Skip to main content

CBSE PHYSICS ADDA: CLASS X SCIENCE (Physics+Chemistry) SUMMATIVE ASS...

1. Why carbon forms strong bonds with other carbon atoms, hydrogen, oxygen, nitrogen or sulphur?
2. Name the part of eye responsible for conversion of light into electrical impulses.
3. a. How does valency vary in a group on going from top to bottom? b. How does atomic size vary in a period on going from left to right?
4. Write one property of hydrogen which makes it resemble with (a) Alkali metals (b) Halogens.
5. a Arrange the following common substances in the increasing order of refractive indices. Ice, Kerosene, Glass, Diamond, Alcohol, Water
b. Is it necessary that optically dense medium possesses greater mass density? Give an example.
6. On reaction with sodium hydroxide, X yielded Ethanoic acid and ethanol.
(a) Give the IUPAC name of X? (b) Name the reaction. (c) Give a chemical reaction for the above reaction.
7. (a) How does the electronic configuration of an atom related to its position in the Modern periodic table? Give one example.
(b) Why nitrogen is more electronegative than phosphorus?
8. A concave length has focal length of 25 cm. At what distance should the object from the lens be placed so that if it forms an image at 20 cm distance from the lens? Also find the magnification produced by the lens.
9. Give an explanation for the formation of a rainbow.
10. How are we able to see distant and near by objects clearly? Which part of eye helps in changing curvature of lens? Why no image is formed at blind spot?
11.(a) Why magnification is taken negative for real images and positive for virtual images? (2)
(b) Why convex mirror is used in rear view mirrors and not concave mirror? (2)
(c) Power of concave lens is 4.5 D. Find its focal length. (1)
12.(a) Find the size, nature and position of image formed when an object of 1 cm is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. (3)
(b) Why does light travels faster in water in comparison to kerosene. (Refractive index of water and kerosene are 1.33 and 1.44 respectively) (2)
13 . a. Which property of carbon leads to formation of large number of compounds? Define it
b. What is the functional group in the following molecules?
i. CH3CH2CH2OH
ii. CH3COOH
c. Which of the following formula represents a saturated hydrocarbon?
CnH 2n, C nH2n+1 Cn H2n+2 Cn H2n-2
d. What happens when methane is burnt in oxygen?
e. Why is the conversion of ethanol to Ethanoic acid an oxidation reaction?
14. a. Give three points to distinguish between alkenes and alkynes.
b. Explain the mechanism of cleaning action of detergents
CLASS X SCIENCE (Physics & Chemistry) SUMMATIVE ASSESSMENT TERM II 2012 SAMPLE PAPER- I (solution)

1. Due to small size and presence of four valence electrons, carbon forms strong bonds with other carbon atoms, hydrogen, oxygen, nitrogen or sulphur.
2. Retina.
3. a. Valency remains same on moving from top to bottom in a particular group. This is because the outermost electronic configuration of all the elements in a group remains same.
b. Atomic size decreases on moving from left to right in a period. This is because the number of shells remains same but the nuclear charge increases. Due to this, electrons are pulled closer to the nucleus and decrease the atomic size.
4. Resemblance with alkali metals:
Hydrogen has the same outermost electronic configuration as that of alkali metals.
Resemblance with halogens:
Hydrogen exists as diatomic molecule as halogens.
5. a. In the order of increasing refractive indices:
Ice; Water; Alcohol; Kerosene; Glass; Diamond
b. No, it is not always necessary. For e.g., kerosene and turpentine oil having high refractive index are optically dense than water. But its mass density is less than that of water. That is why oil floats on the surface of water.
6. (a) X is ethyl Ethanoate.
(b) It is Saponification reaction.
(c) CH3COOC2H5 + Na OH -----------à C2H5OH + CH3COOH 1
7. (a) By the electronic configuration of an atom we can identify the group and period to which the atom belongs.
For example - Oxygen has atomic number 8. Its electronic configuration is 2,6. As it can gain two electrons to complete its octet, so it belongs to group 16 Also, since the electrons are filled in two shells. So. Oxygen belongs to the second period.
(b) As we move down the group, electro negativity of elements decreases as the atomic size increases. Nitrogen is placed above phosphorus in group 15. So, nitrogen is more electronegative than phosphorus.
8. A concave lens always forms a virtual, erect image on the same side of the object.
v = -20 cm, f = -25 cm, u = ?
1/v -1/u = 1/f
1/u = 1/(-20) – 1/(-25)
1/u = -1/100
u= -100cm
Thus object distance is 100 cm.
Magnification= v/u
= -20/ (-100)
= + 0.5 1
Thus image is erect, virtual and is half of the size of object.
9. Formation of rainbow:
The rainbow is formed in the sky when sun shines and it is raining at the same time. The raindrops in the atmosphere act like many small prisms.
As white light enters and leaves these raindrops the various colours present in white light are refracted by different amount due to which an arch of seven colours called rainbow is formed in the sky.
Diagram
10. (a) Accommodation
(b) Ciliary muscles.
(c) No image is formed at the blind spot because no nerve cells are present there to carry the information of image to the brain.
11. (a) Distance measured upward and perpendicular to the principal axis are taken as positive.
Distances measured down ward and perpendicular to the principal axis are taken as negative.
So magnification for a real image is taken negative and for a virtual image it is taken as positive.
(b) A convex lens forms a virtual and small sized image of the object and concave mirror forms real and inverted image. 1
Thus by using convex mirror we can get view of wider field which is not possible in case of a concave mirror. 1
(c) Power = 1/Focal length
Or Focal length = 1/Power
f= 1/4.5
f = 0.22 cm 1
12. (a) Object distance,
u = -15 cm
Image distance, v= ?
Focal length, f = -10 cm
Mirror formula,
1/v + 1/u = 1/f
1/v + 1/ (-15) = 1/(-10)
1/v = -1/10 + -/15
1/v = -3+2/30
1/v = -(1/30)
Thus the position of image is formed on left hand side in front of the concave mirror at a distance of
30 cm, its nature will be real and inverted.
Size of image m= -v/u
m= - [(-30)/(-15)]
m= -2.
Thus the size of image is 2 cm and image is real and inverted.
(b) Refractive index of a medium= Speed of light in air/ Speed of light in medium.
So, speed of light in medium= speed of light in air/ refractive index
Thus speed of light will be maximum in medium having lowest refractive index. Therefore speed of light is more in water in comparison to kerosene.
13. (a) Catenation
The property of carbon atom to link with other carbon atoms to form large molecules is called as catenation.
b. i. Functional group is –OH group.
ii. Functional group is –COOH group.
C. Cn H2n+ 2 represents a saturated hydrocarbon.
d. Methane burns in oxygen with formation of carbon dioxide and water.
CH4 + 2O2 -----------à CO2 + 2H2O
e. Conversion of ethanol to Ethanoic acid is an oxidation reaction because oxygen is added to ethanol and Ethanoic acid is formed.
CH 3CH2 OH + O2 -----------à CH3 COOH + H2 O
Ethanol Ethanoic acid
14. a. No. Alkene
(1) Unsaturated hydrocarbon containing a double bond between two carbon atoms are known as alkenes.
(2) General formula is CnH2n
Example: Ethene, Propene
Alkynes: Unsaturated hydrocarbons containing a triple bond between two carbon atoms are known as alkynes.
(2) General formula is CnH2n -2
(3) Example: Ethyne, Propyne
b. A detergent molecule consists of two ends –
(a) hydrocarbon tail which is hydrophobic (water repelling) and
(b) polar head which is hydrophilic (water attracting or loving).
When a detergent is dissolved in water, the hydrocarbon tail aligns itself towards the dirt and ionic part aligns itself towards the water. The molecules gather together as clusters, called micelles.
When water is agitated, the dirt suspended in the micelles is easily rinsed away. Thus, the cloth gets cleaned.

Comments

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination(Solved)

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination Law of conservation of mass Law of constant proportion Empirical formula 1. If 10 grams of CaCO 3 on heating gave 4.4g of CO 2 and 5.6g of CaO, show that these observations are in agreement with the law of conservation of mass.(Based on Law of conservation of mass) Solution:  Mass of the reactants = 10g ;  Mass of the products = 4.4 + 6.6g = 10g Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. 2. 1.375 g of cupric oxide was reduced by heating and the weight of copper that remained was 1.098g.  In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition . The weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state? Solution: in first experiment: Copper oxide = 1....

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get d...

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Structure of Matter class07

The language of chemistry class 7 Basic science soled questions. Q. What is chemistry? Ans: Chemistry is a branch of science in which we study the composition, structure, properties and change of matter. Q. What is matter? Ans: Anything that is around us having some mass and occupies space is called matter. Q. What are the different classifications of matter? Ans: Matters are classified into element, compound and mixture. Q. What is element? Ans: Elements are the purest form of substance (matter) made up of same kinds of particles. The smallest unit of element is an atom. eg. Hydrogen, oxygen, carbon. Q. What is compound?            See full post   The language of chemistry class 7 Basic science Class 07 Basic sciences Chapter the Language of Chemistry A. Answer these questions 1. What does the formula of a substance tell us? Answer: Formula of a compound or an element represents the name and number of atoms present ...

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page ...