09 March 2016

CBSE 2015 – 2016 Class 09 SA2 Question Papers

Class 9 Original Question Paper March 2016 for 2015-16

9th Maths Question paper(visitor) 2015-16-1 
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9th Maths Question paper(visitor) 2015-16-2
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9th Maths Question paper(visitor) 2015-16-3       
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9th Maths Question paper(visitor) 2015-16-4     
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IX Original English Question paper SA2 March 2016 

9th English_2016_SA2 Question papers - 1
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IX Original Paper with OTBA March 2016 session 2015-16

9th Science question paper by visitor 2015-16-1
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9th Science question paper by visitor 2015-16-2
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02 February 2016

10th Our Environment CBSE Board solved Questions

Class 10 our environment board solved Questions 2016

1. What is the function of ozone in upper atmosphere?

Ans: It shields the surface of the earth from ultraviolet rays from the Sun.

2. Why should biodegradable and non biodegradable waste be discarded in two separate dustbins?
Ans: So that the time and energy required in segregation may be saved and waste may be disposed off quickly

3. The following organisms form a food chain. Which of these will have the highest concentration of non-biodegradable chemicals ? Name the phenomenon associated with it. Insects, Hawk, Grass, Snake, Frog.
Ans: Hawk ; Biomagnifications

4.. Write the full name of the group of compounds mainly responsible for the depletion of ozone layer.

Ans: Chloroflurocarbons

5. List two examples of natural ecosystem.
Ans: Forests ,Ponds,Lakes

6. Differentiate between biodegradable and non-biodegradable substances with the help of one example each.
List two changes in habit that people must adopt to dispose non-biodegradable waste, for saving the environment.

Ans: Biodegradable substances – Any substance that can be broken down into simpler substances by biological process is called biodegradable. Ex. – Human Excreta/ Vegetable peels, etc.

Non-biodegradable substances – Any substance that can can’t be broken down into simpler substances by nature or decomposers. Ex. – Plastic/ glass (or any other) (any one)

Habits that people must adopt to dispose non-biodegradable waste, for saving the environment:

- Use of separate dustbins for biodegradable and non biodegradable waste,

- Reuse of things such as poly-bags, etc.,

- Recycle of waste

- Use of cotton /jute bags for carrying vegetables

7. What will be the amount of energy available to the organisms of the 2nd trophic level of a food chain, if the energy available at the first tropic level is 10,000 joules?

Ans: 10% of 10,000 joules = 1000 J

8. ‘‘Energy flow in food chains is always unidirectional.’’ Justify this statement.

Explain how the pesticides enter a food chain and subsequently get into our body.

Ans: The flow of energy is unidirectional. The energy that is captured by the autotrophs does not revert back to the solar input and the energy which passes to the herbivores does not come back to autotrophs.

Therefore, in food chain the energy moves progressively through the various trophic levels it is no longer available to the previous level. Thus, The flow of energy in food chain is unidirectional.

The pesticides enter a food chain and subsequently get into our body in following way:

(i) Pesticides used for crop protection when washed away into the soil or water bodies absorbed by plants.

(ii) On consumption they enter our food chain and being non – biodegradable these chemicals get accumulated progressively and enter our body.

9. What is ozone ? How and where is it formed in the atmosphere? Explain how does it affect an ecosystem.

Ans: Ozone (O3) is a molecule formed by three atoms of oxygen.

Ozone is formed at at the higher levels of the atmosphere by action of UV radiation on oxygen (O2) molecule.

The higher energy UV radiations split apart some moleculer oxygen (O2) into free oxygen (O) atoms. These atoms then combine with the molecular oxygen to form ozone as shown—

O2 ⎯⎯UV⎯→ O + O then, O + O2 ⎯⎯⎯→ O3 (Ozone)

Ozone shields the surface of the earth from ultraviolet (UV) radiation from the Sun. This radiation is highly damaging to organisms, for example, it is known to cause skin cancer in human beings.

10. Construct an aquatic food chain showing four trophic levels.

Ans: Hydrilla →Scorpio→ Fish → Crane

11. To protect the food plants from insects, an insecticide was sprayed in small amounts but it was detected in high concentration in human beings. How did it happen?

Ans: As insecticides are not degradable, these get accumulated progressively at each tropic level. As human beings occupy the top level in any food chain, the high concentration of insecticide gets accumulated in their bodies. This phenomenon is known as biological magnification

12. Why do most food chains have 3- 4 steps only?
Ans: The loss of energy at each step is so great that very little usable energy remains after four trophic levels. Therefore, food chains generally consist of only three or four steps.

13. What will be impact on ecosystem if bacteria and fungi are removed from the Environment?
Ans: The decomposers bacteria and fungi secrete enzymes that breakdown organic remains into simpler soluble substances that are absorbed by saprophytes. They clean the earth from organic matter hence Environmental pollution created and food chain disturbed if bacteria and fungi are removed from the Environment.

14. Why is a pond self sustaining unit while an aquarium may not be?
Ans: A pond contains microorganisms that break-down the dead remains and waste products of organisms into simple inorganic substances that are used by other organism.

Since, pond is a natural ecosystem while an aquarium is created by man and is artificial, thus a pond self sustaining while an aquarium may not be.

15. Which level shows the maximum biological magnification? Why?
Ans: The top tropic level in any food chain has the maximum concentration of chemicals. From the soil, these are absorbed by the plants along with water and minerals, and from the water bodies these are taken up by plants and animals.


CBSE HIGHER ORDER THINKING SKILLS (HOTS) QUESTIONS :

10th Our Environment CBSEHOTS Questions

1) Write any two ways of energy flow through an ecosystem.

2) Differentiate between biodegradable and non biodegradable with respect to the effect of biological processes on them and the way they affect our environment.

3) Which level shows the maximum biological magnification? Why?

4) Why is a pond self sustaining unit while an aquarium may not be? Justify the answer.

5) Arrange grasshopper, frog, grass, eagle and snake in the form of food chain.

6) If 1000 KJ energy is available at producer level, how much energy will be available at first carnival level?

7) Why do most food chains have 3-5 steps only?

8) Select the biodegradable items from the list given below-

Polythene bags, old clothes, wilted flowers, pencil shavings, glass bangles, bronze statue, vegetable peels.

9) What will be impact on ecosystem if bacteria and fungi are removed from the Environment?

10) Express your feelings on the picture given down below. What will happen if all? Carnivores are eliminated from the environment? What measures will you take to save? Tiger?

Further Study Links by JSUNIL

10th Science – Chapter: Our Environment (Physics)
10th Science - Our Environment: Summative assignments
10th Chapter: Our Environment Questions Paper for FA-III
10th Science: Chapter: Our Environment HOTS Questions
10th Chapter: Our Environment: Gist of Lesson

21 November 2015

Numerical on work and energy for class 9 [with solution]

  Class 9 Chapter Work and energy Solved Numerical problems

1. A force of 10N causes a displacement or 2m in a body in its own direction. Calculate the work done by force. 20j

Solution: the work done by force = F x S = 10 N x 2m = 20 J

2. How much force is applied on the body when 150 joule of work is done in displacing the body through a distance of 10m in the direction of force?(15 N)

Solution: W = F x S Þ  F = w/s = 150/10 = 15 N

3. A body of 5kg raised to 2m find the work done(98j)

Solution: T he work done to raise a body = PE = mgh = 5kg x 9.8 x 2 = 98 joule

4. A work of 4900j is done on road of mass 50 kg to lift it to a certain height. Calculate the height through which the load is lifted. (10m)

Solution: work done on road to lift = mgh Þ 4900 = 50 x 9.8 h Þ h = 10m

5. An engine work 54,000J work by exerting a force of 6000N on it. What is the displacement of the force . (9m)

Solution: S = W/F = 54,000J/6000N = 9m

6. A force of 10N acting on a body at an angle of 60 deg. with the horizontal direction displaces the body through a distance of 2m along the surface of a floor. Calculate the work done.

Now let the force or pulling act on the body makes an angle of 30 deg. with the horizontal. What is the value of the force to displace the body through 2m along the surface of the floor? (Cos 60 =1/2. and Cos 30 = √3/2) [ ans. 10 J, 10√3 N]

Solution: w = F S cos Q = 10 x 2 x Cos 60 = 20 x ½ = 10 N

w = F S cos Q = 10 x 2 x Cos 30 = 20 x √3/2 = 10√3 N

7. A force of 5N acting on body at angle of 30 deg. with the horizontal direction displace it horizontally through of distance of 6 m . Calculate the work done. (15√3 J)

Solution: w = F S cos Q = 5 x 6 x √3/2 = 15√3 J

8. A body of mass 2kg is moving with a speed of 20m/s Find the kinetic energy. (400J)

Solution: KE = 0.5 mv^2 = 0.5 x 2 x 20 x 20 = 400 J

9. A moving body of 30kg has 60 J of KE. Calculate the speed.

Solution: KE = 0.5 mv^2  Þ 60 = 0.5 x 30 x v^2
Þ 60 = 15 v^2 Þ 60/15Þ V = 2m/s

10. A hammer of mass 1kg falls freely from a height of 2 m .Calculate (I) The velocity and (II) The KE. of the hammer just before it touches the ground. Does the velocity of hammer depend on the mass of hammer? (6.29m-2 , 19.6 J )

Solution: PE = mgh = 1 x 9.8 x 2 = 19.6 j

PE = KE = 0.5 mv^2 Þ 19.6 = 0.5 x 1 xv^2 Þ 39.6 = v^2 Þ v = 6.29 m/s

No, velocity of hammer does not depend on the mass of the mass of the hammer as v = u + at

11. Calculate the energy posses by a stone of mass 10kg kept at a height of 5m If 196 x10^2 J of energy were used to raise a 40kg boy above the ground, how high would he be raised? (50m)

Solution: The energy posses by a stone of mass 10kg kept at a height of 5m = PE

PE = mgh = 10 x 5 x 9.8 = 490 j

PE = mgh  => 196 x10^2 J = 40 x 9.8 x h => h =  50m

12. Calculate the change that should be affected in the velocity of a body to maintain the same KE , if mass of the body is increased to 4 times (half the original velocity)

Solution: New kE/Original KE = [½ x 4m x V^2]/[ ½ x m x v^2]

¼ = (V/v)^2   Þ   ½ = (V/v) Þ V = v/2

New velocity will be half the original velocity

13. A machine does 192 J of work in 24 Sec. What is the power of the machine? (8w)

Solution: p = w/t = 192 J / 24Sec = 8w

14. A weighting 50kg runs up a hill rising himself vertically 10m in 20Sec. Calculate power. given g = 9.8m-1 (245w)

Solution: p = w/t = mgh/t = (50 x 9.8 x 10) /20 = 245 w

15. A rickshaw puller pulls the rickshaw by applying a force of 100 N. If the rickaw moves with constant velocity of 36 kmh-1. Find the power of rickshaw puller. (1000w)

Solution: Force = 100 N

Velocity = 36 k m / h = 36 × 5 /18 = 10 m / s

Power = Force x Velocity Power = 100 × 10

Power = 1000 Watt

Therefore power of rickshaw puller = 1000 Watt

16. A athlete weighing 60kg runs up a staircase having 10 steps each of 1m in 30 sec. Calculate power (g = 9.8ms-1 ) (ans: 200W)

Solution: h = 10 x 1m = 10m

Work done = potential energy = mgh = 60 x 9.8 x 10 = 5880 J

Power = w/t = 5880/30 = 196w

17. The heart does 1.5 J of work in each heartbeat. How many times per minute does it beat if its power is 2watt? (80 times)

Solution: Total work = p x t =120 J ,

Number times heartbeat in 1 min. = total work done / work done in each beat

                                                       = 120/1.5 = 80 times

18. Calculate the time taken 60 w bulb to consume 3000 J of energy . (50sec. )

Solution: Power = 60 W and Energy consumed = 3000 J

We know that Power=Energy/Time Taken
Time Taken = Energy Consumed/Power = 3000/60 = 50 sec.

19. A horse exert a force of 200N to pull the cart. If the horse cart system moves with velocity 36kmh-1 on the level road., then find the power of horse in term of horse power (1hp=746W) 

Solution: velocity = 36kmh-1 = 10m/s
w = f x s = 2000 x 10 = 2000j

P = w/t = 2000j/1sec = 2000 w

746W = 1hp 

So, 2000 w = 2000/746 = 2.68 h.p 

20. An electric kettle of 500W is used to heat water everyday for 2 hours . Calculate the number of unit of electrical energy consumed y it in 10 days.

Solution: E = Pt = 500 w x 10 x 2h = 10000wh = 10kwh = 10 unit

21. Calculate the cost of using a 2kwh immersion rod for heating water in a house for one hour each day for 60 days if the rate is Rs. 1.50 per unit kWh. (Rs. 180)

Solution: E = Pt = 2 kw x 60 x 1h = 120kwh = 120 unit

The cost of using a 2kwh immersion rod for heating water = 120 x 1.5 = Rs. 180

22. In an experiment to measure his power, a student records the time taken by him in running up a flight of steps on a staircase.

Use the following data to calculate the power of the student :

Number of steps = 28 ; Height of each step = 20 cm ; Time taken = 5.4 s.

Mass of student = 55 Kg ; Acceleration due to gravity = 9.8 m s−2

Solution. Power =w/t = mgh/t=55×9.8×(28×0.20) / 5.4 = 559J

23. A bullet of mass 15 g has a speed of 400 m/s. What is its kinetic energy ? The bullet strikes a thick target and is brought to rest in 2 cm, calculate the average net force acting on the bullet. What happens to kinetic energy originally in the bullet?

Solution. K.E = ½ mv^2 = 0.5 × 0.015 kg × (400 x 400) = 1200 J.

Work done = Change in K.E.
As final velocity = 0. Because change in KE = Kf−Ki = 1200 J

Therefore, F × d = 1200. (where F is the average force)

F=1200 / 2 ×10^−2 = 6×10^4N.

The kinetic energy is eventually converted to heat energy.

24. The power of a heart which beats 72 times in a minute is 1.2kW. Calculate the work done by heart for each beat. ( 1kJ)
Solution: P = 1200W and T = 60s

W = P x T = 1200 x 60 = 72000J

In 72 times heart beats 72000j energy used

In 1 time = 72000/72 = 1000J

Work done by the heart in every beat is 1KJ

25. Whenn loading a truck, a man lifts boxes of 100 N each through a height of 1.5 m.

(a) How much work does he do in lifting one box ?

(b) How much energy is transferred when one box is lifted ?

(c) If the man lifts 4 boxes per minute, at what power is he working ? (g = 10 m s−2)

Solution. (a) Work done in lifting one box = F × d = 100 × 1.5 = 150 J.

(b) W = E = 150 J.

(c) Power = Work done / Time = (150×4)/ 60=10W        By jsuniltutorial.weebly.com

Numerical Test papers Download

Solved CBSE Test papers:
Class 9 Work, Energy ,Power_ Solved Questions - 01

Class 9 Work, Energy ,Power_ Solved Questions - 02

Class 9 Work, Energy ,Power_ Solved Questions - 03

Class 9 Work, Energy ,Power_ Solved Questions - 04

Class 9 Work, Energy ,Power_ Solved Questions - 05

Class 9 Work, Energy ,Power_ Solved Questions - 06          Download Files

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