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Sunday, November 29, 2015

Biology class IX Ch-07. Diversity in Living Organisms solved Questions New

Diversity in Living Organisms solved CBSE Paper-5

Oswal Question Bank Diversity in Living Organisms        Download File

Very Answer type Questions [1 mark each]  

1.State the phylum to which centipede and prawn belong. (Board 2013)

Answer: Arthropoda.         

2. In which kingdom would you place an organism which is single called, eukaryotic and photo-
synthetic? (Board 2013)

Answer:  Protista.

3.Rewrite the scientific name correctly : (i) panthera tigris (ii) periplaneta Americana. (Board 2012., 

Answer:  (i) Panthers tign's  (ii) Periplaneta americana.

4.Name the term which is used for the following

(i) The left and right halves of the body have the same design.

(ii) Animal tissue defferentiate from the three embryonic germ layers. (Board 2012, 501005; 47023)

Answer:  (i) Bilaterally symmetrical. (ii) Triploblastic.       

5.Name the phylum to which Prawn and Centipede belong. (Board 2012, SC-1008; 47001,) 

Answer:  Arthropoda.

6. Mention the function and location of notochord.

Answer:  Functionn—lt provides a place for muscle to attach for ease of movement.

Location—runs along the back of the animal separating the nervous tissue from the gut.

7. State the phylum to which Antedon (feather star ] and Asterias (Starfish) belong. (Board 2012,  501010)

Answer:  Echinodermata.

8.Mention any one characteristic feature of saprophytes. (Board 2012, 47005)

Answer:  They have cell wall made of tough compare sugars called chitin.

9.Which in your opinion is more basic characteristics for classifying organism - the place where they  live in or the kind of cells they are made of ? (Board 2012, 501006)

Answer:  The kind of cells the organisms are made of.

10. Write the name of phylum to which Hydra and Sea anemone belong.- (Board 2012, 47017)

Answer:  Coelenterata.

11. Name the phylum to which Leech and Earthworm belong. (Board 2012, 47004 }

Answer:  Annelida.

12. Name a division of plants which can be classified as cryptogarnae. (Board 2012, 47010}

Answer:  Pteridophyta / Thallophyta / Bryophyta. (Any one)

13. State the system in poriferans that heips in circulating water throughout the body to bring in food and oxygen. (Board 2012, SC-1011}

Answer:  Canal System

14. Write an example of the species which belong to this phylum and lives in colonies.

Answer:  Corals.

15. Name the phylum to which Octopus and Unio belong ?  (Board 2012, 4700}

Answer:  Mollusca.

Diversity in Living Organisms solved CBSE Paper-1
Diversity in Living Organisms solved CBSE Paper-2
Diversity in Living Organisms solved CBSE Paper-3
Diversity in Living Organisms solved CBSE Paper-4
Diversity in Living Organisms NCERT Solution
9th Diversity in Living Organisms Study Notes
9th 9th Diversity in Living World Notes                            
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Sunday, November 22, 2015

9th Sound Test Paper Solved

1.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?
Ans: 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

2. If the tension in the wire is increased four times how will the velocity of wave in a string varies?
Answer: velocity of the wave in string is directly proportional to the square root of the tension thus if tension is increased 4 times the velocity will be doubled.

3. What is Echo. Explain the conditions that have to be satisfied to hear an echo?
Ans: reflection of sound wave from a large obstacle is called an echo.

The most important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 second after the original sound dies off and the obstacle is at least at a distance of 17 m

4. Why do echoes produced in an empty auditorium usually decrease when it is full of audience?
Ans: When the hall is empty there is no obstacles in between to reflect the sound other than the walls. When the hall is full of audiences the sound produced undergoes multiple reflection from the people and so it overlaps with the sound produced. Hence the listener is not able to distinguish between the original sound and the echo.

5. A girl claps and hears the echo after reflection from cliff which is 660 m away. If the velocity of sound is 330 m s-1, calculate the time taken for hearing the echo. 

Ans: v x t = 2 d Þ t = 2d/v = 2x660/300 = 4s

6. Explain how is the principle of echo used by the dolphin to locate small fish as its prey.
Ans: Dolphins are aquatic animals which send out ultrasonic sound to communicate with each other. They have a sound sensing system which enables them to find animals under water with great accuracy due to the echo of the ultrasonic sound produced by them.

7. Explain by some experiment that sound waves require medium for their propagation.
Ans: An electric bell is suspended inside an airtight glass bell jar connected to a vacuum pump. As the electric bell circuit is completed, the sound is heard. Now if the air is slowly removed from the bell jar by using a vacuum pump, the intensity of sound goes on decreasing and finally no sound is heard when all the air is drawn out. We would be seeing the hammer striking the gong repeatedly. This clearly proves that sound requires a material for its propagation.

8. Distinguish between loudness and intensity of sound.
Ans: Intensity depends on the energy per unit area of the wave and it is independent of the response of the ear, but the loudness depends on energy as well as on the response of the ear.

9. Give two practical applications of the reflection of sound waves.

Ans: (i) In stethoscope the sound of patient’s heartbeat reaches the doctor’s ears by multiple reflections in the tubes.

(ii) Megaphones are designed to send sound waves in particular direction are based on the reflection of sound.

10. Why are longitudinal waves called pressure waves?

Ans: Sound waves travels in the form of compression and rarefactions, which involve change in pressure, and volume of the air. Thus they are called pressure waves.

11. Sound travels faster on a rainy day than on a dry day. Why?
Ans: Sound travels faster on rainy day because the velocity of sound increases with increase in humidity. On rainy day humidity is more thus velocity of sound is also more.

12. How moths of certain families are able to escape capture?
Ans: Moths of certain families can hear high frequency sounds (squeaks) of bat as they have sensitive hearing equipment. Thus they get to know when a bat is near by and hence able to escape its capture

Extra score 

Solution : Class 09 Sound solved Numerical Test Paper-1       Download File

Class 09 Sound solved CBSE Test Paper-2

Class 09 Sound solved CBSE Test Paper-3

Class 09 Sound solved CBSE Test Paper-4                              Download File

Thursday, October 15, 2015

9th CBSE Science Chap 3 Atoms and Molecules Self study Questions

Solved problems:Class 9 _Atoms and Molecules 

1. Q. 5 g of calcium combine with 2 g of oxygen to form a compound. Find the molecular formula of the compound. (Atomic mass of Ca = 40 u; O = 16 u) 

Ans. Number of moles in 5g of calcium = mass / molar mass    =5/40    =  0.125 
Number of moles in 2g of oxygen = mass / molar mass         =  (2 / 16)  =  0.125

Now we will calculate the simplest ratio of the element by dividing the number of moles of each element by the smallest value. 

Since number of moles of each element is 0.125, therefore calcium and oxygen are present in a ratio of 1 : 1. Thus the empirical formula of the compound is CaO.

For calculating the molecular formula, we need the molecular mass of the compound. However, because a compound with the formula CaO is known,

Therefore the molecular formula of the compound is CaO.

. Q. (i) Name the body which approves the nomenclature of elements and compounds.
(ii) The symbol of sodium is written as Na and not as S. Give reason.
(iii) Name one element which form diatomic and one which form tetra atomic molecules. [CBSE 2011 ]

Ans. (i) IUPAC (International Union of Pure and Applied Chemistry)
(ii) Latin name of sodium is Natrium. The first two letters (Na) of this name represents the symbol of sodium.
(iii) Oxygen forms diatomic molecules and phosphorus forms tetra atomic molecules.

3. Q. Calcium and Oxygen are combined in the rates of 5:4 by mass to form calcium oxide.  What mass of Oxygen gas would be required to react with 2.5 g of calcium ? 

Ans:  Calcium and oxygen combine in the rates of 5:4 by mass to form calcium oxide
let x gram of oxygen is required to react with 2.5 g of calcium to form calcium oxide
Therefore 5/4 = 2.5 / x 

     5 x = 2.5 x  4 
       x = (2.5 x 4 ) / 5   
So, x = 2
Therefore 2 grams of oxygen is required to react with 2.5 grams of calcium to form calcium oxide .

Problems (based on mole concept)

4.Q. a. Calculate the number of moles in 81g of aluminium

Solution : Atomic mass of Al= 27gm
     27g  of  aluminium = 1 mole of aluminium  
     81g  of aluminium   = 1/27 x 81=3 moles of aluminium  
OR, Use formula , Number of moles = given mass/atomic mass
Self: ii) 4.6g sodium   iii) 5.1g of Ammonia  iv) 90g of water v) 2g of NaOH

5. Q. Calculate the mass of 0.5 mole of iron

Solution: Atomic mass of iron = 55.9 g
Mass of the 1 mole of iron   = 55.9 g
Mass of the 0.5 mole of iron   = 0.5 x 55.9 g = 27.95 g
Or, Using formula: mass = atomic mass x number of moles
FOLLOW UP: Find the mass of 2.5 mole of oxygen atoms

6.Q.Calculate the number of molecules in 11g of CO2

Solution: gram molecular mass of CO2 = 44g
 44g of CO2  = 6.023 x 1023  molecules
1 g of CO2  = ( 6.023 x 1023 ÷ 44 g ) molecules
11g of CO= ( 6.023 x 1023 ÷ 44 g ) x 11 = 1.51 x 1023 molecules
FOLLOW UP: Calculate the number of molecules in 360g of glucose

7. Q. Calculate the mass of glucose in 2 x 1024 molecules

Solution: Gram molecular mass of glucose = 180g
Mass of glucose  [180 x 2 x 1024 ] / [6.023 x 1023  ] = 597.7g
FOLLOW UP: Calculate the mass of 12.046 x 1023 molecules in CaO.

8. Q. Calculate the number moles for a substance containing 3.0115 x 10^23 molecules in it.
Solution: Number of moles = Number of molecules/Avogadro Number

                       = [3.0115 x 1023 ] / [6.023 x 1023  ]      = 0.5 moles

9. Q. Calculate the mass of 18.069 x 1023 molecules of SO2
Mass of a substance =  gram molecular mass x number of particles/ 6.023 x 1023
Gram molecular mass SO2 = 64g
6.023 x 1023 molecules of SO2     = 64 gm
1           molecules of SO2     = 64/(6.023 x 1023 )  gm
18.069 x 1023  molecules of SO2   = [ 64/(6.023 x 1023 )  x  18.069 x 1023  ]gm = 192 g

Other Download study links:  Chap 3 Atoms and Molecules class9

Class 9 Atoms and molecules NCERT Solutions-1
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Class 9 Atoms and molecules NCERT Solutions-2
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Solved Problems (based on mole concept)
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Chapter based Multiple Choice Questions
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Class 9 SA2: Atoms and molecules solved CBSE Test Papers New

Class 9 Atoms and molecules solved CBSE Test Paper-1

Class 9 Atoms and molecules solved CBSE Test Paper-2

Class 9 Atoms and molecules solved CBSE Test Paper-3

Class 9 Atoms and molecules solved CBSE Test Paper-4

Class 9 Atoms and molecules solved CBSE Test Paper-5         Download Files

Friday, October 2, 2015

Floatation_Class-9 important solved Questions from previous year CBSE Board Paper

Fully Solved Questions from previous year board Paper
Question: 1. What causes buoyant force (or up thrust) on a boat?

Answer: Buoyant force is due to volume of boat and density of water.

Question: 2. what is the cause of buoyant force?

Answer: : The pressure difference on lower part and upper part of a body cause of buoyant force .

Question: 3. An Object of volume V is immersed in a liquid of density P. Calculate the magnitude of buoyant force acting on the object due to the liquid

Answer: Vd g

Question: 4. why does a block of wood held under water rise to the surface when released?

Answer: This happens because as upward force is greater than the weight.

Question: 5. What happens to the buoyant force as more and more volume of a solid object is immersed in a liquid. When does the buoyant force become maximum?

Answer: The buoyant force exerted by a fluid is equal to the weight of fluid displaced. Therefore the buoyant force increases as more and more volume of a solid object is immersed in a liquid.
Buoyant force is maximum when a body is completely immersed in a fluid.

Question: 6. Explain why, a piece of glass sinks in water but it floats in mercury.

Answer: The density of glass is more than water, but less then mercury, Therefore glass float on mercury and sinks in water.

Question: 7. When is the pressure on the ground more when a man is walking or when a man is standing ? Explain. 

Answer :  When a man is walking is weight is alternately on one feet or the other, when he is standing, his weight is divided on both feet. Therefore pressure is more on the ground when he is walking (as surface area is less).

Question: 8.  A solid weighs 80 g in air, 64 g in water. Calculate the relative density of solid. When kept in water, state if the object would float or sink?

Answer:  Weight loss in water = (80 -  64) g = 16
Relative density of solid = (weight in air/ Weight loss in water) = 80/16= 5
Since relative density of water is 1 which is less than that of solid. Hence Solid Sink

Question: 9. why density varies?

Answer: The density of a material depends on two things.
(i) The individual mass of each atom or molecule of substance  (ii) How tightly the atoms are packed

Question: 10. An object is suspended with a string . The string is stretched. When the object is completely immersed in water, the extension of the thread decreases. Why?

Answer : This happen because when the object is completely immersed in water it experience an upward force that cause weight loss.
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