Wednesday, February 29, 2012

chemistry adda: IX Structure of atoms CBSE EXAM Some important que...

chemistry adda: IX Structure of atoms CBSE EXAM Some important que...: Q. A naturally occurring sample of :- (i)69.2% of 53Cu& 30.8% of 65Cu. Find the average atomic mass of a naturally occuring sample of coppe...Read full post

Friday, February 24, 2012

9th Physics Solved Numerical Floating bodies ( Gravitation) Term-2

Q. a cube of mass 1kg with each side of 1cm is lying on the table. find the pressure exerted by the block on the table. take g=10 m/s2

Ans: Pressure is given as force/area 

so, Force, F = mg = 1000 X 1000 gm.cm/s2

and   area, A = 1x1 cm2 = 1 cm2


Thus, the pressure exerted would be

P = (1000 X 1000) / 1 or P = 1 x 10 pa

Q. The mass of a solid iron cube of side 3cm is to be determined usig a spring balance. If the of iron is approximately 8.5 g/cm3, the best suited spring balance for determining weight of the solid would be of 
1. range 0-250gwt ; least count 1gwt             
2. range 0-250gwt ; least count 5gwt
3. range 0-1000gwt ; least count 5gwt           
4.  range 0-1000gwt ; least count 10gwt

Ans: Edge=3 cm ,  Density=8.5 g/cm3


Mass=  density x volume = 8.5 x(3x3)=229.5gwt

Therefore second spring balance of range 0-250 gwt with least count 5gwt will be suitable.

Q. The density of turpentine oil is 840 Kg/ m3. What will be its relative density. (Density of water at 4 degree C is 10 cube kg minus cube)


Ans:Relative Density = Density of Substance/ Density of water at 4 0c

Density of turpentine oil = 840 kg/ m3 ( given).

Density of water at 4 0c = 1000 kg/ m3Relative density of turpentine oil = Density of turpentine oil / 

Density of water at 4 0c  = (840 / 1000 ) kg m-3/ kg m-3     = 0.84

Since, the relative density of the turpentine oil is less than 1, therefore it will float in water.

Q. A solid body of mass 150 g and volume 250cm3 is put in water . will following substance float or sink if the density of water is 1 gm-3?

Ans: The substance will float if its density is less than water and will sink if its greater.

so, density of solid body is   d = mass/volume 

or d = 150/250 = 0.6 gm/cm3     which is less than the density of water (1 gm/3).

So, the solid body will float on water.

Q.  A body weighs 50 N in air and when immersed in water it weighs only 40 N. Find its relative density.


Ans: the relative density would be ratio of the density of the body with respect to air and the density of the body with respect to water.

so,  F1 = 50 N   F2 = 40 Nso,  F1/F2 = 50/40  or relative masses m1/m2 = 5/4 anddensity = mass/volume and as volume remains constant, Relative density =  d1/d2 = 5/4

Q.  A ball of relative density 0.8 falls into water from a height of 2m.  find the depth to which the ball will sink ?


Ans: Speed of the ball

V  = Ö2gh    = Ö 2x10x2  = 6.32 m/s

Buoyancy force by water try to stop the ball.

Buoyancy force = weight of displaced water = dx Vxg

where d = density of water V = volume of the ball  ,  g = 10 m/s2deceleration of the body by

buoyancy force, a = (dVg)/ m

where m= d'V            d' = density of block
 
a = dVg/(d' V) = dg/d' =(d/d')*g =g/(0.8)= 10/0.8

(Given, d'/d = 0.8)= 12.5 m/s2

Net deceleration of ball,a' = a-g = 2.5 m/s2Final speed of ball v' = 0  

Use v' = v2 + 2a's      s= depth of ball in the water

=> 40 = 0 + 2x2.5xs    => s = 8m

Q. Equal masses of water and a liquid of relative density 2 are mixed together. Then, the mixture has a relative density of (in g/cm3)  a)2/3 b)4/3 c)3/2 d)3

Ans: The masses of two liquids are equal, let it be m.

Let the relative densities of water and liquid be ρ1 and ρ2 respectively.


The volume of the two be V1 and V2, of water and liquid respectively.


The volume of the mixture would be, V = V1 + V2 (1)


also, volume = mass/density


thus,


2m/ρ (V) = m/ρ1 (V 1 ) + m/ρ2 (V 2 )


here ρ1 = 1,ρ2 = 2 and ρ is the relative density of the mixture.


now, 2/ρ = 1/ρ1 + 1/ρ2


by substituting the values, we ρ/2 = 2/3


or, the relative density of the combined liquid will be, ρ=4/3


More Questions are solved Here:  http://jsuniltutorial.weebly.com/

Sure shoot MCQ for class 9 science term-2 ClicK Here
IX Thrust and Pressure, Archimedes’ Principle, Relative Density
CBSE Class 9 - Science - Chapter 10: Flotation: Notes and Quest
MCQ: Flotation: Thrust, Pressure, Buoyancy and Density
Thrust and Pressure, Archimedes’ Principle, Relative Density key point
Notes : Flotation: Thrust, Pressure, Buoyancy and Density
Physics Flotation Term-II Class IX  Buoyant force Detail Study
9th Physics Solved Numerical Floating bodies

Tuesday, February 21, 2012

CBSE PHYSICS:Work, Energy and Power IX Physics assignment

CBSE PHYSICS: CBSE Work, Energy and Power IX Physics assignment
1. Under what conditions work is said to be done?
2. Derive the formula for work done by a constant force
3. Give few examples where energy is possessed by a body due to its change in shape.
4. State and prove the law of conservation of energy.
5. Is it possible that force is acting on a body but still work done is zero? Explain.
6. A rocket of mass 3x106 kg takes off from a launching pad and acquires a vertical velocity of 1km/s at an altitude of 25 km. calculate (a) the potential energy and (b) the kinetic energy. (g = 9.8m/s2)
7. If a man lifts a load up with the help of a rope such that it raises the load of mass 50kg to a height of 20m in 100 sec. Find the power of man
8. A ball is dropped from a height of 5m. Find the velocity of the ball just before it reaches the ground. Do you require the value of mass to find the velocity?
9. Two persons A and B do same amount of work. The person A does that work in t 1 sec and the person b in t2sec. Find the ratio of power delivered by them.
10. Why do our hands become warm when rubbed against each other? Explain.
11. The kinetic energy of a body of mass 15 kg is 30J. What is its momentum?
12. Give an example for each of the following energy conversion: (1) electrical energy to kinetic energy. (2) Chemical energy to electrical energy (3) sound energy to electrical energy
13. Two bodies have same momentum. Which will have greater kinetic energy- heavier body or lighter body?
14. An electric bulb of 60w is used for 6h per day .Calculate the units of energy consumed in one day by the bulb.
15. A boy of mass 50kg runs up to a stair case of 45 steps in 9s. If the height of a step is 15cm, find his power. (g= 10m/s2)
16. Two particles of masses 1g and 2g have equal momentum. Find the ratio between their kinetic energies?
17. What will be the work done by the string, when a stone is tied to a string and whirled in a circle?
18. A locomotive exerts a force of 7500N and pulls a train through 1.5 km. How much work is done by locomotive?
19. What work a boy of mass 50kg will do in order to increase running speed from 9km/h to 18km/h.
20. The speed of a moving body is halved. What is the change in its K.E.?
21. State the energy changes taking place in the following cases: (1) A car moves up a hilly road.(2) a stone projected vertically upward returns
22. When we cut a log of wood with a saw it becomes warm, why?
23. If an electric iron of 1200W is used for 30 minutes everyday, find electric energy consumed in the month of April.
CBSE Chapter: Structure of an Atom IX Chemistry assignment
Q.1. What are Canal rays?
Q.2. On the basis of Thomson's model of an atom explain how the atom is neutral as a whole.
Q.3 Draw a sketch of Bohr's model of an atom with three shells.
Q.4. Helium atom has atomic mass of 4u and has two protons in the nucleus. How many neutrons does it have?
Q.5. If the K and L shells of an atom are full, then what would be the number of electrons in the atom?
Q.6. If the number of electrons in an atom is 8 and the number of protons is also 8, then;
(i) What would be the atomic number of the atom? (ii) What is the charge on the atom?
Q.7. Na+ ion has completely filled k and L shells. Explain.
Q.8. The average atomic mass of a sample of element X is 16.2 u. What are the percentages of isotopes 16 X8 and 18 X8 in the sample?
Q.9. If Z=3, what would be the valency of the element ? Also, name the element.
Q. 10. Composition of the nuclei of two atomic species X and Y are given as under
X Y
Protons = 6 6
Neutrons = 6 8
Give the mass numbers of X and Y. What is the relation between the two species ?
Q.11. Why do the elements helium, neon and argon have zero valency ?
Q.12. Why do elements which exist as isotopes have fractional atomic masses ?
Q.13. Are mass number and atomic mass of an element equal in all respects?
Q.14. The element ALuminium is written by the symbol 27 13 Al. Write the number of protons, electrons and neutrons present in it.
Q.15. The electronic configuration of an element Z is 2, 8, 6. How many electrons does it require to have a stable configuration?
Q.16. Two atoms A and B have the following composition:
Atom A Atom B
17 protons 17 protons
18 neutrons 20 neutrons
What are their mass numbers ? What is the relation between the species ?Q.17. The composition of two atomic particles is given :
X Y
Protons 8 8
Neutrons : 8 9
Electrons : 8 8
(i) What is the mass number of X ? (ii) What is the mass number of Y ?
Q.18. Briefly describe the features of the Rutherford Model of an atom. what are the drawbacks ?
Q.19. How do isotopes and isobars differ? Write three applications of isotopes.
Q. 20. What observations in scattering experiment led Rutherford to make the following conclusions:
(i) Most of the space in an atom is empty. (ii) Whole mass of an atom is present in its centre.
(iii) Nucleus is positivity charged.
Q.21. Explain why did Rutherford select a gold foil innhis alpha-ray scatering experiment.
Q.22. The atom of an element 'A' has three electrons in the outermost shell. It loses one of these to the atom of another element 'B'. What will be the nature and value of charge on the ion which results from 'A'
Q.23.The atomic numbers of atoms of two elements are 18 and 20 respectively and their mass numbers are 40. What is the name that can be given to such pairs of atoms. Will they have same chemical characteristics?
Q. 24. Give reasons for the following:
(a) Isotopes of an element are chemically similar.
(b) An atom is electrically neutral
(c) Noble gases show least reactivity
(d) Nucleus of an atom is heavy and positively charged.
(e) Ions are more stable than atoms.
Q.25. which of the two will be chemically more reactive; element X with atomic number 17 or element with atomic number 16 ?
Q.26. an unknown species X has 17 protons and 18 electrons. Predict its nature.

Thursday, February 16, 2012

Solved Questions IX Chapter- Sound

Question:  A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 ms–1.

 Ans. (I) Wavelength of sound waves corresponding to 20 Hz, u = u = 344 ms–1, v = 20 Hz
λ1  =  v/n = 334/20  = 17.2 m  λ2 =v/n = 334/20,000  = 0.0172 m

Question: The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Ans. Frequency of sound = 100 Hz = 100s–1   Number of vibrations in 1 s = 100
Number of vibrations in 1 minute, i.e. 60 seconds =100x60 =6000vibrations

Question: Explain how bats use ultrasound to reach a prey.
Ans. Bats produce high pitched ultrasonic waves that are not heard by humans. These ultrasonic waves strike the prey like an insect and it is reflected back in the form of an echo, which is heard by the bat. This enables the bat to hover on the insect and catch it.

Question: A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from sub-marine is 3,625 m.
Ans. Given,   d = 3,625 m; t = 5 s ; u = ?  We know that d = (v x t)/2  Þ  v = 2d/t = (2x362)/5 = 1450 ms–1

Question: A sound wave travels at a speed of 339 ms–1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Ans. Here, u = 339 ms–1, l = 1.5 cm =0.015m
As you know that v=n λ Þ n=v/l =339 ms–1 / 0.015m = 22600 Hz
The sound will not be audible to the human ear as it has frequency higher than 20,000 Hz, which the upper limit of the human audible frequency range

Question: A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top ? Given, g = 10 ms–2 and speed of sound = 340 ms–1.
Ans.   Here, s = 500 m, u = 0 ms–1, g = 10 ms–2   As we know, s = u t  + 1/2gt2 you find t = 10sec
So, the stone reaches the pond in 10 s
Now, given that, Speed of sound = 340 ms–1 by              
Time taken sound to cover 500m = 500/340 = 1.5sec.   
Total time taken = (10 + 1.5) s = 11.5 s
:. The splash will be heard at the top after 11.5 s

Question: Which type of waves is produced when a stone is dropped on the surface of water in a pond ?
Ans. Transverse waves.

Question: What kind of waves are produced when a spring is pulled in the downward direction and released?
Ans. Longitudinal waves.

Question: What is the nature of electromagnetic waves?
Ans. They are transverse in nature.

Question: Give three important characteristics of wave motion.
Ans. 

(1) Wave motion is the disturbance that travels forward through the medium and not the particles of the medium. The particles of the medium just vibrate about their mean positions.
(2) Each particle receives vibrations a litter later than its preceding particle.
(3) The velocity with which wave travels is not the same as the velocity of the particles with which they vibrate about their mean positions.

Question: Differentiate between a periodic wave and a pulse in a tabular form.
Ans. Difference between a periodic wave and a pulse:
Periodic Wave
1. A periodic wave is a continuous wave having a long duration.
2. A periodic wave is a continuous disturbance in the medium.
3. It is formed due to the displacement of every point or particle in a given part of the medium.
Pulse
1. Pulse is a wave having a short duration.
2. A pulse is a single disturbance in the medium.
3. It is formed due to displacement of only a part of the medium from its mean position.

Question:  On what factors does the speed of sound in a gas depends?
Ans. The speed of sound in a gas depends on the following factors:
1. Density. The speed of sound in a gas is inversely proportional to the square root of its density at constant pressure i.e., Speed of sound
µ 1/Ö Density of gas
2. Temperature. The speed of sound in a gas is inversely proportional to the square root of its absolute temperature, i.e., Speed of sound µ 1/Ö absolute temperature of gas
It means that speed of sound increases with the increase in temperature of the gas.
3. Humidity. The speed of sound is directly proportional to humidity i.e., sound travels faster in moist air than in dry air.   Speed of sound µ Humidity
4. Wind. The speed of sound increases when the wind blows in the direction of sound and decreases when the wind blows in the opposite direction

Question:  A stone is dropped into a well 44.1 m deep, and sound of the splash is heard after 3.13 s. Calculate the velocity of the sound in air. (Given, g = 9.8 ms–2)
Ans:  u = 0, s = 44.1 m; a = 9.8 ms–2    Find Time taken by stone to reach 44.1m deep?
As we know, s = u t + 1/2gt2   putting value we get t  = 3sec
The sound takes the remaining time i.e., 3.13 – 3 = 0.13 s to go to up of well through 44.1 m
Speed=d/t= 44.1 m/0.13 = = 339.23 ms–1

Question:  A ship sends out ultrasound that returns from the seabed and is detected after 3.42 s. If the speed of ultrasound through seawater is 1531 m/s, what is the distance of the seabed from the ship ?
Ans. Here, Time between transmission and detection, t = 3.42 sec; u = 1531 m/s
Distance travelled by the ultrasound = 2 × depth of the sea = 2d, (where d is the depth of the sea)
2d = 1531 m/s × 3.42 s = 5236 m       Þ    d = 5236 m/2 = 2618 m or 2.62 km

Question:  A stone is dropped from the top of a tower 500 m light into a pond of water at the base of the tower, when is the splash heard at the top? (Given g = 10ms-2 and speed of sound = 340ms-1).
s = 500m, u = 0m/s, a = g = 10m/s2, t = ?   s = ut + 1/ you find t = 10s
Time taken by sound to cover 500m = 500/340 = 1.47s
Total time taken = 10s+1.47s = 11.47s

Question:  A submarine enters a sonar pulse, which returns from an under water cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff.
t = 1.02s, v = 1531m/s  Now d=s x t
Distance travelled by sonar pulse = 2d = 2x1531x1.02 = 1561.62m
Distance of the cliff = d/2= 1561.62/2 = 780.81m

9th ASSIGNMENT FOR THE SESSION 2011-2012 Science and technology

CBSE PHYSICS: ASSIGNMENT FOR THE SESSION 2011-2012 Class: IX Sub...
JSUNIL TUTORIAL: Section A : Physics
Q.1. Define work energy and power. Give the proper SI units. Hence derive the expression for kinetic and potential energy.
Q.2. State and prove the principle of conservation of energy. Also prove work energy theorem.
Q.3. A boy is moving on a road against a frictional force of 5N. After travelling a distance of 1.5km,he forgot the correct path at a roundabout (figure) of radius 100m. However he moves on a circular path for one and half cycle and then he moves forward upto 2 km. Calculate the workdone by him.
Q.4. Weight of a wooden piece is 6 kg.While floating in water its 1/3 rd volume remains submerged in
water. Calculate the maximum weight that may be placed on it so that it may float completely submerged.
Q.5. When a boat floats in water its one fourth volume remains submerged. The maximum weight that can be placed on boat is 1200 kgf. Calculate the weight of empty boat.
Q.6. Explain the working and application of a SONAR. How defects in ametal block can be dected using ultrasound.
Q.7. A stone is dropped from the top of a tower 500m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g=10m/s2 and speed of sound =340m/s.
Section B : Chemistry
1. Calculate the mass of : a) 0.5 mole of silver b) 0.5 mole of sugar (C12H22O11). (Given atomic
masses are Ag = 108, C = 12, O = 16, H = 1)
2. Calculate the a) the number of atoms in one gram of gold b) the number of molecules in a drop of water weighing 0.05g.(atomic mass of gold = 197u) 1 x 1022 atoms of an element X are found to have a mass of 930mg. Calculate the molar mass of the element X.
3. What is the difference between the atomic mass and gram atomic mass of an element?
4. What are the limitations of Dalton’s atomic model?
5. What are the draw backs of Rutherford’s atomic model?
6. What are the salient features of Bohr’s atomic model?
7. What are the postulates of Dalton’s modified atomic theory?
8. Write the electronic configuration of the isotopes of chlorine and isobars of Ar, Ca.
9. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes of oxygen 16 and 18 in the sample?
10. A Bromine atom is available in the form of two isotopes. 79Br (49.7%) and 81Br (50.3%). Calculate the average atomic mass of Bromine atom. The atomic no. of Bromine is 35.
11. No. of protons, neutrons and electrons in four spaces A, B, X and Y are respectively 6, 6, 6; 7, 7,7; 6, 8, 6; and 9, 10, 10. Give symbolic representation of each species and tell which of them are isotopes and which are isobars.
12. What do you mean by radio isotopes? Give two applications of isotopes.
13. Differentiate between isotopes and isobars. Write the application of the following isotopes:
a) U-235, Co-60, P-32 and I-131.
14. Atomic no. of Na is 11 and mass no. is 23. What type of ion will be formed by it and how will you represent it? Calculate the no. of electrons, neutrons and protons in the ion formed.
15. An ion X-2 contains ten electrons and 8 neutrons. What are the atomic no. and mass no. of element X? Name the element.
Section C : Biology
CHAPTERS - BIOLOGICAL DIVERSITY, HEALTH & DISEASES, OUR ENVIRONMENT
(BIOGEOCHEMICAL CYCLES, GREENHOUSE EFFECT)
1. What is evolution? How is algae different from fungi? Draw diagram of any one member of Protista.
2. How are pteridophytes different from bryophytes? Write any four differences between angiosperms and gymnosperms.
3. Write the differences (any four) between the following groups a) poriferans and coelenterates b) molluscs and arthropods c) flatworms and roundworms d) aves and mammals
4. Explain the different means by which infectious diseases spread in the environment? List the preventive measures of such diseases.
5. Justify, why it is difficult to make antiviral medicines than antibacterial medicines.
6. Explain the different processes involved in oxygen and water cycle operating in nature.
7. Represent schematically the carbon and nitrogen cycle operating in the biosphere. How is carbondioxidefixed in environment?

Monday, February 13, 2012

class VI Question bank and test paper maths By JSUNIL TUTORIAL

6th Area and Perimeter -1
1. Find perimeter of rectangle whose length =6m8dm and breadth =4m6dm

2. Find the cost of fencing rectangle field 24n long and 18m wide at Rs. 6.25 per meter.

3. The length and breadth of rectangle field are in ratio 5:3. If its perimeter is 64m. Find the dimensions.

4. The cost of fencing a rectangle field at Rs. 14.60 per m is 1606. If the width of the field is 23m find its length.

5. The length and breadth of rectangle field are in ratio 3:2. The cost of fencing the field at Rs. 6.50 per m is Rs. 520. Find the dimensions of the field.

6. The cost of putting fence around a square field at Rs. 15 per m is Rs. 432. Find the length of each side of the field.

7. Each side of square is of 21m. Adjacent to this field, there is a rectangular field having its side in ratio4:3. If perimeters of both fields are equal find the dimensions of the rectangular field.

8.Find the circumference of circle if r=14cm

9. Find the diameter of circle if circumference is 66cm

10. Find the diameter of circle if circumference is 75 3/7 cm

11. The diameter of wheel of a car is 70cm. Find the distance travel by wheel of car in 1000 revolutions.

12. Find the circumference of circle if diameter of circle is 10m.

13. The diameter of wheel of a car is 70cm. How many revolutions will it take to make to travel 1.65 km.

14. The diameter of wheel of a car is 70cm. Find the distance travel by wheel of car in 50 revolutions.

15. Find the diameter of circle if circumference is 88 cm.

16. The diameter of wheel of a car is 77cm. How many revolutions will it take to make to travel 1210m km.

17. Find the radius of circle if circumference is 66cm

18. Find the area of a rectangle whose length and breadth 45cm and 16cm respectively. Also find perimeter of rectangle

19. A room is 5m 40cm long and 3m 75cm wide. Find the area of the carpet needed to cover the floor.

20. Find the cost of cultivating a rectangular field 34m long and 18m wide at Rs. 4.50 per sq metre. Also find the cost of fencing the field at Rs. 2.25 per metre.

6th Area and Perimeter -2

21. A room is 9.68m long and 6.3m wide. Its floor is to be covered with rectangular tiles of size 22cm by 10cm. Find the total cost of the tiles at Rs. 5 per tiles.

22. The area of rectangle is 650 sq cm and one of its side is 13cm. Find the perimeter of rectangle.

23. The total cost of flooring a room at Rs. 8.50 per sq metre is Rs. 510. If the length of the room is 8m. Find its breadth.

24. Find perimeter of square field whose side is 6.3m

25. The top of the table measure 2m 25 cm by 1m 20 cm .Find its area in square metre.

26. How many envelop can be made from a sheet of paper 324cm by 172cm . if envelop require a piece of paper of size 18cm by 12cm.

27. A room is 12.5m by 8m .A square carpet of side 8m is laid on the floor .Find the area of carpet that is not carpeted.

28. A garden is 120 m by 96 m . Two cross road each 5 m wide are laid in the middle parallel to length and breadth. find the area of cross road .

29. Find the area of path running outside if the dimension of field is 15m by 10m and breadth of path is 0.5 m .

30 Find the area of path running inside if the square field of 10 m and width of path is 2 m . Find cost of leveling at Rs 2.50 per sq.m

31. (i) How many sq.cm in a square meter.(ii) What is the difference between 3 square meter and 3 meter square

32. A lane, 150m by 9m is paved with bricks of length 22.5cm and breadth 7.5cm. Find the number of bricks required.

33. A room is 13m long and 9m broad. Find the cost of carpeting the room with a carpet 75cm broad at the rate of Rs. 6.50 per metre.

34. The length and breadth of rectangular park are in ratio 5:3 and its perimeter is 128 m. find the area of the park.

35. Two plots have same perimeter. One is a square of side 64m and other is rectangle of length 70m and breadth 58m which field has greater area and by how much.

36. Find the cost of cultivating a rectangular field at Rs. 1.50 per sq metre is Rs. 1833. If breadth of field is 26 m . Find the cost of fencing at Rs. 7.50 per metre.

37. The length and breadth of a play ground are 62m60cm and 25m40cm respectively. Find the cost of turfing at Rs. 2.50 per square meter .How long man take to go three times around the field if he walks at the rate of 2 metre per second.

6th Ratio and Proportional-1

1. Find the ratio of

(i)40 paisa to Rs 4 (II) 24 min. to an hr (III) 125 ml to 2 litres (IV) 4m to 36 cm (V) Dozen to Score

2. Find the ratio the price of coffee to that of tea, when coffee cost rupees 24 per100 gm and tea Rs 80 per kg?

3. Two numbers are in ratio 2:7. If the sum of the numbers are 81. Find the numbers?

4. Divide the number 642 in ratio 1: 2: 3 among A, B, C

5. The ratio of Length and breadth of rectangle is 5:3. Find l and b if perimeter is 256?

6. Which ratio is larger (I) 5:12 or 17:30 (ii) 2:3 or 16:24

7. A bus travel 126 km in 3hrs.and a train travel 315 km in 5 hrs. Find ratio of their speed?

8. Are (I) 7,42,13,78 in proportional? (ii) Find x if 48, 36, x are in proportional?

9. First, second, fourth term of a proportional are 45, 25, 35, Find third term.

10. If 36 , x , x , 16 are in proportional Find value of x

11. Find value of x if 25,35, x are in proportional

12. An electrical pole cast shadow length 20 m at a time when a tree of 6 cm cast shadow of 8 m. Find the height of pole.

13. The length of the side of a triangle are in ratio 1: 3 : 5 .If perimeter of triangle is 90 cm find its largest side .

14. Rohit earns Rs. 7650 and saves Rs. 918 per month Find ratio of

(I) his income to his saving (II) His income to his expenditure (III) his expenditure to his saving

15. Find the ratio of the price of pencil to that of ball pen if pencil cost Rs 96 per score and ball pen cost Rs. 50.40 per dozen.

16. The ratio of the height of Mohit and Rohit is 4: 3 if Rohit is 1.5 m tall. Find the height of Mohit?

17. Neha got 850 marks out of 1000 marks and Suzie got 375 in 500. Express their marks in ratio?

18. The ratio of income to expenditure 9: 8 Find saving, if total income is Rs. 2700

19. Jaya is 1.30m tall and Shakshi is 1m25 cm tall. Find the ratio of their height

20. First ,third, fourth term of a proportional are 45,25,35, Find third term.

6th Ratio and Proportional – 2

21. If 36 , x , x , 16 are in proportional Find value of x

22. Find value of x if 25, 35, x are in proportional

23. An electrical pole cast shadow length 20 m at a time when a tree of 6 cm cast shadow of 8 m. Find the height of pole.

24. The length of the sides of a triangle are in ratio 1: 3: 5 .If perimeter of triangle is 90 cm find its largest side

25. Rohit earns Rs. 7650 and saves Rs. 918 per month Find ratio of

(I) his income to his saving (II) His income to his expenditure (III) his expenditure to his saving

26. Find the ratio of the price of pencil to that of ball pen if pencil cost Rs 96 per score and ball pen cost Rs. 50.40 per dozen

27. (i)Find the ratio of 75 paisa and Rs 3 (i)Find the ratio of 125 gm to 2 kg. ?

28. Neha got 850 marks out of 1000 marks and Sazia got 375 in 500.Express their marks in ratio?

29. The number of boys in school is 500 and that of girls 275. Find the Ratio of

(i) Boys to that of girls (ii) Boys to total num of students (iii) Total number to that of girls.

30. The ratio of the height of Mohit and Rohit is 4 : 3 . if Rihit is 1.5 m tall . Find the height of Mohit 31. Jaya is 1.30m tall and Shakshi is 1m25 cm tall . Find the ratio of their height ?

32 Are 5 : 6 : : 10 : 12 and 12 : 18 : : 14 : 24 equivalent ?

33. Find the value of x : (i) X/5 = 6/15 (ii) 7/8 = 21 /X (iii) 44 : X : : 11 : 4 (iv) 7 : X = 21 : X

34. The ratio of income to expenditure 9 : 8 Find saving, if total income is Rs. 2700

35. Two numb. are in ratio 2 : 7 . If the sum of the num is 81 .Find the numbers?

36.The ratio of boys to girls in a school is 8 : 3 if the num. of boys are 400 find the number of girls.

38. Is 3 : 5 equivalent to 4: 6 ? Why or why not ?

39. The 1st 2nd and 4th term of a proportion al is 6 ,10 ,15 . Find the third term ?

40. First ,third, fourth term of a proportional are 45,25,35, Find third term.

41. If 36 , x , x , 16 are in proportional Find value of x

42. Find value of x if 25,35, x are in proportional

43. An electrical pole cast shadow length 20 m at a time when a tree of 6 cm cast shadow of 8 m. Find the height of pole.

44. The length of the side of a triangle are in ratio 1 : 3 : 5 .If perimeter of triangle is 90 cm find its largest side.

45. Rohit earns Rs. 7650 and saves Rs. 918 per month Find ratio of

(I) his income to his saving (II) His income to his expenditure (III) his expenditure to his saving

46. Find the ratio of the price of pencil to that of ball pen if pencil cost Rs 96 per score and ball pen cost Rs. 50.40 per dozen.

6 th Unitary methods

1. A truck runs 495 km on 36 liter petrol. How many km can it run on 35 liter petrol?

2.12 men can reap a field in 25 days. in how many days can 20 man reap field ?

3. In a camp, there were provisions for 425 men for 30 days. However, 375 men attend the camp. How many provision last?

4. If 48 boxes are needed for 6000 pens. How many boxes will be needed for 1875 pens?

5. A bus travel 36 km in 2/3hr. If it maintains a uniform speed, how long will it takes to cover162 km? How for does it travel in 8 hrs. ?

6.25 bags each weighting 40 kg, cost Rs 2750. Find the cost of 35 bags of wheat if each weighting 50 kg?

7. 24 workers can build a wall in 15 days. How many days will 9 workers take to build a similar wall

Algebraic equations

1. Write each in product form:- (a) 4a3 (b) 10x3y3z2 (c) 7p2q2

2. Write in exponential form:-

(a) 17 X b X b X c X c X c (b) 13 X p X q X q X q X r X r c) 7 X a X a X a X – - – - – - – - 17 times

3. Add: – (a) 17×2 + 2y2 + 3xy, -9×2 + 11y2 – 3xy and 12×2 – 7y2 + 6xy

(b) –a2 + b2 – 2abc, -3a2 – 2b2 + 4abc and 4a2 + 3b2 – 4abc

4. Subtract

(a) 6×2 – 6y2 + 7xy from 0 (b) 7x – 3y from 7x + 3y (c) -6×2 + 7y2 + 2xy from 6×2 – 7y2 – 3xy

5. Identify the like terms in x2y3z; xyz; -y3zx2; -2xyz

6. If A=-2l+6m-3p; B=l-m+2p then find 2B-3A

7. Subtract 7x+2y-7z from -1

8. From the sum of 4x-3y+z ; -5x subtract 3y-4z

9. Subtract the sum of 3l +2m – p; 2l – 4m + p ; – 6l + m from 0

10. From the sum of -a+b-3c; 2a+b-c; 1a-4b+5c; 2a-6b+c subtract –a-b-c

11. Write the following statement algebraically:

(a)Product of 4 and a divided by the difference of 5 and a

(b)Two – third of x subtracted from 5 less than x

(c) 15 less the quotient of x by 3

12. Simplify the expressions (i) 3y (2y – 7) – 3 (y – 4) -63 (ii) 5y (3y 2 + 5y- 11) + 3(y3 – 4y2 + 6)

13. Add the following. (i) ab – bc, bc – ca, ca – ab (ii) a – b + ab , b – c + bc , c – a + ac
Linear Equation

1. The present age of Sahil’s mother is three times the present age of Sahil. .After 5 years their ages will add to 66 years. Find their present ages.

2. Bansi has 3 times as many two-rupee coins as he has five-rupee coins. If he has in all a sum of Rs 77, how many coins of each denomination does he have?

3. The difference between two whole numbers is 66. The ratio of the two numbers is 2 : 5. What are the two numbers?

4. Aman’s age is three times his son’s age. Ten years ago he was five times his son’sage. Find their present ages.

Understanding Elementary Shapes

(I) Fill in the blanks :

  1. An angle whose measure is greater than that of a right angle is _______________________
  2. Three edges meet at a point called a ________________________.
  3. A ___________________________ is larger than a straight angles.
  4. A Polygon with 5 sides is called a ___________________________.
  5. A triangle having all three unequal sides is called a ____________________.


II) Write down the measure of

a) Some acute angles. b) Some obtuse angles. c) What is the measure of a straight angle?

III) What shape is

a) A brick. b) A match box c) A sweet laddu d) A ball e) A die f) A road roller

(IV) Fill in the blanks

1. A cuboid has __________ faces.
 2. Each face has _______ edges.

3. Each face has _________________ vertices.

Saturday, February 11, 2012

CBSE MATH Assignment CBSE IX Mathematics Surface area and...

CBSE MATH STUDY: Assignment CBSE IX Mathematics Surface area and...:
1.The diameter of garden cylindrical roller is 2.8m long . How much area will it cover in 100 revolutions.
2. A rectangular piece of paper 16 cm by 3.5cm is rotated about the longer side. Find the total surface area of the solid so generated.
3. In a hot water heating system, there is a cylindrical pipe of length 35m and radius 4cm find the total radiating surface in the system
4. Find the curved surface area of the right circular cone whose slant height is 10cm and base radius is 7cm.

Thursday, February 9, 2012

sample paper science class 9th sa2 solved



1. Relative density of silver is 10.8. The density of water is 103 Kgm-3. What is the density of silver in SI unit? 
Ans: Relative density of silver = 10.8 Relative density of water = 103 Kgm-3

Picture

Density of silver = Relative density of silver x Relative density of water = 10.8 x 103 Kgm-3

2. Is potential energy a vector or a scalar quantity?
Ans: Potential energy is a scalar quantity.
3. (i) Give any two examples of longitudinal waves.(ii) What is the most essential property of a wave motion and why?
Ans: (a) i. Waves produced in air. ii. When a freely suspended spring is pulled downwards and released, longitudinal waves are produced.
(b) The most essential property of a wave motion is its frequency. Frequency of a wave is its inherent characteristic and does not change by the change in temperature, pressure or change in medium.
4. (a) Calculate the work done in lifting 200 kg of water through a vertical height 6 meter. (Assuming g = 10m/s²) (b) When an object moves on a circular path, what is the work done?
Ans: 20.(a) Given, mass of water (m) = 200 kg      Height (h) = 6 m 1/2
.·.  weight of water (mg) = 200 x 10 N  .·.  work done = mg x h = 200 x 10 x 6J =12000 J
(b) Work done is zero, because displacement is perpendicular to direction of force always.
5.(a) Calculate the power of an engine which can lift 200 Kg of water to store in a tank at a height of 10 m in 4.9 s. Also express in horse power. (Given = 9.8m/s2)  (b) What type of energy is stored in the spring of a watch? (c) What is the work done by the tension in the string of a sample pendulum?
Ans (a) work done (W) = m x g x h = 200 kg x 9.8 m/s2 x 10 m
Power = Work done /Time=[200 kg x 9.8 m/s2 x 10 m]/ 4.9 s= 4000 W÷746 HP=5.36 HP
(b) Elastic potential energy.    (c) Tension acts perpendicular to the displacement of the simple pendulum hence work done is zero.
6. The atomic number of chlorine is 17 and mass number is 35.  a. What would be the electronic configuration of a negatively charged chloride ion, Cl- ?     b. What would be the atomic number and mass number of Cl- ? c. Define valency and calculate the valency of Cl- .
Ans: a. Chlorine atom (Cl) has atomic number of 17. It contains 17 protons and 17 electrons.
Chloride ion (Cl- ) is formed when Cl gains one electron. So, Cl- has 18 electrons and 17 protons.
Therefore, the electronic configuration of Cl- = 2, 8, 8
b. Atomic number of Cl- = Number of protons = 17  Mass number of Cl- will be the same as Cl i.e. 35
c. Valency is defined as the combining capacity of an atom. For a non metallic element, it is equal to eight minus the number of electrons present in the outermost shell.
Here, Cl- has 8 electrons in the outermost shell, therefore valency of Cl- = 8-8 = 0
7. The relative atomic mass of Boron is 10.8 u. Calculate the percentage of its isotopes 10B5and 11B5, occurring in nature.
Ans: Let the percentage of 10B5be y in the sample. The percentage of 11B5will be (100-y)
Average Mass = (10) ×y / 100 + 11×(100-y)/100 = 10.8  Þ y = 20
The percentage of its isotopes 10B5and 11B5, occurring in nature. 20% and 80%
8. Give one example where kinetic energy is transferred from one object to other?
Ans: In the carom, the player provides the kinetic energy to the striker by pushing it with the finger. If the striker collides with other coins, it will slow down dramatically and the coins it collided with will gain speed as the kinetic energy is transferred on to it.
9. Define Avogadro’s constant. Give its value.       
Ans: Avogadro constant is the actual number of particles (atoms, molecules or ions) present in 1 mole of any substance. Its value is 6.022 x 1023
10. (a) Chlorine occurs in nature in two isotopic forms with masses 35 u and 37 u in the ratio of 3:1. Calculate the average atomic mass of chlorine atom on the basis of this data. (b) Give any three uses of three isotopes.
Ans: a) The isotopes of chlorine are in the ratio 3:1. It means that the two isotopes are (3/4x100%)75% and(1/4x100%) 25% respectively.
Average atomic mass of chlorine =[35x75]÷100    + [35x25]÷100    =142/4 =35.5u
b) Three uses of isotopes:  (i) An isotope of cobalt is used in the treatment of cancer. (ii) An isotope of iodine is used in the treatment of goiter. (iii) An isotope of uranium is used as a fuel in unclear reactors.
11. When a person uses deodorant spray, the other person standing at a distance would hear the sound of spraying first and the fragrance of spray would reach him later. Why so?
Ans: The sound of spraying deodorant travels through the vibrations of air layers so it reaches first. But the fragrance of deodorant reaches the other person through actual movement of air particles, therefore takes more time.
12. Calculate the number of molecules of sulphur present in 16 g of solid sulphur.
Ans: Molecular mass of S8 = 8 x 32 = 256 g
The number of molecules of sulphur present in 16 g = [6.022x1023 /256]x16 moles= 3.76 x 1022 molecules
13. Following observations were taken while determining the relative density of a liquid.
Weight of the solid in air = 0.100 kgf                 Weight of the solid in liquid = 0.080 kgf
Weight of the solid in water = 0.075 kgf         Calculate:  (a) the apparent loss in weight of solid in liquid  (b) the apparent loss in weight of solid in water
Ans: (a) The apparent loss in weight of solid in liquid = weight in air - weight in liquid = 0.100 - 0.08 = 0.02 kgf (b) Apparent loss in weight of solid in water = weight in air - weight in water = 0.100 - 0.075= 0.025 kgf
14. Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Ans: Speed of sound through air: 346 m/s Speed of sound through aluminium : 6420 m/s
The ratio of times taken by the sound wave in air and in aluminium=346/L ÷6420/L=346/6420
15.  What is the consequence of two sound waves which arrive at the ear in a time interval shorter than 0.1 s?
Ans: If the time interval between direct and the reflected sound is less than 0.1 s, it mixed  with the direct sounds and the human ear will be unable to distinguish between the two.
16. Two bodies of masses m1 and m2 (m1>m2) have same kinetic energy. They are stopped by applying same retarding force. Which body will stop first?
Ans: If time taken to stop the masses m1 and m2 are t1 and t2 respectively, the retardation produced in each case is, a1=u1/t1      and       a2 = u2/t2
Since F=ma Þ a=F/m Þ t1=u1m1/F and t2 =u2 m2/F  Þ t1/t2 = m1u1/m2u2------------------------(1)
Given that KE1 = KE2 Þ 1/2 m1(u1)2 =1/2 m2(u2)2 Þ m1(u1)2 = m2(u2) Þ  [u1/u2]2 = m2/m1----(ii)
From (i) and (ii), t1/t2 =Öm1/m2
As given ,  m1 > m2  Þt1 > t2 Þ heavier body will take longer to stop than the lighter body.
17. Explain the following: (a) Swimmers are provided with an inflated rubber jacket/tube. Why?
(b) It is easier to swim in sea water than in river water. Why?
Ans:  (a)An inflated rubber jacket has low weight and large volume Hence, it displaces large volume of water. As a result, up thrust due to water increases and the person remains floating and there is no chance of drowning of the swimmer in such case.
(b) It is easier to swim in sea water because density of sea water is more due to its salty nature. Hence, up thrust acting on the swimmer in sea water is more in case of sea water than in fresh water. So, it is easier to swim in sea water.
18. A ship sends out ultrasound produced by transmitter that returns from the sea bed and detected after 3.42 s. If the speed of ultrasound waves through sea water is 1530 m/s, what is the distance of the sea bed from the ship?
Ans: Depth of sea bed d = vt/2 =[1530x3.42]÷2= 2616.3 m
19. (a) What is the amount of work done:  (i) By an electron revolving in a circular orbit of radius r round a nucleus? (ii) By an electron moving with half the speed of light in empty space free of all forces?
Ans: (a) ( i) Here a centripetal force provided by electrostatic force of attraction acts on the electron towards the centre of orbit but motion is along the tangent to the circular orbit at ecah point. As force and displacement are in mutually perpendicular directions at each point, the work done is zero.
(ii) Here, no force of any sort is acting on the electron, so the work done is zero.
20. (a)An electric pump is used to pump water from an underground sump to the overhead tank situated 20 m above. It transfers 2000 kg of water to overhead tank in 15 minutes. Calculate the power of pump.
(b) What do you mean by instantaneous power of a device?
Ans: Power of pump =Total work/Total time=[mgh]/t=[2000x 9.8x 20J]÷ 900t=435.5W
(c) The instantaneous power of a device at a particular instant of time is defined as the rate of doing work by the device at that very instant.
21. If the kinetic energy of the body is increased by 300% then determine the percentage increase in the momentum.
Ans: KE=1/2mv2 Þ2mE=m2v2 Þp=Ö2mE
Now E1= E+300%of E=4E
So, New P= Ö2mx4E= Ö8mE
The percentage increase in the momentum= [(Ö8mE -Ö2mE )/Ö2mE)]100%  =(Ö4-1)100%=100%
22. A stone of mass 2 kg is falling from rest from the top of a steep hill. What will be its kinetic energy after 5 s?
Ans: Here mass of stone m = 2 kg, intial velocity of stone u = 0 and time = 5 s
.·. Velocity of stone after 5 s, v = u + gt = 0 + 9.8 x 5 = 49 m/s
.·. Kinetic energy of stone KE = 1/2mv2 = 1/2 x 2 x( 49)2 = 2401 J
23. (a) Explain the work done by the person in the following conditions. (i) When he is standing at a place holding a suitcase in his hand.(ii) When he is moving holding the suitcase in his hands.
Ans: (i) When the person is standing at a place holding the suitcase, so there is no change in the position of man or suitcase.   So, displacement (s) = 0  W = F x s = F x 0 = 0 .
(ii) When the person is moving holding the suitcase in his hand, he applies force in upward direction and displacement of suitcase is in forward direction that is perpendicular to the direction of force applied. .
So, q= 90o     .·. W = F x s cosq = F x s. cos 90o = 0   Hence work done on the suitcase is Zero.
24. Describe the law of conservation of energy by giving two examples
Ans: According to law of conservation of energy:
“ Energy remains conserve during its transformation from one form to other” or in other words “during transformation of energy, energy is neither created nor destroyed”.
Examples- When we lift a stone to a vertical height h from earth surface, stone gains a potential energy equal to mgh, but we lose some amount of energy.
When a person kicks a ball, it gets some velocity or kinetic energy, the amount of energy gained by ball is equal to amount of energy lost by man.
25.  A compound was found to have the following percentage composition by mass Zn = 22.65%, S = 11.15%, H = 4.88%, O = 61.32%.  The relative molecular mass is 287g/mol. Find the molecular formula of the compound, assuming that all the hydrogen in the compound is present in water of crystallizations.
Solution: Zn : S:O:H = 22.6565:  11.1532 :  61.3210:  4.88/1  = 0.3485: 0.3484 :3.833: 4.88
To obtain an integral ratio, we divide by smallest number here 0.3484
     0.3485/0.3484:   0.3484/0.3484 :  03.833/0.3484 :  4.88/0.3484     = 1 : 1 : 11 : 14
 Empirical formula is Zn SO11H14
Let Molecular formula be (Zn SO11H14)n       RMM for the molecular = n [65 + 32 + (11x16) +14]= 287n
Given, The relative molecular mass is 287g  Þ 287n = 287Þ n = 1 Þ Molecular formula is Zn SO11H14
26. A source of sound produces 20 compressions and 20 rarefactions in 0.2 sec. The distance between a compression and the next (consecutive) rarefaction is 50 cm. find the wavelength, frequency and time period of wave.
Ans: Distance between a compression and the next (consecutive) rarefaction is half a wavelength
  l/2=50     Þl=100cm    No. of waves =20 time taken to complete these waves is 0.2 sec.
Frequency = 20/0.2=200/2=100Hz     Time Period= 1/n =1/100sec =0.01 sec.
27. A flask contains 4.4gm of CO2 calculate (a) How many moles of CO2 gas does it contains
(b)  How many molecules of CO2 gas are present it contain? (c) How many atoms of oxygen and C are present in given sample?
Ans:  1 mole of CO2 =12+2x16=44gm
(a) No. of moles in 44gm  of CO2 gas = 1 mole
No. of moles in 4.4gm  of CO2 gas = (1/44)x4.4=0.1 mole
(b) No. of molecules in 44gm of CO2 gas= 6.023x1023 molecules
No. of molecules in 4.4gm of CO2 gas= (6.023x1023 ÷ 44)x 4.4 molecules= 6.023x1022
(c) No. of atoms of oxygen (2 atoms in CO2 ) = 6.023x1022 x 2 = 1.204x1023  atoms
No. of atoms of oxygen (1 atoms in CO2 ) = 6.023x1022 x 1 = 6.023x1022
28. A mass of 4 kg is dropped from a tower of height 45 m. Calculate 
(i) the potential energy possessed by the body when it is at the top of the tower  (ii) the kinetic energy possessed by the body when it is at a height of 35 m  (iii) the kinetic energy just before hitting the ground   (iv) the velocity of the body just before hitting the ground.
Ans: Mass of the body (m) = 4 kg  Height of the tower (h) = 45 m  Acceleration due to gravity = 9.8m/s2
i) Potential energy possessed by the body when it is at a height of 45 m above the ground
 = mgh   = 4x9.8x45=1764J
ii) Kinetic energy of a moving body when it is at a height of 35 m from the ground is equal to the kinetic energy possessed by the body after covering 10 m.  First we have to calculate the velocity with which the body covers 10 m.  Using the eq. v2-u2=2gS,   and substituting value   u = 0m/s   S= 45-35=10m  We get V2 =196m2/s2         KE = 1/2 x 4 x196 = 392 Joule
iii) According to the law of conservation of energy the kinetic energy of the body just before hitting the ground is equal to potential energy of the body at a height of 45m above the ground.Kinetic energy of the body just before hitting the ground = 1764 J

(IV) 1764 =1/2 mv2 Þ v= Ö882 = 29.69 m/s

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