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

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

Edited By Safeer PP | Updated on Sep 09, 2022 02:54 PM IST | #CBSE Class 10th
Upcoming Event
CBSE Class 10th  Exam Date : 15 Feb' 2025 - 15 Feb' 2025

NCERT exemplar Class 10 Science solutions chapter 10 discusses light’s fundamental properties such as reflection and refraction. These NCERT exemplar Class 10 Science chapter 10 solutions are prepared and reviewed by our highly skilled physics faculty team, focusing on the solutions’ accuracy and comprehensive nature. These Class 10 Science NCERT exemplar chapter 10 solutions give a step forward to the students while they venture into the basics of Physics and, in turn, develop a strong base of light reflection and refraction concepts. The NCERT exemplar Class 10 Science solutions chapter 10 are prepared to accommodate the CBSE Syllabus for Class 10.

This Story also Contains
  1. NCERT Exemplar Class 10 Science Solutions Chapter 10-MCQ
  2. NCERT Exemplar Class 10 Science Solutions Chapter 10-Short Answer
  3. NCERT Exemplar Class 10 Science Solutions Chapter 10-Long Answer
  4. NCERT Exemplar Class 10 Science Solutions Chapter 10 Important Topics:
  5. NCERT Class 10 Exemplar Solutions for Other Subjects:
  6. NCERT Class 10 Science Exemplar Solutions for Other Chapters:
  7. Features of NCERT Exemplar Class 10 Science Solutions chapter 10

NCERT Exemplar Class 10 Science Solutions Chapter 10-MCQ

Question:1

Which of the following can make a parallel beam of light when light from a point source is incident on it?
(a) Concave mirror as well as convex lens
(b) Convex mirror as well as concave lens
(c) Two plane mirrors placed at 90° to each other
(d) Concave mirror as well as concave lens

Answer : A
When a point source is placed at the focus of a concave mirror or a convex lens, light after reflecting from the concave mirror and light after refracting from a convex lens goes as a parallel beam.

Question:2

A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is
(a) – 30 cm
(b) – 20 cm
(c) – 40 cm
(d) – 60 cm

Answer : B
Given,
Size of object, O = + 10.0 mm = + 1.0 cm
Size of image size, I = 5.0 mm = 0.5 cm
Image distance = − 30 cm (as image is real)
Let, object distance = u
Focal length= f
Magnification m = \frac{I}{O}
Magnification is also given as,m=-\frac{v}{u}
Therefore, \frac{I}{O}=\frac{-v}{u}
\frac{0.5}{1.0}=-\frac{30}{u}
Therefore, u=-60\; cm
Focal length is given by \frac{1}{f}=\left ( \frac{1}{v} \right )+\left ( \frac{1}{u} \right )
or, \frac{1}{f}=-\frac{3}{60}
Therefore, f= -20\; cm

Question:3

Under which of the following conditions a concave mirror can form an image larger than the actual object?
(a) When the object is kept at a distance equal to its radius of curvature
(b) When object is kept at a distance less than its focal length
(c) When object is placed between the focus and center of curvature
(d) When object is kept at a distance greater than its radius of curvature

Answer : C
When object is placed between the focus and center of curvature a concave mirror can form an image larger than the actual object.

Question:4

Figure 10.1 shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is
(a) \sqrt{3}/\sqrt{2}
(b) \sqrt{2}/\sqrt{3}
(c) {1}/\sqrt{2}
(d) \sqrt{2}

Answer : A
Refractive index of B with respect to A =\frac{\sin\; i}{\sin\; r}
=\frac{\sin\; 60^{o}}{\sin\; 45^{o}}
=\frac{\sqrt{3}}{\sqrt{2}}

Question:5

A light ray enters from medium A to medium B as shown in Figure 10.2. The refractive index of medium B relative to A will be
(a) greater than unity
(b) less than unity
(c) equal to unity
(d) zero

Answer : A
Since ray of light bends towards the normal, when it goes from medium A to medium B, therefore, medium A is rarer and medium B is denser medium. Speed of light in rarer medium (v1) is greater than the speed of light in denser medium (v2). Hence, nBA > 1.

Question:6

Beams of light are incident through the holes A and B and emerge out of box through the holes C and D respectively as shown in the Figure10.3. Which of the following could be inside the box?

(a) A rectangular glass slab
(b) A convex lens
(c) A concave lens
(d) A prism?

Answer : A
A rectangular glass slab refracts and then re-refracts the rays when incident rays fall perpendicularly at the point of incidence on a rectangular glass slab. Rectangular glass slab causes the lateral displacement of a ray of light following on it. However, incident ray and emergent ray are parallel to each other.

Question:7

A beam of light is incident through the holes on side A and emerges out of the holes on the other side of the box as shown in figure. Which of the following could be inside the box?

(a) Concave lens
(b) Rectangular slab
(c) prism
(d) Convex lens

Answer: D
The incident rays are parallel and emergent rays are diverging. Concave lens is diverging in nature but the answer is convex lens as you can see the ray 10 which is at the top comes last after emerging out of the box, it means it converges first then diverge as shown in figure below.

Question:8

Which of the following statements is true ?
(a) A convex lens has 4 dioptre power having focal length 0.25 m.
(b) A convex lens has – 4 dioptre power having focal length 0.25 m.
(c) A concave lens has 4 dioptre power having focal length 0.25 m.
(d) A concave lens has – 4 dioptre power having focal length 0.25 m

Answer: A
Convex lens has positive power and positive focal length.
\\\ P=\frac{1}{f}\;\\\ \therefore f=\frac{1}{P}=\frac{1}{4}=0.25\; m


Question:9

Magnification produced by a rear-view mirror fitted in vehicles.
(a) is less than one
(b) is more than one
(c) is equal to one
(d) can be more than or less than one depending upon the position of the object in front of it.

Answer: A
Rear view mirror is a convex mirror, which always forms an image whose size is less than the size of the object. Convex mirror always forms smaller image. Therefore, the magnification of rear-view mirror is always less than 1.

Question:10

Rays from sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed so that size of its image is equal to the size of the object?
(a) 15 cm in front of the mirror
(b) 30 cm in front of the mirror
(c) between 15 cm and 30 cm in front of the mirror.
(d) more than 30 cm in front of the mirror.

Answer: B
Here, Focal length of concave mirror, f = -15 cm
Radius of curvature of the mirror, R = 2f = -30 cm.
In case of concave mirror, size of image is equal to the size of the object if the object is placed at the center of curvature. The distance of center of curvature from the mirror = radius of curvature of the mirror. When an object is placed at R, the image formed is of the same size as that of the object. The image is formed at R and it is inverted.

Question:11

A full-length image of a distance tall building can definitely be seen by using
(a) concave mirror
(b) convex mirror
(c) plane mirror
(d) both concave as well as plane mirror.

Answer: B
Field of convex mirror is more than any type of mirror. Convex mirror forms full length of a distant tall object irrespective of the position of the object. However, plane mirror forms full size image of the object if the size of the plane mirror is half the size of the object. Concave mirror forms full size image of the object if the object is far away from it.

Question:12

In torches, search lights and headlights of vehicles the bulb is placed
(a) between the pole and the focus of the reflector
(b) very near to the focus of the reflector
(c) between the focus and center of curvature of the reflector
(d) at the center of curvature of the reflector

Answer: B
Concave mirrors are used in headlight reflectors and search lights. When the source of light is placed at the focus, the reflected light appears like a beam.

Question:13

The laws of reflection hold good for
(a) plane mirror only
(b) concave mirror only
(c) convex mirror only
(d) all mirrors irrespective of their shape

Answer: D
The laws of reflection hold good for all mirrors irrespective of their shape.

Question:14

The path of a ray of light coming from air passing through a rectangular glass slab traced by four students are shown as A, B, C and D in the figure. Which one of them is correct?

(a) A
(b) B
(c) C
(d) D

Answer: B
When light passes from air to glass, it bends towards the normal and when light passes form glass to air it bends away from normal. Glass slab causes the lateral displacement of a ray of light falling on it. However, incident ray and emergent ray are parallel to each other.

Question:15

You are given water, mustard oil, glycerin and kerosene. In which of these media, a ray of light incident obliquely at same angle would bend the most ?
(a) Kerosene
(b) Water
(c) Mustard Oil
(d) Glycerin

Answer: D
The refractive indices of water, kerosene, mustard oil and glycerin are 1.33, 1.44, 1.46 and 1.47 respectively. The ray would bend the most, when it goes from rarer medium (say air) to the denser medium. Since refractive index of glycerin is the highest among all these medium, so glycerin is the most dense medium.

Question:16

Which of the following ray diagrams is correct for the ray of light incident on a concave mirror as shown in figure?

(a) Fig. A
(b) Fig. B
(c) Fig. C
(d) Fig. D

Answer: D
Any ray of light parallel to the principal axis passes through the focus (F) after reflecting from the concave mirror.

Question:17

Which of the following ray diagrams is correct for the ray of light incident on a lens shown in figure ?


(a) Fig. A
(b) Fig. B
(c) Fig. C
(d) Fig. D

Answer: A
A ray of light passing through the focus of a lens travels parallel to the principal axis after refracting through the lens.

Question:18

A girl is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. The order of combinations for the magic mirror from the top:
(a) Plane, convex and concave
(b) Convex, concave and plane
(c) Concave, plane and convex
(d) Convex, plane and concave

Answer: C
Concave mirror forms a magnified (enlarged) image of the object if the object is placed close to the concave mirror (i.e., a distance less than its focal length). Plane mirror always forms the image of the same size as that of the object.

Question:19

In which of the following, the image of an object placed at infinity will be highly diminished and point sized?
(a) Concave mirror only
(b) Convex mirror only
(c) Convex lens only
(d) Concave mirror, convex mirror, concave lens and convex lens.

Answer: D
From ray diagrams we can see in Concave mirror, convex mirror, concave lens and convex lens the image of an object placed at infinity will be highly diminished and point sized.

NCERT Exemplar Class 10 Science Solutions Chapter 10-Short Answer

Question:20

Identify the device used as a spherical mirror or lens in the following cases, when the image formed is virtual and erect in each case.
(a) Object is placed between device and its focus, image formed is enlarged and behind it.
(b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of an object.
(c) Object is placed between infinity and device, image formed is diminished and between focus and optical center on the same side as that of the object.
(d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.

Answer :
(a) Concave mirror is used.

(b) Convex lens.

(c) Concave lens

(d) Convex mirror

Question:21

Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.

Answer:
When light ray enters denser medium from rarer medium it bends towards the normal. Again, when this ray exists the second medium and enters the first, it ends away from the normal. In this case extent of bending of ray at opposite faces is the same, this is the reason why the emergent ray is parallel to the incident ray.

Question:22

A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent, if instead of water, we use liquids like, kerosene or turpentine. Support your answer with reason.

Answer:
The bending of light takes place because of refraction. Refraction dependents on refractive indices of the medium. A pencil dipped in water appears to be bent at the interface of air and water due to the refraction of light. The refraction of light occurs because the speed of light changes when light travels from one medium to another. The pencil will not appear to be bent to the same extent, when it is dipped in kerosene or turpentine. This is because refractive index of kerosene or turpentine is greater than the index of water and hence speed of light (v = c/n) is less in kerosene or turpentine as compared to in water.

Question:23

How is the refractive index of a medium related to the speed of light ? Obtain an expression for refractive index of a medium with respect to another in terms of speed of light in these two media.

Answer:
The refractive index can be seen as the factor by which the speed and the wavelength of the radiation are reduced with respect to their vacuum values.
n = c/v (where n: refractive index, c=speed of light, v: velocity of light in that medium)
Refractive index of one medium in relation to a second medium is given by ratio of speed of light in second medium to speed of light in first medium.
n_{21}=\frac{v_{1}}{v_{2}}

Question:24

Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5. Find out the absolute refractive index of diamond.

Answer :
Absolute refractive index of a medium shows the ratio of speed of light in air to speed of light in that medium. Let us assume c is the speed of light in air, v1 is speed of light in glass and v2 is speed of light in diamond.
Rdg =1.6 ----(1)
Absolute refractive index of glass ,
Rga =1.5 -----(2)
multiplying equation (1) and (2) ,
R_{da}\times R_{ga}=1.6\times 1.5=2.4
So, absolute refractive index of diamond = 2.4

Question:25

A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?

Answer:
The statement is correct.
When an object is placed between F and F2 of a convex lens, its enlarged, inverted and real image is formed beyond 2F2, i.e., on the other side of lens. So, for this we need to place the object between 20cm and 40cm of the lens.
When an object is placed between F and O of a convex lens, its enlarged, erect and virtual image is formed beyond F2, i.e., on the same side of lens. So, for this we need to place the object at a distance less than 20 cm from the lens.

Question:26

Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction, will she move the screen to obtain a sharp image of the building? What is the approximate focal length of this lens?

Answer:
A real image can be obtained on the screen. Therefore, the lens used is convex lens as it forms real as well as virtual image. The distance of the real image formed by a convex lens from the lens decreases as the object distance from the lens increases. Hence, the screen has to be moved towards the lens to obtain the sharp image of the building.
Approximate focal length of the lens = 15 cm. The rays of light from the window pane are considered to come from infinity. These rays of light are focused by the convex lens at its focus (i.e. on the screen).

Question:27

How are power and focal length of a lens related? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light?

Answer:
Power of a lens is reciprocal to its focal length. So, smaller focal length means more power. Out of the given lenses, the lens with 20 cm as focal length has more power than the lens with 40 cm as focal length. The lens with higher power should be used to obtain more convergent light.
The lens of focal length 20 cm or power 5.0 D will be used to have more convergent light. This is because lens of small focal length or large power strongly converges the parallel beam of light.

Question:28

Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence. Show the same with the help of diagram.

Answer:
When two plane mirrors are at right angles to each other, incident ray and reflected ray will always be parallel to each other, whatever may be the angle of incidence. From the diagram put any value of i the deviation will always be 180 degrees. This means that incident ray and reflected ray will always be parallel; irrespective of value of angle of incidence.

Question:29

Draw a ray diagram showing the path of rays of light when it enters with oblique incidence (i) from air into water; (ii) from water into air.

Answer:
(i) Ray diagram showing the path of rays of light when it enters with oblique incidence from air into water

(ii) from a denser medium to a rarer medium the light ray bends away from the normal.

NCERT Exemplar Class 10 Science Solutions Chapter 10-Long Answer

Question:30

Draw ray diagrams showing the image formation by a concave mirror when an object is placed
(a) between pole and focus of the mirror
(b) between focus and centre of curvature of the mirror
(c) at centre of curvature of the mirror
(d) a little beyond centre of curvature of the mirror
(e) at infinity

Answer:
(a) between pole and focus of the mirror

(b) between focus and center of curvature of the mirror

(c) at center of curvature of the mirror

(d) a little beyond center of curvature of the mirror

(e) at infinity

Question:31

Draw ray diagrams showing the image formation by a convex lens when an object is placed
(a) between optical centre and focus of the lens
(b) between focus and twice the focal length of the lens
(c) at twice the focal length of the lens
(d) at infinity
(e) at the focus of the lens

Answer:
(a) between optical center and focus of the lens

(b) between focus and twice the focal length of the lens

(c) at twice the focal length of the lens

(d) at infinity

(e) At the focus of the lens


Question:33

Draw ray diagrams showing the image formation by a concave lens when an object is placed
(a) at the focus of the lens
(b) between focus and twice the focal length of the lens
(c) beyond twice the focal length of the lens

Answer :
(a) at the focus of the lens

(b) between focus and twice the focal length of the lens

(c) beyond twice the focal length of the lens

Question:34

Draw ray diagrams showing the image formation by a convex mirror when an object is placed
(a) at infinity
(b) at finite distance from the mirror

Answer :
(a) at infinity

(b) at finite distance from the mirror

Question:35

The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?

Answer :
The image is obtained on the screen so it is real.
Therefore, magnification, m = –3,
v = 80 cm
Now, m=\frac{v}{u}
Therefore,
-3=\frac{80}{u}
u=\frac{-80}{3}cm
f=\left ( \frac{1}{v} \right )-\left ( \frac{1}{u} \right )=\left ( \frac{1}{80} \right )+\left ( \frac{3}{80} \right )=\frac{1}{20}
f=20\; cm
The lens is convex and image formed at 80 cm from the lens is real and inverted.

Question:36

Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?

Answer:

Using,
m = - v/u = h?/h?
⇒ - v/u = 1/3 h?/h?
⇒ u = - 3v
Using mirror formula,
1/v + 1/u = 1/f, we get
when f = + 20 cm
⇒ 1/v - 1/3v = 1/20
⇒ 3 - 1/3v = 1/20
⇒ 2/3v = 1/20
⇒ 3v = 2 × 20
⇒ v = 2 × 30/3
⇒ v = 40/3 .. (i)
u = - 3v
By putting v value, we get
⇒ u = - 3 × 40/3
⇒ u = - 40 cm.
Object Distance = - 40 cm
If we take f = - 20 cm, value of u will be + 40 cm which is not possible.
m = - v/v
⇒ m = - 40/3/- 40
⇒ m = + 1/3
⇒ m = 0.33 cm
This implies that the nature of the image is erect and diminished.
Nature of Image = Convex Mirror.

Question:37

Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of –50 cm. What is the nature of the lens and its power used by each of them?

Answer:
P = 1 / f, where f is in meter.
The unit of power is Diopter. Lens is convex in the first case and concave in the second case. Power of lens (first student) = +2 diopter
Power of lens (second student) = -2 diopter

Question:38

A student focused the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as under
Position of candle = 12.0 cm
Position of convex lens = 50.0 cm
Position of the screen = 88.0 cm
(i) What is the focal length of the convex lens?
(ii) Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm?
(iii) What will be the nature of the image formed if he further shifts the candle towards the lens?
(iv) Draw a ray diagram to show the formation of the image in case (iii) as said above.

Answer:

Answer :
(i) Focal length = 38 ÷ 2 = 19 cm
(ii) object distance u = 50 – 31 = 19 cm
In this case, object distance = focal length
This means that images is formed at infinity.
(iii) The image formed will be virtual and erect.
(iv)

Question:32

Write laws of refraction. Explain the same with the help of ray diagram, when a ray of light passes through a rectangular glass slab.

Answer:

Laws of refraction:
  1. The incident ray, the normal to any refracting surface at the point of incidence, and the refracted ray all lie in the same plane called the plane of incidence or plane of refraction.
  2. The ratio of the sine of the angle of incidence to the angle of refraction is always constant.
\frac{\sin i}{\sin r}=\mathrm{constant}

\frac{\sin i}{\sin r}=\frac{\mu _{2}}{\mu _{1}}=\frac{v_{1}}{v_{2}}=\frac{\lambda_{1}}{\lambda_{2}}
or
$$ \mu _{1} \sin i=\mu _{2} \sin r $$
\frac{\sin(i)}{ \sin(r)}= \mu _{21}
= refractive index of the second medium with respect to the first medium.
For a glass slab as shown in figure

i = angle of incidence
r1 = angle of reflection
\frac{sini}{sinr_1}=\frac{\mu_g}{\mu_{air}}


NCERT Exemplar Class 10 Science Solutions Chapter 10 Important Topics:

Key topics followed in the NCERT exemplar Class 10 Science chapter 10 solutions cover the understanding of:

  • The phenomenon of reflection and refraction.
  • NCERT exemplar Class 10 Science solutions chapter 10 discusses bending of light called refraction and is governed by Snell’s law.
  • Different types of optical instruments like lens and mirror.
  • Type of image formed and understanding the difference between real and virtual image.

NCERT Class 10 Exemplar Solutions for Other Subjects:

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NCERT Class 10 Science Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 10 Science Solutions chapter 10

These Class 10 Science NCERT exemplar chapter 10 solutions provide a basic understanding of a small range of electromagnetic waves called light. The phenomenon of vision has great importance on human development. Light Reflection and Refraction is one crucial and fundamental building block for students aspiring to study Physics. The problems posted in books such as NCERT Class 10 Science Book, Physics question bank, S. Chand by Lakhmir Singh and Manjit Kaur, et cetera can be easily dealt with by practicing from NCERT exemplar class 10 science chapter 10 solutions Light Reflection and Refraction.

Students can use NCERT exemplar Class 10 Science solutions chapter 10 pdf download for keeping the solutions handy while studying NCERT exemplar Class 10 Science chapter 10 in an offline mode.

Check NCERT Solutions for questions given in the book

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Frequently Asked Questions (FAQs)

1. What kind of image of a convex lens form?

Any convex lens in air can form a real image of a real object, virtual image of a real object, real image of a virtual object but cannot form virtual image of virtual object

2. If a bird in air and fish in water has a physical separation of X meter, what separation will they observe?

The bird will observe the separation of fish less than X meter, and fish will observe the separation of bird more than X meter because of refraction.

3. Will spherical diamond glow?

Any diamond glows because of total internal reflection through multiple surfaces of a diamond. In a spherical diamond, there will be no total internal reflection, and it will not glow.

4. What is the difference between diffused reflection and specular reflection?

The diffused reflection occurs by a rough surface, and specular reflection occurs by plane reflecting surface like a mirror. In diffused reflection, we will see the reflecting body, and in specular reflection, we will see the light source’s image.

5. Is Light Reflection and Refraction a vital chapter?

It is an essential chapter for a student’s journey in science and accounts for approximately 8-10 % marks of the final paper. NCERT exemplar Class 10 Science solutions chapter 10 provides an extra edge to understand and practice Light Reflection and Refraction concepts.

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  6. Final Destination:

    • Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

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sai raja 01 September,2024
show the previous year exams ?

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

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SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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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.

Read Complete Answer
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 .
Read More

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
Read More

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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