CLASS-10 SUBJECT : SCIENCE CHAPTER-10 LIGHT REFLECTION & REFRACTION.

 EVENTS CONVENT HIGH SCHOOL

09/11/2021      CLASS- 10   SESSION 2021-22
SUBJECT :  SCIENCE

CHAPTER-10
LIGHT REFLECTION & REFRACTION

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Question 1 Which one of the following materials cannot be used to make a lens ?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) Clay

Question 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:
(d) Between the pole of the mirror and its principal focus.

Question 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:
(b) At twice the focal length.

Question 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:
(a) Both concave

Question 5 No matter how far you stand from mirror, your image appears erect. The mirror is likely to be
(a) plane
(b) concave
(c) convex
(d) either plane or convex.
Answer:
(d) Either plane or convex.

Question 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:
(c) A convex lens of focal length 5 cm.

Question 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:
A concave mirror gives an erect image when the object is placed between the focus F and the pole P of the concave mirror, i.e., between 0 and 15 cm from the mirror. The image thus formed will be virtual, erect and larger than the object.

Question 8 Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
Answer:
(a) Concave mirrors are used as reflectors in headlights of cars. When a bulb is located at the focus of the concave mirror, the light rays after reflection from the mirror travel over a large distance as a parallel beam of high intensity.

(b) A convex mirror is used as a side/rear-view mirror of a vehicle because

• A convex mirror always forms an erect, virtual and diminished image of an object placed anywhere in front it.
• A convex mirror has a wider field of view than a plane mirror of the same size.

(c) Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.

Question 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:
A convex lens forms complete image of an object, even if its one half is covered with black paper. It can be explained by considering following two cases.
Case I : When the upper half of the lens is covered
In this case, a ray of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure.

Case II: When the lower half of the lens Is covered
In this case, a ray of light coming from the object is refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the given figure.

Question 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 the nature of the image formed.
Answer:
Here : Object distance, u= -25 cm,
Object height, h = 5 cm,
Focal length, f = +10 cm
According to the lens formula, 1f=1ν−1u , we have
⇒ 1ν=1f−1u=110−125=15250orν=25015=16.66cm
The positive value of v shows that the image is formed at the other side of the lens.

The negative value of image height indicates that the image formed is inverted.
The position, size, and nature of image are shown alongside in the ray diagram.

Question 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.
Solution:
Focal length, f = -15 cm, Image distance, ν = -10 cm (as concave lens forms the image on the same side of the lens)
From the lens formula 1f=1ν−1u , we have

Object distance, u = -30 cm
The negative value of u indicates that the object is placed in front of the lens.

Question 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.
Solution:
Object distance, u = -10 cm, Focal length, f = +15 cm, Image distance, ν = ?

Thus, image distance, ν = + 6 cm
Because ν is +ve, so a virtual image is formed at a distance of 6 cm behind the mirror.
Magnification, m=−υu=−6−30=15 (i.e. < 1)
The positive value of m shows that image erect and its value, which is less than 1, shows that image is smaller than the object. Thus, image is virtual, erect and diminished.

CLASS-9 SUBJECT SCIENCE CHAPTER-3 ATOMS AND MOLECULES

 EVENTS CONVENT HIGH SCHOOL

08/11/2021      CLASS- 9   SESSION 2021-22
SUBJECT :  SCIENCE

CHAPTER-3
ATOMS AND MOLECULES

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Question 1. In a reaction 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass carbonate.
Answer.

Class 9 Science NCERT Textbook Page 32
Question 1. In a reaction 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium etkanoate. Show that these observations are in agreement with the law of conservation of mass carbonate.
Answer:

Question 2. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Answer: Ratio of H : O by mass in water is:
Hydrogen : Oxygen —> H2O
∴ 1 : 8 = 3 : x
x = 8 x 3
x = 24 g
∴ 24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

Question 3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer: The postulate of Dalton’s atomic theory that is the result of the law of conservation of mass is—the relative number and kinds of atoms are constant in a given compound. Atoms cannot be created nor destroyed in a chemical reaction.

Question 4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer: The relative number and kinds of atoms are constant in a given compound.

Class 9 Science NCERT Textbook Page 35
Question 1. Define the atomic mass unit.
Answer: One atomic mass unit is equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12. The relative atomic masses of all elements have been found with respect to an atom of carbon-12.

Question 2. Why is it not possible to see an atom with naked eyes?
Answer: Atom is too small to be seen with naked eyes. It is measured in nanometres.
1 m = 109 nm

NCERT Textbook Questions – Page 39
Question 1. Write down the formulae of
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide
Answer: The formulae are

Question 2. What is meant by the term chemical formula?
Answer: The chemical formula of the compound is a symbolic representation of its composition, e.g., chemical formula of sodium chloride is NaCl.

Question 3. How many atoms are present in a
(i) H2S molecule and
(ii) P043- ion?
Answer: (i) H2S —> 3 atoms are present
(ii) P043- —> 5 atoms are present

NCERT Textbook Questions – Page 40
Question 1. Calculate the molecular masses of H2, O2, Cl2, C02, CH4, C2H2,NH3, CH3OH.
Answer: The molecular masses are:

Question 2.Calculate the formula unit masses of ZnO, Na2O, K2C03, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Answer: The formula unit mass of
(i) ZnO = 65 u + 16 u = 81 u
(ii) Na2O = (23 u x 2) + 16 u = 46 u + 16 u = 62 u
(iii) K2C03 = (39 u x 2) + 12 u + 16 u x 3
= 78 u + 12 u + 48 u = 138 u

Class 9 Science NCERT Textbook Page 42
Question 1. If one mole of carbon atoms weigh 12 grams, what is the mass (in grams) of 1 atom of carbon?
Answer:

Question 2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given atomic mass of Na = 23 u, Fe = 56 u)?
Answer:

Questions From NCERT Textbook for Class 9 Science

Question 1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer: Boron and oxygen compound —> Boron + Oxygen
0.24 g —> 0.096 g + 0.144 g

Question 2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Answer: The reaction of burning of carbon in oxygen may be written as:

It shows that 12 g of carbon bums in 32 g oxygen to form 44 g of carbon dioxide. Therefore 3 g of carbon reacts with 8 g of oxygen to form 11 g of carbon dioxide. It is given that 3.0 g of carbon is burnt with 8 g of oxygen to produce 11.0 g of CO2. Consequently 11.0 g of carbon dioxide will be formed when 3.0 g of C is burnt in 50 g of oxygen consuming 8 g of oxygen, leaving behind 50 – 8 = 42 g of O2. The answer governs the law of constant proportion.

Question 3. What are poly atomic ions? Give examples.
Answer: The ions which contain more than one atoms (same kind or may be of different kind) and behave as a single unit are called polyatomic ions e.g., OH–, SO42-, CO32-.

Question 4. Write the chemical formulae of the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Answer: (a) Magnesium chloride
Symbol —> Mg Cl
Change —> +2 -1
Formula —> MgCl2
(b) Calcium oxide
Symbol —> Ca O
Charge —> +2 -2
Formula —> CaO
(c) Copper nitrate
Symbol —> Cu NO
Change +2 -1
Formula -4 CU(N03)2
(d) Aluminium chloride
Symbol —> Al Cl
Change —> +3 -1
Formula —> AlCl3
(d) Calcium carbonate
Symbol —> Ca CO3
Change —> +2 -2
Formula —> CaC03