e^{\frac{1}{2}}.e^C = e^{\frac{1}{2}}.1000 \\ \implies p =(1.648)(1000) = 1648

After 10 years, the total amount would be Rs.1648

Question:22 In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Answer:

Let n be the number of bacteria at any time t.

According to question,

\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)

\\ \implies \int \frac{dn}{n} = \int kdt \\ \implies \log n = kt + C

Now, at t=0, n = 100000

\\ \implies \log (100000) = k(0) + C \\ \implies C = 5

Again, at t=2, n= 110000

\\ \implies \log (110000) = k(2) + 5 \\ \implies \log 11 + 4 = 2k + 5 \\ \implies 2k = \log 11 -1 =\log \frac{11}{10} \\ \implies k = \frac{1}{2}\log \frac{11}{10}

Using these values, for n= 200000

\\ \implies \log (200000) = kt + C \\ \implies \log 2 +5 = kt + 5 \\ \implies (\frac{1}{2}\log \frac{11}{10})t = \log 2 \\ \implies t = \frac{2\log 2}{ \log \frac{11}{10}}

Question:23 The general solution of the differential equation \frac{dy}{dx} = e^{x+y} is

(A) e^x + e^{-y} = C

(B) e^{x }+ e^{y} = C

(C) e^{-x }+ e^{y} = C

(D) e^{-x }+ e^{-y} = C

Answer:

Given,

\frac{dy}{dx} = e^{x+y}

\\ \implies\frac{dy}{dx} = e^x.e^y \\ \implies\int \frac{dy}{e^y} = \int e^x.dx \\ \implies -e^{-y} = e^x + C \\ \implies e^x + e^{-y} = K\ \ \ \ (Option A)

More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.4

There are 23 questions in the NCERT syllabus exercise 9.4 Class 12 Maths. The first 10 questions of NCERT solutions for Class 12 Maths chapter 9 exercise 9.4 is to find the general solutions. Next 12 questions of Class 12 Maths chapter 9 exercise 9.4 is to find particular solutions using the variable separable method. And last one question of Class 12th Maths chapter 6 exercise 9.4 is of objective type and is to find the general solution.

Also Read| Differential Equations Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.4

  • Exercise 9.4 Class 12 Maths can be used for the preparation of the topic 9.5 for the CBSE Class 12 Board Exam.
  • Solving all the exercises given in the NCERT Book will give a good score in the exam.
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Key Features Of NCERT Solutions for Exercise 9.4 Class 12 Maths Chapter 9

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 9.4 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 9.4, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 9.4 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 9.4 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 9.4 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 9.4 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What is the order and degree of differential equations handled in topic 9.5?

The differential equations of order 1 and degree 1 are handled.

2. What type of differential equation is solved in exercise 9.4 Class 12 Maths?

Differential equations of variable separable type are solved in the NCERT solutions for Class 12 Maths chapter 9 exercise 9.4.

3. How many solved examples are given in topic 9.5.1?

Six example problems and their solutions are given under topic 9.5.1.

4. How many questions are there in Class 12 Maths chapter 9 exercise 9.4?

Twenty-three questions are explained in the Class 12th Maths chapter 6 exercise 9.4.

5. How many questions of exercise 9.1 are multiple-choice questions?

One question is of objective type with 4 choices.

6. In topic 9.5, the number of sub-topics discussed is?

Three subtopics are discussed in topic 9.5.

7. In session 9.5 of NCERT Class, 12 Maths book how many methods to solve the first-degree first-order differential equations are discussed?

3 methods are discussed. 

8. Name the subtopics discussed in 9.5

The topics discussed are listed below

Differential equations of variable separable type, homogeneous and linear differential equations. 

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Hello there! Thanks for reaching out to us at Careers360.

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Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

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hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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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)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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

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