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NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

Edited By Vishal kumar | Updated on Jul 19, 2023 11:22 AM IST | #CBSE Class 10th

NCERT Solutions Class 10 Science Chapter 10 – CBSE Free PDF Download

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction -If you are in search of NCERT Class 10 Science Chapter 10 Exercise Solutions- Light Reflection and Refraction, look no further. This article provides comprehensive step-by-step solutions that will assist you in completing exercise questions and assignments. Light class 10 ncert solutions reflection and Refraction, being an essential and straightforward chapter in Class 10, often contains 3-4 questions in board examinations.

This Story also Contains
  1. NCERT Solutions Class 10 Science Chapter 10 – CBSE Free PDF Download
  2. NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction
  3. Topic 10.2 Spherical Mirrors
  4. Topic 10.2.4 Mirror forumula and magnification
  5. Topic 10.3 Refraction of light
  6. Topic 10.3.8 Power of Lenses
  7. NCERT Class 10 Science Chapter 10 Light Reflection and Refraction: Exercise Solutions
  8. NCERT Science Exemplar Solutions Class 10 For All The Chapters:
NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction
NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

In addition to NCERT Solutions, experts at Careers360 have compiled a valuable collection of chapter-wise important formulas. These formulas serve as a valuable resource for students, aiding in revision during exams and enabling them to solve questions from the NCERT textbook or any other reference book.

By reading this article, you will gain access to easy-to-understand, step-by-step solutions for each question in the chapter reflection of light class 10. Alternatively, you can download the class 10 science chapter 10 question answer and important formulas in the form of a PDF, allowing you to access them offline at any time and from anywhere, absolutely free of cost.

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NCERT Solutions for Class 10 Science: Important Formulas and Diagrams + eBook link

To solve all the questions or understand each concept in detail of any chapter, it is essential to know the important formulas associated with that chapter. The subject matter experts at Careers360 have compiled all the important formulas of this light class 10 ncert solutions, which are provided below. Additionally, you can find the chapter-wise important formulas of Class 10 Science by clicking on the link given below and downloading the PDF for free.

  • Law of Reflection:

Angle of incidence (θᵢ) = Angle of reflection (θᵣ)

  • Snell's Law:

n₁sin(θ₁) = n₂sin(θ₂)

Where, n₁ and n₂ are the refractive indices of the initial and final media, respectively, and θ₁ and θ₂ are the corresponding angles of incidence and refraction.

  • Mirror formula:

1/v+1/u=1f

  • Lens Formula:

1/v-1/u=1/f

  • Magnification by a spherical mirror:

m= hi/ho=-v/u

Where:

hi = Height of image and ho=Height of object

By having access to these formulas of ncert class 10 science chapter 10 exercise solutions, you can enhance your understanding of the chapter and confidently tackle various questions and problems. These formulas serve as valuable tools for revision and assist you in applying the concepts effectively.

Class 10 Science ch 10 Question Answer Light Reflection and Refraction - Important Topics

If you're aiming to score well in ncert class 10 science chapter 10 exercise solutions, focusing on these important topics of light reflection and refraction will greatly assist you. By dedicating your study efforts to these specific areas, you can maximize your understanding and performance, ultimately increasing your chances of achieving good marks.

  • Reflection of light: Understanding the laws of reflection of light class 10 ncert solutions, types of reflection, and image formation by plane mirrors.
  • Refraction of light: Grasping the laws of refraction, refractive index, and image formation by lenses.
  • Spherical mirrors: Learning about the types of spherical mirrors, image formation by spherical mirrors, and the mirror formula.
  • Lenses: Familiarize yourself with the types of lenses, image formation by lenses, and the lens formula.
  • Applications of light: Exploring the practical applications of light in photography, optical instruments, and fibre optics.

Additionally, pay attention to these important concepts of class 10 science chapter 10 question answer:

  • Rectilinear propagation of light: Understanding how light travels in a straight line.
  • Real and virtual images: Differentiating between real and virtual images.
  • Magnification: Grasping how the size of an image is affected by the object-to-mirror or object-to-lens distance.
  • Sign convention: Understanding the convention used to represent distances in ray diagrams.

By reading these concepts of class 10th science chapter 10 question answer students can revise the chapter in less time and score good marks in exams.

This chapter has been renumbered as Chapter 9 in accordance with the CBSE Syllabus 2023–24.

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

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NCERT Solutions For Class 10 science ch 10 NCERT Solutions Light Reflection and Refraction

Topic 10.2 Spherical Mirrors

Q.1 Define the principal focus of a concave mirror.

Answer:

It is the point on the principal axis where a beam of light parallel to the principal axis after reflection actually meets.

1635754101788

F represents the focal length

Q.2 The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer:

focal\:length=\frac{Radius }{2}

So,

focal\:length=\frac{20}{2}=10cm

Hence the Focal length of the spherical mirror is 10 cm.

Q.3 Name a mirror that can give an erect and enlarged image of an object.

Answer:

Convex mirror usually gives a virtual and erect image.

The concave mirror gives a virtual and enlarged image only when the object is between pole and focus.

Q.4 Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer :

Convex mirrors are preferred as a rearview mirror in vehicles as:

i) It forms an erect (Rigidly upright or straight.) image of an object hence object becomes easily identified.

ii) It forms a diminished image of the object thus increases the field of vision.

iii) An object that is far away from us is seen closer which helps us to take early decisions while driving.

CBSE Class 10 Science ch 10 Question Answer Light

Topic 10.2.4 Mirror forumula and magnification

Q.1 Find the focal length of a convex mirror whose radius of curvature is 32 cm .

Answer:

As we know for the convex mirror,

Radius (R) = Focus (f) X 2

So,

f=\frac{R}{2}

putting the given values,

f=\frac{30}{2}=15cm

Hence Focus of the convex mirror is 15 cm.

Q. 2 A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

Answer:

Here Given Magnification:

m= -3 (real image)

The object distance :

u =-10cm

Now As we know,

m=-\frac{v}{u}

v = - m\times u

v = - (-3)\times(-10)

v = -30cm

Hence,

The image is 30 cm in front of the mirror.

NCERT Textbook Solutions for Light Class 10 Science Chapter

Topic 10.3 Refraction of light

Q.1 A ray of light traveling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

Answer:

As we know that when the light goes from rare medium to denser medium the light bends towards the normal. So when the light goes from air to water it will bend toward the normal.

Q. 2 Light enters from air to glass having a refractive index 1.50 . What is the speed of light in the glass? The speed of light in a vacuum is 3 \times 108\: m\: s^1 .

Answer:

As we know, from the definition of the refractive index that

Refractive Index :

\mu=\frac{c}{v}

Where = Velocity of light in the medium and

c = Speed of light in vacuum

So putting those given values we get,

1.5 = \frac{3\times10^ {8}}{v}

v=\frac{3\times10^8}{1.5}

v= 2\times10^8m/s

Hence the speed of the light in the glass is 2\times10^8m/s .

Q.3 Find out, from Table 10.3, the medium having highest optical density. Also, find the medium with the lowest optical density.


Material Medium Refractive Index Material Medium Refractive Index
Air
Ice
water
Alcohol
Kerosene
Fused Quartz
Turpentine oil
Benzene
Crown Glass

1.0003
1.31
1.33
1.36
1.44
1.46
1.47
1.50
1.52 		Canada Balsam
Rock Salt
Carbon-Disulphide
Dense Flint Glass
Ruby
Sapphire
Diamond 		1.53
1.54
1.63
1.65
1.71
1.77
2.42

Answer :

As we know optical density is the tendency to hold(absorb) the light. So,

more refractive index = more absorbing power = more optical density.

It can be observed from table 10.3 that diamond and air respectively have the highest and lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density.

Q.4 You are given kerosene, turpentine, and water. In which of these does the light travel fastest? Use the information given in Table 10.3 .

Material Medium Refractive Index Material Medium Refractive Index
Air
Ice
water
Alcohol
Kerosene
Fused Quartz
Turpentine oil
Benzene
Crown Glass

1.0003
1.31
1.33
1.36
1.44
1.46
1.47
1.50
1.52 		Canada Balsam
Rock Salt
Carbon-Disulphide
Dense Flint Glass
Ruby
Sapphire
Diamond 		1.53
1.54
1.63
1.65
1.71
1.77
2.42

Answer:

As we can see from the table:

Refractive index of kerosene = 1.44

Refractive index of terbutaline = 1.47

Refractive index of water = 1.33

We know from the definition of refractive index, that the speed of light is higher in a medium with the lower refractive index.

So, the light travels fastest in water relative to kerosene and turpentine.

speed of light---> water > kerosene > turpentine

Q.5 The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer:

Refractive Index shows the comparison of light speeds in two mediums. Light may travel from rarer to denser medium or from denser to rarer medium. When we say the refractive index of diamond is 2.42, it means light is traveling from rarer to a denser medium, and the speed of light in the air (vacuum) is 2.42 times the speed of light in a diamond. On the other hand, if the light travels from denser to rarer medium, that is, from diamond to air the refractive index will be the reciprocal of 2.42.


Topic 10.3.8 Power of Lenses

Q.1 Define 1 dioptre of power of a lens.

Answer:

A lens whose focal length is 1 metre is said to have the power of 1 dioptre.

Q.2 A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens .

Answer:

Since the image formed by the lens is real and inverted it is formed on the side opposite to the one where the object is placed.

Image position v = 50 cm

Let the object position be u.

Since image formed is inverted and the size image is equal to the size of the object, magnification (m) = -1.

\\m=\frac{v}{u}\\ -1=\frac{v}{u}\\ u=-v\\ u=-50\ cm

The needle is placed 50 cm in front of the lens

Using lens formula we have

\\\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\ \frac{1}{f}=\frac{1}{50}-\frac{1}{-50}\\ \frac{1}{f}=\frac{2}{50}\\ f=25\ cm

Power of the lens P is given by

\\P=\frac{1}{f}\\ P=\frac{1}{25\times 10^{-2}}\\ P=4\ D

The power of the lens P is 4 Dioptre.

Q.3 Find the power of a concave lens of focal length 2 m.

Answer:

The focal length of the lens is f = -2 m. (The focal length of a concave lens is negative)

Power of the lens P is given by

\\P=\frac{1}{f}\\ P=\frac{1}{-2}\\ P=-0.5\ D

The Power of the lens is - 0.5 Dioptre.

NCERT Class 10 Science Chapter 10 Light Reflection and Refraction: Exercise Solutions

Q1. Which one of the following materials cannot be used to make a lens?


(a) Water (b) Glass (c) Plastic (d) Clay

Answer:

Clay cannot be used to make a lens.

(d) is the correct answer.

Q 2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?


(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.

Answer:

The position of the object should be between the pole of the mirror and its principal focus.

(d) is the correct answer.

Q 3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?

(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.

Answer:

An object should be placed at twice the focal length in front of a convex lens to get a real image of the size of the object

(b) is the correct answer.

Q 4 . A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be

(a) both concave.

(b) both convex.

(c) the mirror is concave and the lens is convex.

(d) the mirror is convex, but the lens is concave.

Answer:

The mirror and the lens are both concave.

(a) is the correct answer.

Q 5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.

Answer:

The mirror is likely to be either plane or convex.

(d) is the correct answer.

Q 6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.

Answer:

I would prefer to use a convex lens of focal length 5 cm while reading small letters found in a dictionary because a convex lens gives magnified image if the object is in between focus and radius of curvature and the magnification will be high for shorter focal length

(c) is the correct answer.

Q 7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Answer:

\\\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ v=\frac{fu}{u-f}

For the image to be erect it has to be virtual and therefore as per convention v has to be positive

v > 0

\frac{fu}{u-f}>0

u < 0

f < 0

fu > 0

u - f > 0

u > f

Therefore the object must be placed between the pole and focus of the mirror.

i.e -15 cm < u < 0

The image formed would be virtual and larger than the object.


1635754149924

Q 8.(a) Name the type of mirror used in the following situations.

  • Headlights of a car.

Support your answer with reason.

Answer:

A concave mirror is used for headlights of a car as they can produce parallel beams of high intensity which can travel through large distances if the source of light is placed at the focus of the mirror.

Q 8.b) Name the type of mirror used in the following situations.

  • Side/rear-view mirror of a vehicle

Support your answer with reason.

Answer:

The convex mirror would be used as a side/rear-view mirror of a vehicle as the image produced will be erect, virtual and diminished. A convex mirror also has a larger field of view as compared to a plane mirror of the same size.

Q 8.c) Name the type of mirror used in the following situations.

  • Solar furnace.

Support your answer with reason.

Answer:

The concave mirror would be used in a solar furnace to concentrate the incident rays from the sun on a solar panel.

Q 9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer:

The lens would produce a complete image. Only the brightness of the image would be diminished.

Verification by experiment:

Apparatus required: a convex lens, a screen, and a candlelight

procedure:

1. Place screen behind a lens placed vertically

2. Move the candle to obtain a clear full-length image of a candle on the screen

3. Cover half of the lens with the black paper without disturbing the position of the lens

4. Note the observations

Observation: Size of the image is the same but the brightness reduces

Q 10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and nature of the image formed.

Answer:

Object distance u = -25 cm.

Focal length = 10 cm

Let image distance be v

\\\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{10}=\frac{1}{v}+\frac{1}{-25}\\ \frac{1}{v}=\frac{1}{10}+\frac{1}{25}\\ v=16. 66\ cm

The position of the image is 16.66 cm on the other side of the lens.

Object size O = 5 cm.

Let the image size be I

Magnification is m

\\m=\frac{v}{u}\\ \frac{I}{O}=\frac{16.66}{-25}\\ I=5\times -\frac{16.66}{25}\\ I=-3.33\ cm

The nature of the image is real and its size is -3.33 cm.

The formation of the image is shown in the following ray diagram

1635754193377


Q 11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Answer:

Focal length, f = -15 cm

Image distance, v = -10 cm ( As in case of the concave lens image is formed on the same side as the object is placed)

Let the object distance be u.

As per the lens formula

\\\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\ \frac{1}{u}=\frac{1}{v}-\frac{1}{f}\\ \frac{1}{u}=\frac{1}{-10}-\frac{1}{-25}\\ v=-30\ cm

The object is placed at a distance of 30 cm from the lens.

1635754225039

Q 12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Answer:

Object distance, u = -10 cm

Focal length, f = 15 cm

Let the image distance be v

As per the mirror formula

\\\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{15}=\frac{1}{v}+\frac{1}{-10}\\ v=6\ cm

\\Magnification=-\frac{v}{u}\\ =-\frac{6}{-30}\\ =0.2

The image formed is virtual, erect, diminished and is formed 6 cm behind of the mirror.

Q 13. The magnification produced by a plane mirror is +1. What does this mean?

Answer:

It means the image formed is virtual, erect and of the same size as the object.

Q 14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature, and size .

Answer:

Radius of curvature, R = 30 cm

Focal length, f = R/2 = 15 cm

Object distance, u = -20 cm

Let the image distance be v

\\\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{15}=\frac{1}{v}+\frac{1}{-20}\\ v=8.57\ cm

Since v is positive image is formed behind the mirror.

\\Magnification=-\frac{v}{u}\\ =-\frac{8.57}{-20}\\ =0.428

Object size = 5 cm

\\Image \ size= Object\ size \times Magnification\\ =5\times 0.428\\ =2.14\ cm

A virtual, erect and diminished image of size 2.14 cm would be formed 8.57 cm behind the mirror.

Q 15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharply focussed image can be obtained? Find the size and nature of the image.

Answer:

Object distance, u = -27 cm

Focal length, f = -18 cm

Let the image distance be v

As per the mirror formula

\\\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{-18}=\frac{1}{v}+\frac{1}{-27}\\ v=-54\ cm

The negative sign shows the image is formed in front of the mirror and is real.

Object size = 7 cm

\\Image\ size=magnification\times Object\ size\\ =-\frac{v}{u}\times O\\ =-\frac{-54}{-27}\times 7\\ =-14\ cm

A real, inverted and magnified image of size 14 cm is formed in front of the mirror.

Q 16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?

Answer:

\\f=\frac{1}{P}\\ f=\frac{1}{-2}\\ f=-0.5\ m\\ f=-50\ cm

The lens is concave because it's focal length is negative and is equal to -50 cm.

Q 17 . A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer:

\\f=\frac{1}{P}\\ f=\frac{1}{1.5}\\ f=0.6667\ m\\ f=66.67\ cm

The less is convex because its focal length is positive and therefore is converging.

NCERT Solutions for Class 10 for other subjects

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction - Topics

Apart from the important topics we mentioned above in this article, there are a total of 16 topics of class 10 science ch 10 ncert solutions which are listed below:

10.1 Reflection of Light

10.2 Spherical Mirrors

10.2.1 Image Formation by Spherical Mirrors

10.2.2 Representation of Images Formed by Spherical Mirrors Using ray Diagrams

10.2.3 Sign Convention for Reflection by Spherical Mirrors

10.2.4 Mirror Formula and Magnification

10.3 Reflection of Light

10.3.1 Reflection through a Rectangular Glass Slab

10.3.2 The Refractive Index

10.3.3 Reflection by Spherical Lenses

10.3.4 Image Formation by Lenses

10.3.5 Image Formation in Lenses using Ray Diagrams

10.3.6 Sign Convention for Spherical Lenses

10.3.7 Lenses Formula and Magnification

10.3.8 Power of a lens

Key features of chapter 10 science class 10 exercise answers Light Reflection and Refraction

Easy to understand: The class 10 science chapter 10 question answer are designed to be easy and thorough, allowing students to gain a deeper understanding of the solutions and concepts of light reflection and refraction.

Score well in board exams: By utilizing the light reflection and refraction class 10 exercise answers, students can enhance their understanding of the chapter light class 10 and improve their performance in the CBSE board exams. The ncert class 10 science chapter 10 exercise solutions provide clarity on important topics and offer effective strategies for solving questions.

Comprehensive exercise questions: The NCERT solutions of chapter 10 science class 10 provide answers to the exercise questions given in the textbook, ensuring that students have a clear understanding of the concepts. Additionally, there are additional exercise questions included, allowing students to practice and reinforce their knowledge.

Expert guidance: The NCERT solutions for reflection of light class 10 Science are prepared by subject matter experts. In case of any doubts or queries, students can directly reach out to the experts for clarification and guidance, ensuring a thorough understanding of the topics covered in the chapter.

Offline Benefits: Class 10 science ch 10 NCERT Solutions are available in the form of PDF, students can download them and use them offline.

How to use NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

  • Firstly, it is essential to know the complete NCERT syllabus and clear all your concepts related to the given chapter.

  • Afterwards, practice the class 10th science chapter 10 question answer on your own. If you find it difficult to solve, then take help from the Light Reflection And Refraction Class 10 Science solutions.

  • Practice from previous year question papers or sample question papers to have a look at what type of questions were asked previously.

Also, Check NCERT Books and NCERT Syllabus here:

NCERT Solutions for Class 10 Science- Chapter Wise


NCERT Science Exemplar Solutions Class 10 - Chapter Wise

NCERT Science Exemplar Solutions Class 10 For All The Chapters:

Frequently Asked Questions (FAQs)

1. What are the important topics of Class 10 Science Chapter 10 Light Reflection and Refraction?

Centre of Curvature, Principal Axis, Principal Focus, Focal Length, Images Formed by Spherical Mirrors, Mirror formula & Magnification, Refractive index, Refraction & Laws of refraction, Lens Formula & Magnification, Power of a lens.

2. How can I get the PDF of NCERT solutions for Class 10 Chapter 10 pdf?

NCERT solutions class 10 Science chapter 10 pdf download button available. By downloading you can save the webpage to check solutions offline.

3. What is the weightage of Chapter 10 Light Reflection and Refraction in CBSE class 10 board final exam?

6-7 marks. All the formulas and raydiagrams of the chapter are important. To get more problems of Light Reflection and Refraction refer to CBSE previous year papers, NCERT book and NCERT exemplar for Class 10.

4. What are the benefits of referring the Caerres360 NCERT Solutions for Class 10 Science Chapter 10?

Advantages of using Careers360 NCERT Solutions for reflection of light class 10:

  • The in-depth explanation for better understanding of concepts.

  • Aligned with the CBSE syllabus.

  • Authentic and precise for improving problem-solving skills.

  • Free PDF downloads for easy access by students.

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Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

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Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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sadafreen356 03 July,2024
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Popular CBSE Class 10th 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

 
Read More

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
Read More

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.
Read More

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.
Read More

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
Read More

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
Read More

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