10th maths formative assessment- iii_fa-3 Download
10th science formative assessment- iii_fa-3 Download
CBSE Guidelines for the Scheme of CCE for Class X
We would like to bring to your notice the CBSE guidelines for the scheme of Continuous and Comprehensive Evaluation that will be followed for the students of the present Class X.
Evidence of Assessments (EAs)The CBSE has decided to issue uniform certificates to all the students appearing in class X in March 2011 Examination irrespective of the fact whether they are assessed by the Board or by the School in Summative Assessment-II (SA-II).
To ensure the reliability, validity and fairness of assessment, the Board is collecting marks/grades assigned at school level through specially designed software. Besides the collection of marks/grades, the Board is simultaneously doing a random collection and verification of the ‘Evidence of Assessments (EAs)’ conducted at school level under Formative assessment and Summative assessment.
These ‘Evidence of Assessments (EAs)’ and the marks awarded will be verified by the subject experts appointed and empanelled by the Board at the Regional level. The objective is to assess the ‘Practices of School Based Assessment’. The experts would submit their report on these ‘Evidence of Assessments (EAs)’ vis-a-vis award of marks. Their report will help in verifying the school based assessments as well as provide feedback to the Board about implementation of CCE scheme at grassroots level.
The schools which will be shortlisted by the Regional Office will be requested to send the ‘Evidence of Assessments (EAs)’ for SA-1, FA3 & FA4.
This material (FA tasks and SA answer sheets) is being collected initially in five subjects: Hindi, English, Mathematics, Science and Social Science. Please ensure that these ‘Evidence of Assessments (EAs)’ sent to the Board comprise of the performance of five students taken from each of the three categories i.e. top, middle and bottom levels of achievements. Thus, your school if selected, should be submitting fifteen Evidence of Assessments (EAs) (FA tasks and SA answer sheets) for every assessment i.e. 15 for FA (FA3, FA4 taken together) and 15 for SA1. This is applicable only to one subject which will be communicated to you by the Regional Offices.
The school is required to send details about all the fifteen FA tasks in the subject as directed by Regional Office which should include the break-up of marks as well as the parameters for assessment. If the nature of the Evidence of Assessments (EAs) is such that it cannot be posted or transported such as oral testing, seminars, group discussion, model, chart etc. a brief write-up as already prepared by the teacher on the assignment may please be sent along with marks awarded. This should include the details of the task assigned to students individually or in groups as well as the strategies adopted and the parameters used for assessment. The schools should send only those tasks that have been taken into account for arriving at the grades for FA3 and FA4.
Question Paper used and its marking scheme in the subject should also be attached with SA answer sheet irrespective of whether you have used the question paper sent by the Board or your own question paper.
The Board would initiate similar exercise for the co-scholastic areas shortly. You are requested to keep the ‘Evidence of Assessments (EAs)’ for the Co-scholastic areas also in safe custody.
After you are informed by the Regional Office, please ensure that all the ‘Evidence of Assessments (EAs)’ in the subject indicated by the Regional Office are sent through speed post or hand delivered (at school’s cost) to the concerned Regional Office latest by 20th March 2011 along with the checklist for collection of evidence.
SELECTION OF THE SAMPLES
i.Arrange all the students of class X (all sections) in increasing orde
r of marks calculated for SA-1.
ii.Divide the total number of students in three groups as follows:
Top one-third students,Middle one-third students, and Bottom one-third students.For example: If there are 100 students in Class X in a School, the three groups may consist of 33, 33 and 34 students after they have been arranged in increasing order of marks.
iii. Pick up the top five students from the first group, the last five students from the third group and any five students from the middle group.
iv. This set of fifteen students is your sample for whom ‘Evidence of Assessments (EAs)’ are to be sent.
i. Arrange all the students of class X (all sections) in increasing order of marks calculated for FA3 & FA4 taken together.
ii. Divide the total number of students in three groups as follows:
a. Top one-third students,
b. Middle one-third students, and
c. Bottom one-third students.For example: If there are 100 students in Class X in a School, the three groups may consist of 33, 33 and 34 students after they have been arranged in increasing order of marks.
iii. Pick up the top five students from the first group, the last five students from the third group and any five students from the middle group.
iv. This set of fifteen students is your sample for whom ‘Evidence of Assessments (EAs)’ are to be sent.
Note : Samples of 15 students selected for Summative Assessment and 15 students selected for Formative Assessments may differ depending on the performance of students.
Source: CBSE Circular No. 12, 2011
Point to remember:-
1. Species maintain continuity by reproduction.
2. Variation in different degrees occur.
3. Variants have unequal chances of survival.
4. In sexually reproducing individuals, for the same trait genes have two copies. Unidentical
copies bear dominant and recessive nature.
5. Traits separately inherited, yields new combinations of traits in progeny.
6. The sex of human progeny is determined by the sex chromosomes inherited from father.
7. Variation in combination with geographical isolation might lead to speciation.
8. Evolutionary relationship among living organisms become visible in hierarchy of
9. Study of living forms as well as dead remains of organisms emerges as evolutionary
10. DNA changes accumulated over a time span has resulted in evolution of complex organs like eye as exhibited by structure of eye in different animals.
11. Features like feathers have evolved as a function of change in allocated task.
12. Evolution is not progress from lower to higher forms.
1 mark questions :-
1. Which of the two exhibit greater variation : asexual or sexual reproduction?
2. Why all the variants don't have equal chances of surviving?
3. What is heredity?
4. At which place on earth did humans appeared first?
5. Give the name of plant studied by Mendel?
6. Name the acid which is information source for making proteins in cells.
7. What is the chemical nature of enzymes?
8. Define gene.9. Which one of the two is shorter in length 'X' or 'Y'?
Chapter - 9 Heredity and Evolution read more............
Points to remember:-
1. Reproduction is essential for continuity of living organisms.
2. It involves creation of a DNA copy along with the formation of additional cellular apparatus during cell division.
3. Basically it is of two types – Asexual and sexual. Some plants like Rose and Banana
are prolifrated by vagetative propogation.
4. There are examples of asexual reproduction where new generation arises from a single cell or single individual as in fission, fragmentation, regeneration, budding, spore formation.
5. Sexual reproduction involves two individuals to produce off spring.
6. Variation occurs due to DNA copying mechanism and sexual reproduction.
7. Reproduction in flowering plants involve transfer of pollen grains from anther to stigma – pollination. Gametes fertilize to form zygote.
8. Changes occur in boys and girls at puberty, beard and moustache in boys and growth
in breast region in girls apart from other changes. These changes indicate sexual
maturation and biological preparedness for reproduction.
9. Sexual reproduction takes place by fusion of both male and female gametes resulting in the formation of zygote which gives rise to offspring.
10. Awareness regarding family planning and sex related communicable diseases (STDs) help individual to maintain normal reproductive health.
11. Condoms, oral pills, copper T are some of the contraceptives to avoid pregnancy. 1 mark questions :-
1. Why simply copying of DNA in a dividing cells not enough to maintain continuity of life?
2. How does plasmodium undergo fission?
3. How spirogyra reproduces by fragmentation?
Full Study Chapter - 8 How Do Organisms Reproduce Click here
POINTS TO REMEMBER:
1) ECOLOGY – The study of the interaction of living organisms with each other and
their surroundings is called ecology.
2) ENVIRONMENT – Everything that surrounds organisms and influences its life.
a) Biotic components of environment --- The living organisms .e.g.
Plants and animals.
b) Abiotic components of environment --- The nonliving components
like water, air, light, etc.
It is the sum total of all external conditions and influences that affect the life and the development of organisms i.e.,it includes all the biotic and abiotic factors.
3) ECOSYSTEM – All interacting organisms in an area together with the nonliving
constituents of environment. (Functional unit of an environment)
Functions of ecosystem:-
a) Flow of energy
b) Cycling of nutrients(bio-geo chemical cycles).
4) PRODUCERS --- They make the energy from sunlight available to the rest of the
5) CONSUMERS --- Animals can not manufacture their own food. They are called
6) BIODEGRABLE ---- Substances that are broken down by the action of bacteria or
other saprophytes. e. g. – Paper.
7) NONBIODEGRABLE --- Substances that are not broken down by the action of
bacteria or saprophytes. e. g. Plastic.
8) FOOD CHAIN ---- The process of one organism eating the other.
GRASS -> GRASSHOPPER -> FROG -> SNAKE
Importance of food chain
1.It helps in transfer of food energy from one organism to another.
2.It is a pathway for the flow of energy.
3. It helps in understanding the interdependence amongst different organisms.
Trophic levels are the various steps in the food chain .
Producers --- first trophic level
Herbivores --- second trophic levels
Carnivores --- third trophic level
Top carnivores --- fourth trophic level
9) FOODWEB ----- It is a network of food links between populations in a community.
10) FLOW OF ENERGY -Ten percent law Energy available at each successive trophic level is 10 % of the previous level. (Lindeman 1942)
The law states that, “ only a 10 % amount of the total available energy is transferred from one trophic level to the next. The rest 90% of energy is used up or lost to the surrounding.”
11) BIOLOGICAL MAGNIFICATION --- Progressive accumulation of nonbiodegradable waste at various trophic levels of food chain.
12. Differences between food chain & food web
1. The process of eating & being eaten to transfer food energy.
2. It forms a part of food web.
3. It has limited populations.
1. It is a system of interconnected food chains.
2. It contains many food chains.
3. It has several populations of different species.
13. HOW DO OUR ACTIVITIES AFFECT THE ENVIRONMENT?
Two major problems:-
a) Depletion of ozone layer
b) Disposal of wasteGood ozone and bad ozone
The thick blanket of ozone layer in the atmosphere which forms a protective cover & prevents UV radiation from reaching the earth‘s surface. This is called good ozone and is found in the stratosphere. Ozone when present in the troposphere act as a pollutant . This harmful ozone is called bad ozone.
14. Depletion of ozone layer : The ozone layer protects all the organisms from the harmful ultra violet rays.
Ozone depleting substances: CFCs, N2O, CH4 , CCl4
These chlorine containing compounds are used in:
Aerosols, solvents, refrigerants and fire extinguishers.
Ozone hole developed over Antartica.
15. Effects of ozone depletion on human health- Due to depletion of ozone layer UV radiations reaches the earth and cause:
i) skin cancer ii) increased chances of cataract
iii) suppression of immune system . Managing the garbage we produce
16. Reasons for large production of garbage
i) Improvement in life style
ii) Changes in attitude (more things become disposable)
iii) Population explosion iv) Changes in packaging(non -biodegradable)
More topics to read
FORMATIVE ASSIGNMENT - III 10th Our Environment
10th Our Environment HIGHER ORDER THINKIN SKILLS (HOTS) QUESTIONS
10th physics Chapter - 15 Our Environment POINTS TO REMEMBER
1. Determine the mass of 6.022 X 10^23 number of N2 molecules.
2. Calculate the number of particles in-
(i) 8 g of O2 molecules (ii) 2.5 mol of calcium atoms.
3. What is the mass of 2.5 mol of Methane?
4.Find the mass of one molecule of water.
5.Calculate the number of water molecules and number of oxygen and hydrogen atoms
in a drop of water containing 0.03 mol of water.
6. Calculate the actual mass of one atom of carbon if 12 grams of Carbon contain one mole of Carbon.
7. Calculate mass of Nitrogen (N2) which contains same number of molecules as are present in 4.4 grams of Carbon-di-oxide (CO2).
8.Atomic mass of gold is 197 u. How many moles of gold are present in an ornament containing 88.65 grams of gold?
9. How many moles of SO2 have same mass as 3 moles of oxygen?
10. A glass of water contains 5 mol of water. How many molecules of water arepresent?
11. What is the mass of a formula unit of Na + C l - ?
12. How many atoms of Silver are present in a silver wire weighing 5.4 grams?
13. Calculate the ratio of molecules present in 16 g of methane and 16 g of oxygen.
14. Convert into mole. (a) 12 g of oxygen gas (b) 20 g of water (c) 22 g of carbon-dioxide.
15. Determine the number of bromide ion in 0.2 mole of Mg Br2.
th Mole concept numerical problems solved-9
9th Mole concept numerical problems solved-10
Class 9 Atoms and molecules solved CBSE Papers New links
More Problems based on mole concept
9th Atoms and Molecules -
1. Mole Concept and Problems based on mole concept
2. Numerical Problems based on mole concept 9th Chemistry
3. Numerical based on mole concept by ChemistryAdda
1a) Charged particles in matter :-
Atoms have three types of sub atomic particles. They are electrons, protons and neutrons.
Electrons are negatively charged (e-), protons are positively charged (p+) and neutrons have no charge (n).
The mass of an electron is 1/2000 the mass of a hydrogen atom. The mass of a proton is equal to the mass of a hydrogen atom and is taken as 1 unit. The mass of a neutron is equal to the mass of a hydrogen atom and is and is taken as 1 unit.
b) Discovery of sub atomic particles :-
In 1900, J.J.Thomson discovered the presence of the negatively charged particles called electrons in the atom.
In 1886, E.Goldstein discovered new radiations in gas discharge and called them canal rays. These rays were positively charged. This later led to the discovery of the positively charged particles called protons in the atom.
In 1932 Chadwick discovered the presence of particles having no charge in the atom called neutrons.
a) Thomson’s model of an atom :-
According to Thomson an atom is similar to a Christmas pudding. The pudding had positive charge and the electrons having negative charge were like plums on the pudding.
He proposed that :-
i) An atom consists of a positively charged sphere and the electrons are embedded in it.
ii) The negative and positive charges are equal in magnitude So the atom as a whole is electrically neutral.
Download full notes and many CBSE 9th Chemistry test paper
1. A 15-kg rock falls from a cliff at a height of 140 m to a cliff that is 36 m above the ground. How fast is the rock falling when it strikes the lower cliff? (45 m/s)
2. A 2.00-kg block of ice is sliding across a frozen pond at 3.00 m/s. A 12-N force is applied in the direction of motion. After the ice block slides 17.0 m, the force is removed. The work done by the applied force is
3. A 5.0-kg ball on the end of a string is whirled at a constant speed of 1.0 m/s in a horizontal circle of radius 5 m. What is the work done by the centripetal force during one revolution?
4. A woman stands on the edge of a cliff and throws a stone vertically downward with an initial speed of 15 m/s. The instant before the stone hits the ground below, it has 2000 J of kinetic energy. If she were to throw the stone horizontally outward from the cliff with the same initial speed of 15 m/s, how much kinetic energy would it have just before it hits the ground?
5. Mike is cutting the grass using a human-powered lawn mower. He pushes the mower with a force of 100 N directed at an angle of 20° below the horizontal direction. Calculate the work that Mike does on the mower in pushing it 5 m across the yard.
6. The kinetic energy of a car is 7000 J as it travels along a horizontal road. How much work is required to stop the car in 20 s?
7. A 1500-kg car travels at a constant speed of 30 m/s around a circular track that is 100 m across. What is the kinetic energy of the car?
8. Use the work-energy theorem to find the force required to accelerate an electron (m =9.11 × 10−31 kg) from rest to a speed of 1.50 × 109 m/s in a distance of 1.25 m.
9. A 1000-kg elevator moves upward with constant speed through a vertical distance of 50m. How much work was done by the tension in the cable?
10. A 15-kg block is lifted vertically 10 meters from the surface of the earth. To one significant figure, what is the change in the gravitational potential energy of the block?
11. An engineer is asked to design a playground slide such that the speed a child reaches at the bottom does not exceed 4.0 m/s. Determine the maximum height that the slide can be.
12. A warehouse worker uses a forklift to lift a crate of pickles on a platform to a height 5 m above the floor. The combined mass of the platform and the crate is 100 kg. If the power expended by the forklift is 2000 W, how long does it take to lift the crate?
Answers: 1) (45 m/s) 2) 204 J, 3) 0 J, 4) 2000 J, 5) 470 J, 6) -7000 J, 7) 6.75x105 J, 8) 8.20x10-13 N, 9) 4.9x105 J, 10) 1000 J, 11) 0.816 m, 12) 2.45 s
1. When a body falls freely towards the earth, then its total energy
(c) remains constant
(d) first increases and then decreases
2. A car is accelerated on a leveled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car
(a) does not change
(b) becomes twice to that of initial
(c) becomes 4 times that of initial
(d) becomes 16 times that of initial
3. In case of negative work the angle between the force and displacement in degree is
(a) 0 (b) 45 (c) 90 (d ) 180
4. An iron sphere of mass 10 kg has the same diameter as an aluminum sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
(c) potential energy
(d) kinetic energy
5. A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g =10 m s–2)
(a) 6 ×103 J
(b) 6 J
(c) 0.6 J
6. Which one of the following is not the unit of energy?
(b) newton metre
(d) kilowatt hour
7. The work done on an object does not depend upon the
(b) force applied
(c) angle between force and displacement
(d) initial velocity of the object
8. Water stored in a dam possesses
(a) No energy
(b) Electrical energy
(c) kinetic energy
(d) Potential energy
9. A body is falling from a height h. After it has fallen a height h/2 , it will possess
(a) only potential energy
(b) only kinetic energy
(c) half potential and half kinetic energy
(d) more kinetic and less potential energy
10. How are Joule (J) and ergs (erg) related?
(a) 1J = 107 erg (b) 1erg = 10 -7J (c) 1J = 10-7 erg (d) None
1. (c) 2. (a) 3. (d) 4. (a) 5. (d) 6. (c) 7. (d) 8. (d) 9. (c) 10.(b)
10. A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?
Ans: Initial velocity = u, then v = 3 u
Initial kinetic energy = 1/2 m u2
Final kinetic energy (K.E.) =1/2 m v′2 =1/2m (3u)2 =9 x (1/2 m u2 )
(K.E) initial : (K.E) final=1:9
11. Avinash can run with a speed of 8 m s–1 against the frictional force of 10 N, and Kapil can move with a speed of 3 m s–1 against the frictional force of 25 N. Who is more powerful and why?
Ans: Power of Avinash PA = FA x vA = 10 × 8 = 80 W
The power of Kapil Pk = Fk x vk = 25 × 3 = 75 W
So, Avinash is more powerful than Kapil.
12. A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km he forgot the correct path at a round about of radius 100 m. However, he moves on the circular path for one and half
cycle and then he moves forward upto 2.0 km. Calculate the work done by him.
Ans: F = 5 N
W = F.S = 5 × [1500 + 200 + 2000] = 18500 J.
13. Can any object have mechanical energy even if its momentum is zero? Explain.
Ans: Yes, mechanical energy comprises both potential energy and kinetic energy. Momentum is zero which means velocity is zero. Hence, there is no kinetic energy but the object may possess potential energy.
14. Can any object have momentum even if its mechanical energy is zero? Explain.
Ans: No. Since mechanical energy is zero, there is no potential energy and no kinetic energy. Kinetic energy being zero, velocity is zero. Hence, there will be no momentum.
15. The power of a motor pump is 2 kW. How much water per minute the pump can raise to a height of 10 m? (Given g = 10 m s–2)
p=w/Dt=mgh/Dt = [(m x10 x10)/60]
m = 1200kg
Q16. If you push vigorously against a brick wall, how much work do you do on the wall? Explain.
Ans:There is no work done on the wall as there is no displacement of the wall.
Work = W = F S CosQ
F - Force applied S - Displacement Q - Angle between F and S
Work will be zero if F or S = 0 or angle Q = 90
In the current situation the wall is not moving and S is therefore zero. It follows that Work is zero, that is no work is done.
Q 17. 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds at 20 m/s. Find the change in momentum of the ball as a result of the collision with the side walk (in kg m/s).
Ans: Mass of the rubber ball = 0.2 kg = m
Initial velocity = 30 m/s = v1
Final velocity after collision = -20m/s = v2
Change in momentum = mv1 - mv2= m(v1 - v2)
= 0.2[30 - (-20)] = 10 kg m /s
Q.18. A 2.0 kg block is thrown upward from a point 20m above Earth's surface. At what height above Earth's surface will the gravitational potential energy of the Earth block system have increased by 500J?
Ans. Mass of the block m = 2.0 kg
P.E. of the block at a height of 20 m is = mgh = 2 x 9,8 x 20 = 392 J
Increase in P.E. = 500 J
Let the new height be h
New P.E. = mgh
mgh -392 = 500
mgh = 500 + 392 = 892 J
therefore h = 892/2x9.8 = 45.51m
Q.19. A person pushes a 10 kg cart a distance of 20 meters by exerting a 60 Newton horizontal force. The frictional resistance force is 50 Newtons. How much work is done by each force acting ont he cart? How much kinetic energy does the cart have at the end of the 20 meters if it started from rest:
Mass of cart = 10 kg Distance moved = 20 m
Horizontal force = 60 N Frictional force = 60 N
Frictional resistance force = 50 N Net force = 60 - 50 = 10 N
Work done by the horizontal force = 60 x 20 = 1200 J
Work done by the frictional force = - 50 x 20 = -1000 J
(It is negative because force and displacement are opposite to each other)
Net work done = 1200 - 1000 = 200 J = Change in the K.E. of the cart.
If the cart started from rest, the initial K.E. = 0
Final K.E. - Initial K.E. = 200 J
So Final K.E. = 200J
Q 20. Assuming an efficiency of 25% for the muscle system in the process of converting food energy into mechanical work, how much energy would be used by a person of mass 75 kg (weight 165lb) in the process of climbing four flights of stairs for a total height of 15 metes? Find the answer in joules and convert to dietary calories (1 dietary calorie = 4186 Joules)
Efficiency ? = 25%
Mass of the person = 75 kg
Height climbed = h = 15 m
Work done = gain in P.E. = mgh = 75 x 9.8 x 15 = 11025 J
Since the efficiency of the muscle system
= 25%, the food energy required to do this work
= (100 x 11025)/25 = 44100 J
4186 Joules = 1 dietary calorie
44100 Joules = 44100 / 4186 = 10.5 dietary calories.
The potential energy of a 0.2 kg particle moving along the x axis is given by:
U(x) = 8x2 - 2x4 where U is in Joules and x in meters.
When the particle is at x = 10 m, what will be its acceleration?
m = 0.2 kg
P.E. U(x) = 8x2 - 2 x4
here U is in Joules and x in meters
Intensity T = -du/dx
=- -d[8x2 - 2 x4]/dx = -(16x -8x)
The intensity is equal to the force per unit mass which is the same as acceleration
a = -(16x - 8x)|x = 10 m = -(160 -80) = -80 m/s2
AREAS RELATED TO CIRCLES CLASS X
1. The radius of the circle is 3 m. What is the circumference of another circle, whose area is 49 times that of the first?
2. Two circles touch externally. The sum of their areas is 130 p sq. cm and the distance between their centres is 14 cm. Find the radii of the circles.
3. A wire when bent in the form of an equilateral triangle encloses an area of 121 √3 cm2 . If the same wire is bent in the form of a circle, find the area of the circle.
4. The area enclosed between the two concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, calculate the radius of the inner circle.
5. A wheel of diameter 42 cm, makes 240 revolutions per minute. Find :(i) the total distance covered by the wheel in one minute. (ii) the speed of the wheel in km/hr.
6. An arc of length 20pcm subtends an angle of 144° at the centre of the circle. Find radius of circle.
7. The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. Find the area of the sector.
8. In the given figure, the length of the minor arc is 7/24 of the circumference of the circle. Find : (i) <AOB(ii) If it is given that the circumference of the circle is 132 cm, find the length of the minor arc AB and the radius of the circle.
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