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NCERT Solutions for Exercise 6.1 Class 12 Maths Chapter 6 - Application of Derivatives

NCERT Solutions for Exercise 6.1 Class 12 Maths Chapter 6 - Application of Derivatives

Edited By Ramraj Saini | Updated on Dec 03, 2023 05:23 PM IST | #CBSE Class 12th
Upcoming Event
CBSE Class 12th  Exam Date : 15 Feb' 2025 - 15 Feb' 2025

NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.1

NCERT Solutions for Exercise 6.1 Class 12 Maths Chapter 6 Application of Derivatives 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 6.1 Class 12 Maths chapter 6 surrounds by questions related to the topic rate of change quantities. The Class 12th Maths chapter 6 exercise 6.1 are curated by Mathematics faculties according to the CBSE pattern. All these NCERT solutions given in Class 12 Maths chapter 6 exercise 6.1 give a more deep knowledge on the concept of rate of change of quantities that is discussed in the NCERT book. There are 6 exercises in Class 12 NCERT chapter applications of derivatives. Other than Exercise 6.1 Class 12 Maths, the following exercise, are also present in the NCERT syllabus of this chapter.

This Story also Contains
  1. NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.1
  2. Access NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1
  3. Application of Derivatives Class 12 Chapter 6 Exercise 6.1
  4. More About NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1
  5. Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1
  6. Key Features Of NCERT Solutions for Exercise 6.1 Class 12 Maths Chapter 6
  7. NCERT Solutions Subject Wise
  8. Subject Wise NCERT Exemplar Solutions

12th class Maths exercise 6.1 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 6 Exercise 6.1

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Application of Derivatives Class 12 Chapter 6 Exercise 6.1

Question:1 b) Find the rate of change of the area of a circle with respect to its radius r when
r = 4 cm

Answer:

Area of the circle (A) = \pi r^{2}
Rate of change of the area of a circle with respect to its radius r = \frac{dA}{dr} = \frac{d(\pi r^{2})}{dr} = 2 \pi r
So, when r = 4, Rate of change of the area of a circle = 2 \pi (4) = 8 \pi
Hence, Rate of change of the area of a circle with respect to its radius r when r = 4 is 8 \pi

Question:2 . The volume of a cube is increasing at the rate of 8 cm^3 /s . How fast is the surface area increasing when the length of an edge is 12 cm?

Answer:

The volume of the cube(V) = x^{3} where x is the edge length of the cube
It is given that the volume of a cube is increasing at the rate of 8 cm^3 /s

we can write \frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt} ( By chain rule)

\frac{dV}{dt} = 8 = \frac{dV}{dx}.\frac{dx}{dt}

\frac{dx^{3}}{dx}.\frac{dx}{dt} = 8 \Rightarrow 3x^{2}.\frac{dx}{dt} = 8

\frac{dx}{dt} = \frac{8}{3x^{2}} - (i)
Now, we know that the surface area of the cube(A) is 6x^{2}

\frac{dA}{dt} = \frac{dA}{dx}.\frac{dx}{dt} = \frac{d6x^{2}}{dx}.\frac{dx}{dt} = 12x. \frac{dx}{dt} - (ii)

from equation (i) we know that \frac{dx}{dt} = \frac{8}{3x^{2}}

put this value in equation (i)
We get,
\frac{dA}{dt} = 12x. \frac{8}{3x^{2}} = \frac{32}{x}
It is given in the question that the value of edge length(x) = 12cm
So,
\frac{dA}{dt} = \frac{32}{12} = \frac{8}{3} cm^2/s

Question:3 The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Answer:

Radius of a circle is increasing uniformly at the rate \left ( \frac{dr}{dt} \right ) = 3 cm/s
Area of circle(A) = \pi r^{2}
\frac{dA}{dt} =\frac{dA}{dr}.\frac{dr}{dt} (by chain rule)
\frac{dA}{dt} =\frac{d \pi r^{2}}{dr}.\frac{dr}{dt} = 2\pi r \times 3 = 6\pi r
It is given that the value of r = 10 cm
So,
\frac{dA}{dt} = 6\pi \times 10 = 60\pi \ cm^{2}/s
Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60\pi \ cm^{2}/s

Question:4 An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Answer:

It is given that the rate at which edge of cube increase \left ( \frac{dx}{dt} \right ) = 3 cm/s
The volume of cube = x^{3}
\frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt} (By chain rule)
\frac{dV}{dt} = \frac{dx^{3}}{dx}.\frac{dx}{dt} = 3x^{2}.\frac{dx}{dt} = 3x^{2}\times 3 = 9x^{2} cm^{3}/s
It is given that the value of x is 10 cm
So,
\frac{dV}{dt} = 9(10)^{2} = 9\times 100 = 900 \ cm^{3}/s
Hence, the rate at which the volume of the cube increasing when the edge is 10 cm long is 900 \ cm^{3}/s

Question:5 A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Answer:

Given = \frac{dr}{dt} = 5 \ cm/s

To find = \frac{dA}{dt} at r = 8 cm

Area of the circle (A) = \pi r^{2}
\frac{dA}{dt} = \frac{dA}{dr}.\frac{dr}{dt} (by chain rule)
\frac{dA}{dt} = \frac{d\pi r^{2}}{dr}.\frac{dr}{dt} = 2\pi r \times 5 = 10\pi r = 10\pi \times 8 = 80\pi \ cm^{2}/s
Hence, the rate at which the area increases when the radius of the circular wave is 8 cm is 80\pi \ cm^{2}/s

Question:6 The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Answer:

Given = \frac{dr}{dt} = 0.7 \ cm/s
To find = \frac{dC}{dt} , where C is circumference
Solution :-

we know that the circumference of the circle (C) = 2\pi r
\frac{dC}{dt} = \frac{dC}{dr}.\frac{dr}{dt} (by chain rule)
\frac{dC}{dt} = \frac{d2\pi r}{dr}.\frac{dr}{dt} = 2\pi \times 0.7 = 1.4\pi \ cm/s
Hence, the rate of increase of its circumference is 1.4\pi \ cm/s

Question:7(a) . The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rate of change of

the perimeter of rectangle

Answer:

Given = Length x of a rectangle is decreasing at the rate (\frac{dx}{dt}) = -5 cm/minute (-ve sign indicates decrease in rate)
the width y is increasing at the rate (\frac{dy}{dt}) = 4 cm/minute
To find = \frac{dP}{dt} and at x = 8 cm and y = 6 cm , where P is perimeter
Solution:-

Perimeter of rectangle(P) = 2(x+y)
\frac{dP}{dt} = \frac{d(2(x+y))}{dt} = 2\left ( \frac{dx}{dt} + \frac{dy}{dt} \right ) = 2(-5+4) = -2 \ cm/minute
Hence, Perimeter decreases at the rate of 2 \ cm/minute

Question:7(b) The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rates of change of the area of the rectangle.

Answer:

Given same as previous question
Solution:-
Area of rectangle = xy
\frac{dA}{dt} = \frac{d(xy)}{dt} = \left ( x\frac{dy}{dt} + y\frac{dx}{dt} \right ) = \left ( 8\times 4 + 6 \times (-5) \right ) = (32 -30) = 2 \ cm^{2}/minute
Hence, the rate of change of area is 2 \ cm^{2}/minute

Question:8 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Answer:

Given = \frac{dV}{dt} = 900 \ cm^{3}/s
To find = \frac{dr}{dt} at r = 15 cm
Solution:-

Volume of sphere(V) = \frac{4}{3}\pi r^{3}
\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})} {dr}.\frac{dr}{dt} = \frac{4}{3}\pi\times 3r^{2} \times \frac{dr}{dt}

\frac{dV}{dt}= 4 \pi r^{2} \times \frac{dr}{dt}
\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^{2}} = \frac{900}{4\pi \times(15)^{2}} = \frac{900}{900\pi} = \frac{1}{\pi} \ cm/s
Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is \frac{1}{\pi} \ cm/s

Question:9 . A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Answer:

We need to find the value of \frac{dV}{dr} at r =10 cm
The volume of the sphere (V) = \frac{4}{3}\pi r^{3}
\frac{dV}{dr} = \frac{d(\frac{4}{3}\pi r^{3})}{dr} = \frac{4}{3}\pi \times 3r^{2} = 4\pi r^{2} = 4\pi (10)^{2} = 4\pi \times 100 = 400\pi \ cm^{3}/s
Hence, the rate at which its volume is increasing with the radius when the later is 10 cm is 400\pi \ cm^{3}/s

Question:10. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Answer:

Let h be the height of the ladder and x be the distance of the foot of the ladder from the wall
It is given that \frac{dx}{dt} = 2 \ cm/s
We need to find the rate at which the height of the ladder decreases (\frac{dh}{dt})
length of ladder(L) = 5m and x = 4m (given)
By Pythagoras theorem, we can say that
h^{2}+x^{2} = L^{2}
h^{2} = L^{2} - x^{2}
h = \sqrt{L^{2} - x^{2}}
Differentiate on both sides w.r.t. t
\frac{dh}{dt} = \frac{d(\sqrt{L^{2} -x^{2}})}{dx}.\frac{dx}{dt} = \frac{1}{2}\frac{-2x}{\sqrt{5^{2}-x^{2}}}.\frac{dx}{dt} = \frac{-x}{\sqrt{25-x^{2}}}\frac{dx}{dt}
at x = 4

\frac{dh}{dt} = \frac{-4}{\sqrt{25-16}}\times 2 = \frac{-4}{3} \times 2 =\frac{-8}{3} \ cm/s
Hence, the rate at which the height of ladder decreases is \frac{8}{3} \ cm/s

Question:11. A particle moves along the curve 6y = x^3 + 2 Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Answer:

We need to find the point at which \frac{dy}{dt} = 8\frac{dx}{dt}
Given the equation of curve = 6y = x^3 + 2
Differentiate both sides w.r.t. t
6\frac{dy}{dt} = \frac{d(x^{3})}{dx}.\frac{dx}{dt} +0
= 3x^{2}.\frac{dx}{dt}
\frac{dy}{dt} = 8\frac{dx}{dt} (required condition)
6\times 8\frac{dx}{dt}= 3x^{2}.\frac{dx}{dt}
3x^{2}.\frac{dx}{dt} =48\frac{dx}{dt} \Rightarrow x^{2} = \frac{48}{3} = 16
x = \pm 4
when x = 4 , y = \frac{4^{3}+2}{6} = \frac{64+2}{6} = \frac{66}{6} = 11
and
when x = -4 , y = \frac{(-4)^{3}+2}{6} = \frac{-64+2}{6} = \frac{-62}{6} = \frac{-31}{3}
So , the coordinates are
(4,11) \ and \ (-4,\frac{-31}{3})

Question:12 The radius of an air bubble is increasing at the rate of 1 /2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer:

It is given that \frac{dr}{dt} = \frac{1}{2} \ cm/s
We know that the shape of the air bubble is spherical
So, volume(V) = \frac{4}{3}\pi r^{3}
\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})}{dr}.\frac{dr}{dt} = 4\pi r^{2}\times\frac{1}{2} = 2\pi r^{2} = 2\pi \times (1)^{2} = 2\pi \ cm^{3}/s
Hence, the rate of change in volume is 2\pi \ cm^{3}/s

Question:13 A balloon, which always remains spherical, has a variable diameter \frac{3}{2}( 2x +1) Find the rate of change of its volume with respect to x.

Answer:

Volume of sphere(V) = \frac{4}{3}\pi r^{3}
Diameter = \frac{3}{2}(2x+1)
So, radius(r) = \frac{3}{4}(2x+1)
\frac{dV}{dx} = \frac{d(\frac{4}{3}\pi r^{3})}{dx} = \frac{d(\frac{4}{3}\pi (\frac{3}{4}(2x+1))^{3})}{dx} = \frac{4}{3}\pi\times 3\times\frac{27}{64}(2x+1)^{2}\times 2
= \frac{27}{8}\pi (2x+1)^{2}

Question:14 Sand is pouring from a pipe at the rate of 12 cm 3 /s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Answer:

Given = \frac{dV}{dt} = 12 \ cm^{3}/s and h = \frac{1}{6}r
To find = \frac{dh}{dt} at h = 4 cm
Solution:-

Volume of cone(V) = \frac{1}{3}\pi r^{2}h
\frac{dV}{dt} = \frac{dV}{dh}.\frac{dh}{dt} = \frac{d(\frac{1}{3}\pi (6h)^{2}h)}{dh}.\frac{dh}{dt} = \frac{1}{3}\pi\times36\times3h^{2}.\frac{dh}{dt} = 36\pi \times(4)^{2}.\frac{dh}{dt}
\frac{dV}{dt} = 576\pi.\frac{dh}{dt}

Question:15 The total cost C(x) in Rupees associated with the production of x units of an
item is given by C ( x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000

Find the marginal cost when 17 units are produced.

Answer:

Marginal cost (MC) = \frac{dC}{dx}
C ( x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000
\frac{dC}{dx} = \frac{d(.007x^{3} - .003x^{2}+15x+400)}{dx} = 3\times .007x^{2} - 2\times.003x+15
= .021x^{2} - .006x + 15
Now, at x = 17
MC = .021(17)^{2} - .006(17) + 15
= 6.069 - .102 + 15
= 20.967
Hence, marginal cost when 17 units are produced is 20.967

Question:16 The total revenue in Rupees received from the sale of x units of a product is
given by R ( x) = 13 x^2 + 26 x + 15

Find the marginal revenue when x = 7

Answer:

Marginal revenue = \frac{dR}{dx}
R ( x) = 13 x^2 + 26 x + 15
\frac{dR}{dx} = \frac{d(13x^{2}+26x+15)}{dx} = 13\times2x+ 26 = 26(x+1)
at x = 7
\frac{dR}{dx} = 26(7+1) = 26\times8 = 208
Hence, marginal revenue when x = 7 is 208

Question:17 The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π (B) 12π (C) 8π (D) 11π

Answer:

Area of circle(A) = \pi r^{2}
\frac{dA}{dr} = \frac{d(\pi r^{2})}{dr} = 2\pi r
Now, at r = 6cm
\frac{dA}{dr}= 2\pi \times 6 = 12\pi cm^{2}/s
Hence, the rate of change of the area of a circle with respect to its radius r at r = 6 cm is 12\pi cm^{2}/s
Hence, the correct answer is B

Question:18 The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x^2 + 36x + 5 . The marginal revenue, when x = 15 is
(A) 116 (B) 96 (C) 90 (D) 126

Answer:

Marginal revenue = \frac{dR}{dx}
R ( x) = 3 x^2 + 36 x + 5
\frac{dR}{dx} = \frac{d(3x^{2}+36x+5)}{dx} = 3\times2x+ 36 = 6(x+6)
at x = 15
\frac{dR}{dx} = 6(15+6) = 6\times21 = 126
Hence, marginal revenue when x = 15 is 126
Hence, the correct answer is D

More About NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1

18 questions and their solutions are discussed in exercise 6.1 Class 12 Maths solutions. Out of these eighteen questions, 2 are multiple-choice questions. NCERT book gives 6 solved examples before Class 12th Maths chapter 6 exercise 6.1. Solving these examples helps in understanding the questions in NCERT solutions given in Class 12 Maths chapter 6 exercise 6.1. After going through the examples try to solve the exercise alone and if any confusion arises refer to the solutions given here. The solutions given here are in detail and is beneficial for the CBSE Class 12 Mathematics board exam.

Also Read| Application of Derivatives Class 12 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1

  • Solving the Exercise 6.1 Class 12 Maths helps in the preparation of exams like JEE main
  • The concepts explained in Class 12th Maths chapter 6 exercise 6.1 are useful for higher studies in the field of Mathematics, Science and Engineering
  • The application of derivatives holds a good weightage for various board exams also.
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Key Features Of NCERT Solutions for Exercise 6.1 Class 12 Maths Chapter 6

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 6.1 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 6.1, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 6.1 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 6.1 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 6.1 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 6.1 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. How many exercise are in Class 12th chapter Application of Derivative?

Including miscellaneous, there are 6 exercises in the unit Applications of Derivatives

2. What is the number of multiple-choice questions present in the NCERT Solutions for Class 12 Maths chapter 6 exercise 6.1?

Two multiple-choice questions are discussed in the NCERT exercise 6.1

3. What number of solved examples are there in the NCERT topic rate of change of quantities?

There are six solved examples

4. What is the meaning of dy/dx

dy/dx means the rate of change of y with respect to x

5. When dy/dx is positive?

dy/dx is positive if y increases as x increases

6. What is the topic discussed in the NCERT Class 12th Maths book after exercise 6.1?

The topic number 6.3 increasing and decreasing function is coming after exercise 6.1. 

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quartely question paper for mathematics cbse

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

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

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.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
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How to Apply:
Get the Medhavi app by visiting the Google Play Store.
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Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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.

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Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

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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kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

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
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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

 
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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
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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
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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 .
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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
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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.
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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.
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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
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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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