chemistry class 9th:Is matter around us pure solved questions

Q.1: Which separation techniques will apply for the separation of the following?

(a) Sodium chloride from its solution in water.

(b) Ammonium Chloride from a mixture containing Sodium Chloride and Ammonium Chloride.

(d) Different pigments from an extract of flower petals.

(e) Butter from curd.

(f) Oil from water.

(g) Tea leaves from tea.

(h) Iron pins from sand.

(i) Wheat grains from husk.

(j) Fine mud particles suspended in water.

Ans: (a) Crystallization or Evaporation. (b) Sublimation. (c) Centrifugation or Sedimentation. (d) Chromatography. (e) Centrifugation. (f) Separating funnel. (g) Hand-picking. (h) Magnetic separation. (i) Winnowing. (j) Centrifugation.

Q.2: Write the steps you would use for making tea. Use the words – solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

Ans: Take the solvent, water, in a kettle. Heat it. When the solvent boils, add the solute, milk. Milk and water forms a solution. Then pour some tea leaves over a sieve. Pour slowly hot solution of milk over tea leaves. Colour of tea leaves goes into solution as filtrate. The remaining tea leaves being insoluble remains as residue. Add requisite sugar which dissolves and the tea is ready.

Q.3: Explain the following giving examples:

(a) Saturated solution, (b) Pure substance, (c) Colloid, (e) Suspension.

Ans:

(a) Saturated Solution – a solution in which no more of the solid (solute) can be dissolved at a given temperature is called a saturated solution. Suppose 50 gm of a solute is the maximum amount that can be dissolved in 100 gm water at 298 K. Then 150 gm of solution so obtained is the saturated solution at 298 K.

(b) Pure Substance – A pure substance consists of a single of matter or particles and can not be separated into other kind of matter by any physical process. Pure substances always have the same colour, taste and texture at a given temperature and pressure. For example, pure water is always colourless, odorless and tasteless and boils at 373 K at normal atmospheric pressure.

(c) Colloid – Colloids are heterogeneous mixtures the particle size is too small to be seen with a naked eye, but it is big enough to scatter light. The particles are called the dispersed phase and the medium in which they are distributed is called the dispersion medium. Colloids are useful in industry and daily life.

A colloid has the following characteristics:

(1) It is a heterogeneous mixture.

(2) The size of particles of a colloid lies between 1 – 100 nm and can not be seen by naked eyes.

(3) The particles of colloid can scatter a beam of light passing through it and make the path visible.

(4) The particles of colloid can not be separated from the mixture by filtration. The process of separation of colloidal particles is known as ‘centrifugation’.

(5) They do not settle down when left undisturbed. In other words colloids are quite stable e.g. smoke, milk, fog, cloud etc.

(d) Suspension –

A ‘suspension’ is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium.

A suspension has the following characteristics:

(1) It is a heterogeneous mixture.

(2) The size of particles of a suspension is greater than 100 nm and is visible to naked eyes.

(3) The particles of suspension can scatter a beam of light passing through it.

(4) The particles of a suspension settle down when left undisturbed.

(5) The particles of a suspension can be separated from its mixture by filtration.

Q.4: Classify each of the following as a homogeneous or heterogeneous mixture: soda water, wood, air, soil, vinegar, filtrated tea.

Ans: Homogeneous mixture – soda water, air, vinegar, filtered tea.

Heterogeneous mixture – wood, soil.

Q.5: How would you confirm that a colourless liquid given to you is pure water?

Ans: Every liquid has a characteristic boiling point at 1 atmospheric pressure. If the given colourless liquid boils exactly at 373 K at 1 atmospheric pressure, then it is pure water. If the boiling point is different then the water is contaminated.

Q.6: Which of the following materials fall in the category of a ‘pure substance’?

(a) Ice (b) Milk (c) Iron (d) Hydrochloric acid (e) Calcium oxide (f) Mercury (g) Brick (h) Wood (i) Air.

Ans: (a), (c), (d), (e) and (f) are pure substances.

Q.7: Which of the following will show “Tyndall effect”?

(a) salt solution (b) milk (c) copper sulphate solution (d) starch solution.

Ans: (b) and (d) are colloids and will show Tyndall Effect.

Q.8: Classify the following into elements, compounds and mixtures.

(a) sodium (b) soil (c) sugar solution (d) silver (e) calcium carbonate (f) tin (g) silicon (h)

coal (i) air (j) soap (k) methane (l) carbon dioxide (m) blood

Ans: Elements – sodium, silver, tin, silicon.

Compounds – calcium carbonate, methane, carbon dioxide.

Mixtures – soil, sugar solution, coal, air, soap, blood.

Q. 9. In the formation of sodium chloride from its constituent elements, do the properties of its elements change. Explain.

Ans: Sodium is a very reactive metal that reacts exothermically with water. If we were to ingest a pinch of sodium, it would burn our intestines. Chlorine is a greenish yellow gas with a characteristic irritating odour and pungent taste. When sodium and chlorine combine to form sodium chloride, the properties of the elements are totally changed. Sodium chloride is a white substance totally safe to be ingested and is used to add flavour to our food.

Q. 10. Briefly describe how to separate, i) Sulphur from a mixture of sulphur and sand. ii) Black CuO from a mixture of CuO and ZnO.

Ans:

Answer: i) Add a solvent to the mixture of sulphur and sand. Sulphur dissolves in carbon disulphide while sand does not. When filtered, the residue is sand. The filtrate is kept open, carbon disulphide evaporates and the sulphur crystals form.
ii) Add a solvent to the mixture of CuO and ZnO that dissolves only one component e.g. sodium hydroxide. When sodium hydroxide is added to the mixture, ZnO dissolves. Filter to obtain the residue of CuO.

science que-ans/IX-X

IX- CHEMISTRY

CBSE Chemistry IX Matter in our surrounding Chemistry solved questions

Q.1 Camphor disapperars without leaving any residue. Explain?
Ans. Camphor disappears without leaving any residue because of sublimation, as it changes its state
directly from solid to gas without changing to liquid.
Q.2 Why do we feel cool when we touch a piece of ice?
Ans. We feel cool because the temperature of ice is 0 deg C and our body’s temp is higher than 0 deg C.
Q.3 Convert the following:-
a) 573 K
Ans. 573-273= 300 deg C
b) 36 deg C
Ans. 273+36 = 309 K
c) 373 deg C
Ans. 373+273= 646 K
Q.4 Both the process of evaporation and boiling involves the change of state from liquid to gas but still
they are different from each other. Justify.
Ans. Evaporation is diffeent from boiling as evaporation is a natural phenomenon, occurs on the surface
of the water and also it is a slow process whereas boiling is an artificial phenomenon , it occurs in the bulk
of the liquid and is a fast process.
Q.5 Why is Kelvin considered as the best scale for measuring the temperature?
Ans. Kelvin is the best scale for measuring the temperature because it has no max temp. and it can
measure temperature to any extent.
Q.6 How can you show that evaporation causes cooling?
Ans. When we put some acetone on our hand , after sometime we will feel coolness on our hand because
the action absorbs kinetic energy from our hand & evaporates and evaporation causes cooling.
Q.7 What is the significance of boiling point & melting point of a substance?
Ans. The significance of boiling point and melting point is that it shows the purity of the substance.
Q.8 When we put CuSO4 in water, after sometime we find the soln. turns blue. Why? Also, on heating it,
what change will occur?
Ans. The solution turns blue because of diffusion. On heating the solution nothing will happen

Q.9 How can you justify that table is a solid?
Ans. When we apply force on the table it doesn’t change its shape because its intermolecules are tightly
packed , this shows that table is solid.
Q.10 Arrange the following in decreasing order of force of attraction: Water, Salt, Oxygen,
Also, state the reason.
Ans. Salt, Water, Oxygen
Salt has the maximum force of attraction as it is a solid, followed by water as it is a liquid. Oxygen
is a gas so its force of attraction is the least.
Q. 11 State characteristics of matter demonstrated by :
a) diffusion
Ans. Diffusion involves movement of different particles so that they become intermixed uniformly .
b) Brownian motion
Ans.It is thwe zig-zag movement of the small particles suspended in a liquid or a gas .
Q.12 When an incense stick (agarbatti) is lighted in one corner of a room, its fragrance spreads in the
whole room quickly.
Which characteristic of the particles of matter is illustrated by this observation?
Ans. Particles of matter are constantly moving.
Q.13 The boiling point of alcohol isn 78 deg C. What is this temperature on Kelvin scale?
Ans. K= Deg C + 273 = 78 + 273 = 351 K
Q. 14 The Kelvin scale temperature is 0 K. What is the corresponding Celsius scale temperature?
Ans. – 273 deg C
Q. 15 What is Latent Heat of Fusion?
Ans. The latent heat of fusion (or melting) of a solid is the quanity of heat in joules required to convert 1
kilogram of the solid (at its melting point ) to liquid, without any change in temperature.
Q. 16 Define latent heat of Vaporisation?
Ans. The latent heatm of vaporisation of a liquid is the quantity of heat in joules required to convert 1
kilogram of the liquid (at its boiling point ) to vapour or gas, without any change in temperature.

X Force and laws of motion solved questions: NCERT / CBSE Textbook Exercise Questions Solved

Q.1: Which of the following has more inertia?
(a) A rubber ball and a stone of the same size. (b) A bicycle and a train. (c) A five rupee coin and a one rupee coin.
Ans: (a) stone (b) train (c) five rupee coin.
Q.2: In the following example, try to identify the number of times the velocity of ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team.”
Also identify the agent supplying the force in each case.
Ans: The velocity of the ball changes three times. First time, the velocity changes when the football player of one team kicks the ball. Second time the velocity changes when another player of the same team kicks the football. Third time the velocity changes when the goalkeeper of the opposite team kicks the football.
The agent supplying the force in each case, have been underlined.
Q.3: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Ans: Before shaking the branches the leaves are at rest. When the branches are shaken, they come in motion at once while the leaves tend to remain at rest due to inertia of rest. As a result leaves get detached from the branches and fall down.
Q.4: Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Ans: When a moving bus brakes to a stop, the lower part of our body in contact with bus comes to rest while the upper part of our body tends to keep moving due to inertia of motion. Hence we fall forwards. When the bus accelerates from rest, the lower part of our body comes into motion along with the bus while the upper part of body tends to remain at rest due to inertia of motion and as a result which we fall backwards.
Q.5: If action is always equal to reaction, explain how a horse can pull cart.
Ans: The horse pulls the cart with a force (action) in the forward direction. Since every action has an equal and opposite reaction so, the cart also pulls the horse with an equal force (reaction) in the backward direction. As a result of which the two forces get balanced. But while pulling the cart the horse also pushes the ground with its feet in the backward direction. The reaction of the earth of the earth makes it forward direction along with the cart. This is how the horse applies force and pulls the cart.
Q.7: Explain why it is difficult for a fireman to hold a hose, which ejects large amount of water at a high velocity.
Ans: Water is ejected with a large forward force (action). As we know by Newton’s third law of motion that every action has an equal and opposite reaction so, because of this action fireman experiences a large backward force or reaction. That is why he feels difficulty in holding the hose.
Q.9: From a rifle of mass 4 kg a bullet of mass 50 gm is fired with an initial velocity of 35 ms-1. Calculate the initial recoil velocity of the rifle.
Ans:
According to the law of conservation of momentum,
Total momentum after firing = Total momentum before firing,
Or, m1v1 + m2v2 = m1u1 + m2u2
Or, 0.05 x 35 + 4v2 = 0 + 0
Or, v2 = – 0.44 ms-1 The negative sign indicates the direction of recoil (backward).
Q.10: Two objects of masses of 100 gm and 200 gm are moving in along the same line and direction with velocities of 2 ms-1 and 1 ms-1 respectively. They collide and aftercollision, the first object moves at a velocity of 1.67 ms-1. Determine the velocity of the second object.Ans: By the law of conservation of momentum, m1v1 + m2v2 = m1u1 + m2u2
Or, 0.1 x 1.67 + 0.2 v2 = 0.1 x 2+ 0.23 x 1 Or, v2 = 1.165 ms–1.
It will move in the same direction after collision.
NCERT / CBSE Textbook Exercise Questions Solved

Q.1: An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with the non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Ans: Yes, an object may travel with a non-zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the up thrust and the velocity of air. The net force on the drop is zero.
Q.2: When a carpet is beaten with a stick, dust comes out. Explain, why?
Ans: When a carpet is beaten with a stick it comes into motion at once. But the dustparticles continue to be at rest due to inertia and get detached from the carpet.
Q.3: why is it advised to tie any luggage kept on the roof of a bus with a rope?
Ans: Due to sudden jerks or due to the bus taking sharp turns on the road, the luggage may fall down from the roof because of its tendency to continue to be either at rest or in motion in the same direction (inertia of motion). To avoid this, it advised to tie the luggage kept on the roof of a bus with a rope.
Q.5: a truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400 m in 20 sec. Find its acceleration. Also find the force acting on it if its mass is 7 metric tones.
Ans: Here, u = 0, s = 400 m, t = 20 s
We know, s = ut + ½ at2 Or, 400 = 0 + ½ a (20)2 Or, a = 2 ms–2
Now, m = 7 MT = 7000 kg, a = 2 ms–2 Or, F = ma = 7000 x 2 = 14000 N Ans.
Q.6: A stone of 1 kg is thrown with a velocity of 20 ms-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?Ans:Here, m = 1 kg, u = 20 ms-1 v = 0, s = 50 m
Since, v2 – u2 = 2as, Or, 0 – 202 = 2a x 50, Or, a = – 4 ms-2
Force of friction, F = ma = – 4N
Q.7: An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) The net accelerating force;
(b) The acceleration of the train; and
(c) The force of wagon 1 on wagon 2.
Ans: Total mass, m = mass of engine + mass of wagons
Or, m = 8000 + 5 x 2000 = 18000 kg.
(a) The net accelerating force, F = Engine force – Frictional force
Or, F = 40000 – 5000 = 35000 N
(b) The acceleration of the train, a = F ÷ m = 35000 ÷ 18000 = 1.94 ms–2.
(c) The force of wagon 1 on wagon 2
= The net accelerating force – (mass of wagon x acceleration)
= 35000 – (2000 x 1.94) = 31111.2 N Ans.
Q.10: Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at constant velocity. What is the force of friction that will be exerted on the cabinet?
Ans: The cabinet will move with constant velocity only when the net force on it is zero.
Therefore, force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.
Q.11: Two objects each of mass 1.5 kg are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms–1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Ans: Here, m1 = m2 = 1.5 kg, u1 = 2.5 ms–1 u2 = –2.5 ms–1
Let v be the velocity of the combined object after collision.
By the law of conservation of momentum,
Total momentum after collision = Total momentum before collision,
Or, (m1 + m2) v = m1u1 + m2u2
Or, (1.5 + 1.5) v = 1.5 x 2.5 +1.5 x (–2.5) [negative sign as moving in opposite direction]
Or, v = 0 ms–1 Ans.
Q.12: According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Ans: The logic is that Action and Reaction always act on different bodies, so they can not cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so the truck does not move.
Q.13: A hockey ball of mass 200 gm travelling at 10 ms–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms–1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Ans: Change in momentum = m (v – u) = 0.2 (–5 – 10) = –3 kg ms–1.
(The negative sign indicates a change in direction of hockey ball after it is struck by hockey stick. Magnitude of change in momentum = 3 kg ms–1).
Q.16: An Object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1 to 8 ms–1 in 6 sec. Calculate the initial and final momentum of the object. Also find the magnitude of the force exerted on the object.
Ans: Here, m = 100 kg, u = 5 ms–1, v = 8 ms–1, t = 6 sec.
Initial momentum, p1 = mu = 500 kg ms–1
Final momentum, p2 = mv = 800 kg ms–1
The magnitude of the force exerted on the object, F = (p2 – p1) ÷ t = (800 – 500) ÷ 6 = 50 N

Q.17. A constant force acts on an object of mass 5kg for a duration of 2sec. It increases the object velocity from 3ms- 1 to 7ms-1 find the magnitude of force of the applied force. Now if the force was applied for a duration of 5s. What would be the final velocity?

Ans: F = ma = m(v-u)/t = 5(7-3)/2 = 10N

Now, same force 10N applies for 5 sec then, final velocity will be v

So, F = ma = m(v-u)/t Þ 10 = 5 (v-3)/5 Þ v = 13m/s

Q.18. Prove that impulse is equal to change in momentum of the body?

Ans: Impulse of a force is a measure of total effect of the force. It is given by the time for which force acts on the body Þ Impulse = force x time =Ft

By 2nd law of motion, F = Change in momentum/time

Þ F = d P/t Þ F x t = Change in momentum Þ impulse = Change in momentum

Q. 19. A car weighing 2000 kg and moving with a speed of 20m/s is stopped in 10sec on applying brakes. Calculate the retardation and the retarding force?

Ans: Let’s first calculate the retardation = (v-u)/t = (0-20)/10= -2 m/s2

Retarding Force = ma = 2000kg x – 2m/s2 = -4000N

Q.20. The change in momentum of a body in 0.01 sec is 10kg m/s. Find the force acting on this body.

Ans: Force = change in momentum/time taken = 10/0.01 = 1000 N

Q. 21. A railway wagon of mass 2500kg moving a velocity 36 km/hr has a head-on collision with a stationary wagon of mass 3000kg, if two wagons move together after collision calculate their common velocity after collision.

Ans: Using, law of conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2

250o x 36 + 3000 x 0 + 2500xv + 300v Þ v = 16.36km/h

Q.22. Two balls of masses 50gm and 100gm are moving along the same line and direction with velocity 3 m/s and 1.5 m/s respectively. They collide and after collision the first ball moves with a velocity 2.5 m/s, determine the velocity of second ball.

Using law of conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2

50 x 3 + 100 x 2.5 = 50 x 2.5 + 100xv2 Þ v2 = 2.75m/s

Q.23. A bullet of mass 20gm is fired from a gun of mass 8kg with a velocity of 400 m/s, calculate the recoil velocity of gun

Ans: Using conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2

0 + 0 = 0.02 x 400 + 8 x v2 Þ v2 = -1m/s

Q. 24.Calculate the acceleration produced when a force of 350 Newton acts on a body of mass 500kg.
Ans: the acceleration produced = 350/500m/s2

Q. 25. A body of mass 25kg has a momentum of 125kg m/s, calculate velocity of body ?

Ans: We have momentum = mass X velocity p = m v OR, p/m = v v = (125kg m/s)/ 25 kg v = ( 125/25 )m/s v = 5 m/s

Q.26. Why is it easier to stop a tennis ball than a cricket ball moving with same speed ?

Ans: Momentum is product of mass and velocity of object. Here Mass of cricket ball is more than of tennis ball. So the momentum of tennis ball is less than that of cricket ball. That’s why it is easy to stop a tennis ball than a cricket ball moving with same speed.

Q.27. A boy of 40kg runs from left to right at 5m/s & catches a ball of 200g coming from right to left at 10m/s .After taking the catch , find the speed of the boy with the ball in his hand.

Ans: Using conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2

40 x 5 + 0.2 x 10 + 40 v1 + 0 Þ v1 = 5.05m/s

Q.28. What is impulse of a Force?
Ans: Impulse of a Force is the product of force and the time for which the force
acts. According to Newton’s 2nd Law,
F = ma or F = m(v-u)/t or Ft = mv – mu
i.e. Impulse of a Force=Change in momentum.
Impulse of a Force is defined as the Change in momentum produced by the force

Q. 29.State the Law of conservation of momentum and prove it for an isolated system.
Ans: The sum of momenta of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. i.e. In an isolated system, the total momentum remains conserved.
Proof: Suppose two objects (two balls A and B, say) of masses m A and m B are traveling in the same direction along a straight line.
Let the velocities uA and uB, respectively and Let uA > uB.
There are no other external unbalanced forces acting on them.
The two balls collide with each other for a time t,
Then, the force on ball B by ball A(action)= the force on ball A(reaction) by ball B

Q.30. A revolver of mass 500 g , fires a bullet of mass 10g with a speed of 100 m/s. Find (i)momentum of the bullet. (ii) Initial momentum of revolver and bullet as a system. (iii) Recoil velocity of the revolver.
Ans: Before the bullet was shot the system of bullet and revolver was at rest.

Mass of the revolver = 500 g = 0.5 Kg Mass of the bullet = 10 g = .01 Kg

Velocity with which the bullet is fired = 100 ms-1

(i) Momentum of the bullet after being shot = .01 X 100 = 1 Ns

(ii) Initial momentum of the system = 0

(iii) Let v1 be the recoil velocity of the revolver.

Final momentum of the system is, momentum of the bullet + momentum of the revolver = 1 + 0.5v1

Applying momentum conservation, 0 = 1 + 0.5v1 Or, v1 = -1/0.5 = -2 ms-1

So, the velocity of the revolver is -2 ms-1, the negative sign indicates that the direction of velocity of the revolver is opposite to the direction of the bullet.

Rotatory Motion : In rotatory motion, the particles of the body describe concentric circles about the axis of

motion.

Vibratory Motion : In vibratory motion the particles move to and fro about a fixed point. Example Simple

Pendulum

Distance : The distance covered by a moving object is the actual length of the path followed by the object.

Distance is a scalar quantity. SI unit of distance is Metre .

Displacement is the shortest distance covered by a moving object from the point of reference (initial

position of the body), in a specified direction.

Note: But the displacement when the bus moves from A  B and then from B A is zero. SI unit of

displacement is metre. Displacement is a vector, i.e., the displacement is given by a number with proper

units and direction.

☼ When a body covers equal distances in equal intervals of time then the body is said to describe uniform

motion.

☼ When a body moves unequal distances in equal intervals of time or vice-versa, then the body is said to

describe non-uniform motion.

☼ Speed can be defined as the distance covered by a moving object in unit time

Speed = distance / time = s/t where S is the distance covered and t is the time taken. SI unit of speed is

m/s or m s

-1

. Speed is a scalar quantity.

Uniform Speed : An object is said to be moving with uniform speed if it covers equal distances in equal

intervals of time

Non-uniform : An object is said to be moving with variable speed or non-uniform speed if it covers equal distances in unequal intervals of time or vice-versa.

Average speed : When we travel in a vehicle the speed of the vehicle changes from time to time depending upon the conditions existing on the road. In such a situation, the speed is calculated by taking the ratio of the total distance travelled by the vehicle to the total time taken for the journey. This is called the average speed.

Instantaneous speed : When we say that the car travels at an average speed of 60 km/h it does not mean that the car would be moving with the speed of 60 km/h throughout the journey. The actual speed of the car may be less than or greater than the average speed at a particular instant of time.

The speed of a moving body at any particular instant of time, is called instantaneous speed.Velocity is defined as the distance travelled in a specified direction in unit time. The distance travelled in a specified direction is displacement.

Therefore, velocity can be defined as the rate of change of displacement .

Velocity is defined as the distance covered by a moving object in a particular direction in unit time or speed

in a particular direction.

DISTANCE TRAVELLED IN A SPECIFIC DIRECTION

VELOCITY= ——————————————————————–

TIME TAKEN

V=S/t [WHERE ‘S’ IS THE DISTANCE COVERED AND ‘t’ IS THE TIME TAKEN]

☼ SI unit of velocity is m/s (metre/second). [ SI unit of distance is metre and that of time is second] ☼

Velocity is a vector quantity.

Acceleration : When the train starts from rest its speed increases from zero and we say that the train is

accelerating. After sometime the speed becomes uniform and we say that it is moving with uniform speed that means the train is not accelerating. But as the train is nearing Mysore it slows down, which means the train is accelerating in negative direction. Again the train stops accelerating when it comes toa halt at Mysore .

Acceleration is defined as the rate of change of velocity of a moving body with time

☼ Acceleration = Rate of change of velocity with time

= CHANGE IN VELOCITY/ TIME; a=( v-u)/t

☼ The SI unit of velocity is m/s and time is s

☼ Acceleration is a vector quantit

SI UNIT OF ACCELERATION IS =m/s 2

Positive Acceleration If the velocity of an object increases then the object is said to be moving with positiveacceleration. Example: A ball rolling down on an inclined plane.

Negative Acceleration : If the velocity of an object decreases then the object is said to be moving withnegative acceleration. Negative acceleration is also known as retardation or deceleration. Example: (1) Aball moving up an inclined plane. 2) A ball thrown vertically upwards is moving with a negative accelerationas the velocity decreases with time

Zero Acceleration : If the change in velocity is zero, i.e., either the object is at rest or moving with

uniform velocity, then the object is said to have zero acceleration. Example: a parked car, a train movingwith a constant speed of 90 km/hr

Uniform Acceleration : If the change in velocity in equal intervals of time is always the same, then theobject is said to be moving with uniform acceleration. Example: a body falling from a height towards thesurface of the earth

v = (125kg m/s)/ 25 kg v = ( 125/25 )m/s v = 5 m/s

Q.26. Why is it easier to stop a tennis ball than a cricket ball moving with same speed ?

Q.27. A boy of 40kg runs from left to right at 5m/s & catches a ball of 200g coming from right to left at 10m/s .After taking the catch , find the speed of the boy with the ball in his hand.

Ans: Using conservation of momentum, m1u1 + m2u2 = m1v1+ m2v2

40 x 5 + 0.2 x 10 + 40 v1 + 0 Þ v1 = 5.05m/s

Mass of the revolver = 500 g = 0.5 Kg Mass of the bullet = 10 g = .01 Kg

Velocity with which the bullet is fired = 100 ms-1

(i) Momentum of the bullet after being shot = .01 X 100 = 1 Ns

(ii) Initial momentum of the system = 0

(iii) Let v1 be the recoil velocity of the revolver.

Final momentum of the system is, momentum of the bullet + momentum of the revolver = 1 + 0.5v1

Applying momentum conservation, 0 = 1 + 0.5v1 Or, v1 = -1/0.5 = -2 ms-1

So, the velocity of the revolver is -2 ms-1, the negative sign indicates that the direction of velocity of the revolver is opposite to the direction of the bullet.

Rotatory Motion : In rotatory motion, the particles of the body describe concentric circles about the axis of

motion.

Vibratory Motion : In vibratory motion the particles move to and fro about a fixed point. Example Simple

Pendulum

Distance : The distance covered by a moving object is the actual length of the path followed by the object.

Distance is a scalar quantity. SI unit of distance is Metre .

Displacement is the shortest distance covered by a moving object from the point of reference (initial

position of the body), in a specified direction.

Note: But the displacement when the bus moves from A  B and then from B A is zero. SI unit of

displacement is metre. Displacement is a vector, i.e., the displacement is given by a number with proper

units and direction.

☼ When a body covers equal distances in equal intervals of time then the body is said to describe uniform

motion.

☼ When a body moves unequal distances in equal intervals of time or vice-versa, then the body is said to

describe non-uniform motion.

☼ Speed can be defined as the distance covered by a moving object in unit time

Speed = distance / time = s/t where S is the distance covered and t is the time taken. SI unit of speed is

m/s or m s

-1

. Speed is a scalar quantity.

Uniform Speed : An object is said to be moving with uniform speed if it covers equal distances in equal

intervals of time

Non-uniform : An object is said to be moving with variable speed or non-uniform speed if it covers equal distances in unequal intervals of time or vice-versa.

Therefore, velocity can be defined as the rate of change of displacement .

Velocity is defined as the distance covered by a moving object in a particular direction in unit time or speed

in a particular direction.

DISTANCE TRAVELLED IN A SPECIFIC DIRECTION

VELOCITY= ——————————————————————–

TIME TAKEN

V=S/t [WHERE ‘S’ IS THE DISTANCE COVERED AND ‘t’ IS THE TIME TAKEN]

☼ SI unit of velocity is m/s (metre/second). [ SI unit of distance is metre and that of time is second] ☼

Velocity is a vector quantity.

Acceleration : When the train starts from rest its speed increases from zero and we say that the train is

Acceleration is defined as the rate of change of velocity of a moving body with time

☼ Acceleration = Rate of change of velocity with time

= CHANGE IN VELOCITY/ TIME; a=( v-u)/t

☼ The SI unit of velocity is m/s and time is s

☼ Acceleration is a vector quantit

SI UNIT OF ACCELERATION IS =

Positive Acceleration If the velocity of an object increases then the object is said to be moving with positive

acceleration. Example: A ball rolling down on an inclined plane.

Negative Acceleration : If the velocity of an object decreases then the object is said to be moving with

negative acceleration. Negative acceleration is also known as retardation or deceleration. Example: (1) A

ball moving up an inclined plane. 2) A ball thrown vertically upwards is moving with a negative acceleration

as the velocity decreases with time

Zero Acceleration : If the change in velocity is zero, i.e., either the object is at rest or moving with

uniform velocity, then the object is said to have zero acceleration. Example: a parked car, a train moving

with a constant speed of 90 km/hr

Uniform Acceleration : If the change in velocity in equal intervals of time is always the same, then the

object is said to be moving with uniform acceleration. Example: a body falling from a height towards the

surface of the earth

CBSE Important Questions For Science Class IX(physics)

CBSE ADDA Science class 9th

Q.1. what is persistence of hearing? (1)

Q.2. what does the slope of speed-time graph indicate? (2)

Q.3. what is one Newton? (1)

Q.4. why will a sheet of paper fall lower than one that is crumpled into a ball? (2)

Q.5. A stone is dropped from the edge of a roof.

How long does it take to fall 4.9m?
How fast does it move at the end of that fall?
How does it move at the end of 7.9m?
What is its acceleration after 1 second and after 2 second? (4)

Q.6. A boy and a girl do the same work in 5 min and 10 min respectively. Which of these has more power and why? (1)

Q.7. (a) What name has been given to the force with which two objects lying a part attract each other?

(b) What happens to the gravitational force between two objects when the distance between them is:
(i) doubled
(ii) half (2)

Q.8. (a) what is the condition for a force to do work on a body?
(b) A boy throws a rubber ball vertically upwards. What type of work, positive or negative is done.
(i) By force applied by boy?
(ii) By the gravitational force of earth? (3)

Q.9. name the characteristics of sound which depends on:
(i) Amplitude
(ii) waveform

Q.10. give one use of ultrasound in industry and one in hospitals? (1)

Q.11. A sonar station picks up a return signal after 3seconds. How far away is object? (speed of sound in water=1440m/s) (1)

Q.12. Name that part of ear which vibrates when outside sound fall on it? (1)

Q. 13. When the passenger m oving in a bus , the road side trees appear to b e moving ———————————- (a) Back ward (b)For ward

Q. 14. In uniform motion the object covers ———— (a) equel distance (b) un equal distance

Q. 15. Where objects cover un equal distances in equal interval of time then the motion is called —————- (a) uniform motion (b) non uniform motion

Q. 16. The average speed = =A car accelerate = (a) Total time taken /total distance traveled (b) Total distance traveled/total time taken

Q. 17. v = (a) t/s (b) s/t

Q. 18. An object travels 16m in 4sec and another 16 m in 2 sec. what is the average speed of the object———– (a) 5.33 ms-1 (b) 5. 34 ms-1

Q. 19. If we specify , its direction of motion along with its speed , the quantity that specifies with the aspects is called————- (a) velocity (b) motion

Q. 20. Automobiles are fitted with a device that show the distance traveled . such a device is known ———- (a) odometer (b) Speedo meter

Q. 21. The velocity of an object can be ————– (a) non uniform or variable (b) uniform or variable

Q. 22. Average velocity = (a) initial velocity + final velocity/ 2 (b) initial velocity + final velocity/1

Q. 23. Average velocity = (a) placement/ total time taken (b) Displacement/ total time taken

Q. 24. Acceleration = (a) change in velocity/ time taken (b) change in motion / time

Q. 25. S = (a) ut + 1/2at² (b) ut + 1 at²

Q. 26. A car accelerate uniformly from 18 km h^-1 to 36 km h^-1 in 5 sec . calculate the acceleration————- (a) 1ms^-2(b) 1 ms^-3

Q. 27. A train starting from rest attaind a velocity of 72 km h^-1 in 5 minutes . Assuming that the acceleration is uniform. Find the distance traveled by the train for attaing this velocity ——————————— (a) 2km (b) 5 km

Q . 28. Define SI unit of force.(1)

Q . 29. A body is moving with a uniform speed of 15 m/s on a circular track. Why is its motion called acceleration motion.(1)

Q . 30. What is the distance travelled and displacement of a body as its moves from A to B via C.(2)

Q . 31. A motor boat starting from rest on a lake accelerates in a straight line at a constant rate of 2m/s2 for 4 seconds. How far does the boat travels during this time.(2)

Q . 32. Name the force which keeps various planets in their orbits. Write down the formula for the force which exists between two bodies of mass m1 and m2 separated by a distance R from each other.(2)

Q . 33. State Newton’s second law of motion. A ball is thrown on a smooth surface with a force and it stops after some time. Is it supporting First law of motion. Give reasons for your answers.(2)

Q . 34. Velocity-time graph of a body is given above. Answer the following questions (3)

Which part of the graph shows uniform acceleration.
Which part of the graph shows uniform retardation.
Calculate the total distance travelled by the body.
Which part of the graph shows zero acceleration.(3)

Q . 35.

Derive the first equation of motion v=u + at graphically.
Write the equation describing the principle of conservation of momentum.
Why does the Gun recoil when a bullet is fired from it. (5)

9TH Motion Numerical

1. An airplane accelerates down a runway at 3.20 m/s2for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
2. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall?
3. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.
4. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2 Determine the time for the feather to fall to the surface of the moon.
5. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is the distance that the sled travels?
6. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.
7. An engineer is designing the runway for an airport. Of the planes that will us e the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s.Assuming this minimum acceleration, what is the minimum allowed length for the runway ?
8. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration).
9. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.
10. If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?
11. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel ofthe rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).
12. A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reac hes its peak.(Hint: the time to rise to the peak is one-half the total hang-time.)
13. The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.
14. A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)
15. A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.
16. It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.
17. A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.
18. A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration (assume uniform) of the dragster.
19. With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.9TH Motion Numerical
Solutions
1. . d = 1720 m
2. a = 8.10 m/ s2
3.V =-25.5 m/s (- indicates direction)
4. a = 11.2 m/s2, d = 79.8 m
5. t = 1.29 s
6. a = 247 m/s2
7. d = 400 m
8. a = 0.712 m/s2
9. d = 704 m
10. d = 28.6 m
11. vi = 7.17 m/s
12. vi = 5.03 m/s
13. hang time = 1.03 s
14. a = 1.62*105m /s2
15. vi = 30.6 m/s , d = 47.9 m
16. t = 8.69 s
17. a = -1.08*16m /s2
18. d = -57.0 m
19. vi = 47.6 m /s
20. a = 2.86 m/s2
21. t = 30. 8 sa = 15.8 m/s2
22. vi = 42.3 m/s , vi = 94.4 mi/hr
23.

SCIENCE-IX-X

FOR

ANIMAL KINGDOM

If you are given a specimen, what are the steps that you would follow to classify it?

Answer

There is a certain common fundamental feature that helps in classification of living organisms. The features that can be used in classification are as follows.

(i)LEVEL OF CLASSIFICATION

CELLULAR LEVEL

TISSUE LEVEL

ORGAN LEVEL

(ii)BODY CAVITY PRESENT

ABSENT

(iii)TYPE OF BODY SYMMETRY RADIAL

BILATERAL

(iv)TYPE OF COELOM DEVELOPMENT ACOELOM

PSEUDOCOELOM

TRUE COELOM

(v) TRUE COELOM ENTEROCOELOM

SCHIZOCOELOM

On the basis of above features, we can easily classify a specimen into its respective category.

Question 3:

How useful is the study of the nature of body cavity and coelom in the classification of animals?

Answer

Coelom is a fluid filled space between the body wall and digestive tract. The presence or absence of body cavity or coelom plays a very important role in the classification of animals. Animals that possess a fluid filled cavity between body wall and digestive tract are known as coelomates. Annelids, mollusks, arthropods, echinodermates, and chordates are examples of coelomates. On the other hand, the animals in which the body cavity is not lined by mesoderm are known as pseudocoelomates. In such animals, mesoderm is scattered in between ectoderm and endoderm. Aschelminthes is an example of pseudocoelomates. In certain animals, the body cavity is absent. They are known as acoelomates. An example of acoelomates is platyhelminthes.

Distinguish between intracellular and extracellular digestion?

Answer

Intracellular digestion		Extracellular digestion
1.	The digestion of food occurs within the cell.	1.	The digestion occurs in the cavity of alimentary canal.
2.	Digestive enzymes are secreted by the surrounding cytoplasm into the food vacuole.	2.	Digestive enzymes are secreted by special cells into the cavity of alimentary canal.
3.	Digestive products are diffused into the cytoplasm.	3.	Digestive products diffuse across the intestinal wall into various parts of the body.
4.	It is a less efficient method.	4.	It is a more efficient method of digestion.
5.	It occurs in unicellular organisms.	5.	It occurs in multicellular organisms.

What is the difference between direct and indirect development?

Answer

	Direct development		Indirect development
1.	It is a type of development in which an embryo develops into a mature individual without involving a larval stage.	1.	It is a type of development that involves a sexually-immature larval stage, having different food requirements than adults.
2.	Metamorphosis is absent.	2.	Metamorphosis involving development of larva to a sexually-mature adult is present.
3.	It occurs in fishes, reptiles, birds, and mammals.	3.	It occurs in most of the invertebrates and amphibians.

What are the peculiar features that you find in parasitic platyhelminthes?

Answer

Taenia (Tapeworm) and Fasciola (liver fluke) are examples of parasitic platyhelminthes.

Peculiar features in parasitic platyhelminthes are as follows.

1. They have dorsiventrally flattened body and bear hooks and suckers to get attached inside the body of the host.

2. Their body is covered with thick tegument, which protects them from the action of digestive juices of the host.

3. The tegument also helps in absorbing nutrients from the host’s body.

What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?

Answer

The phylum, Arthropoda, consists of more than two-thirds of the animal species on earth. The reasons for the success of arthropods are as follows.

i. Jointed legs that allow more mobility on land
ii. Hard exoskeleton made of chitin that protects the body
iii. The hard exoskeleton also reduces water loss from the body of arthropods making them more adapted to terrestrial conditions.

Question 8:

Water vascular system is the characteristic of which group of the following:

(a) Porifera (b) Ctenophora (c) Echinodermata (d) Chordata

Answer

Water vascular system is a characteristic feature of the phylum, Echinodermata. It consists of an array of radiating channels, tube feet, and madreporite. The water vascular system helps in locomotion, food capturing, and respiration.

Question 9:

“All vertebrates are chordates but all chordates are not vertebrates”. Justify the statement.

Answer

The characteristic features of the phylum, Chordata, include the presence of a notochord and paired pharyngeal gill slits. In sub-phylum Vertebrata, the notochord present in embryos gets replaced by a cartilaginous or bony vertebral column in adults. Thus, it can be said that all vertebrates are chordates but all chordates are not vertebrates.

Question 10:

How important is the presence of air bladder in Pisces?

Answer

Gas bladder or air bladder is a gas filled sac present in fishes. It helps in maintaining buoyancy. Thus, it helps fishes to ascend or descend and stay in the water current.

What are the modifications that are observed in birds that help them fly?

Answer

Birds have undergone many structural adaptations to suit their aerial life. Some of these adaptations are as follows.

(i) Streamlined body for rapid and smooth movement

(ii) Covering of feathers for insulation

(iii) Forelimbs modified into wings and hind limbs used for walking, perching, and swimming

(iv) Presence of pneumatic bones to reduce weight

(v) Presence of additional air sacs to supplement respiration

Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?

Answer

The numbers of eggs produced by an oviparous mother will be more than the young ones produced by a viviparous mother. This is because in oviparous animals, the development of young ones takes place outside the mother’s body. Their eggs are more prone to environmental conditions and predators. Therefore, to overcome the loss, more eggs are produced by mothers so that even under harsh environmental conditions, some eggs might be able to survive and produce young ones. On the other hand, in viviparous organisms, the development of young ones takes place in safe conditions inside the body of the mother. They are less exposed to environmental conditions and predators. Therefore, there are more chances of their survival and hence, less number of young ones is produced compared to the number of eggs.

Segmentation in the body is first observed in which of the following:

(a) Platyhelminthes (b) Aschelminthes (c) Annelida (d) Arthropoda

Answer

The body segmentation first appeared in the phylum, Annelida (annulus meaning little ring).

Match the following:

(a) Operculum	(i) Ctenophora
(b) Parapodia	(ii) Mollusca
(c) Scales	(iii) Porifera
(d) Comb plates	(iv) Reptilia
(e) Radula	(v) Annelida
(f) Hairs	(vi) Cyclostomata and Chondrichthyes
(g) Choanocytes	(vii) Mammalia
(h) Gill slits	(viii) Osteichthyes

Answer

	Column I		Column II
(a)	Operculum	(viii)	Osteichthyes
(b)	Parapodia	(v)	Annelida
(c)	Scales	(iv)	Reptilia
(d)	Comb plates	(i)	Ctenophora
(e)	Radula	(ii)	Mollusca
(f)	Hairs	(vii)	Mammalia
(g)	Choanocytes	(iii)	Porifera
(h)	Gill slits	(vi)	Cyclostomata and Chondrichthyes

Prepare a list of some animals that are found parasitic on human beings.

Answer

S. No.	Name of organism	Phylum
1	Taenia solium	Platyhelminthes
2	Fasciola hepatica	Platyhelminthes
3	Ascaris lumbricoides	Aschelminthes
4	Wuchereria bancrofti	Aschelminthes
5	Ancyclostoma	Aschelminthes

PLANT KINGDOM

What is the basis of classification of algae?

Answer

Algae are classified into three main classes – Chlorophyceae, Phaeophyceae, and Rhodophyceae. These divisions are based on the following factors:

(a) Major photosynthetic pigments present

(b) Form of stored food

(d) Number of flagella and position of insertion

Class I – Chlorophyceae

Common name – Green algae

Major pigments – Chlorophylls a and b

Stored food – Starch

Cell wall composition – Cellulose

Flagella number and position – 28; equal and apical

Class II – Phaeophyceae

Common name– Brown algae

Major pigments – Chlorophylls a and c, and fucoxanthin

Stored food – Mannitol and laminarin

Cell wall composition – Cellulose and algin

Flagella number and position – 2; unequal and lateral

Class III – Rhodophyceae

Common name – Red algae

Major pigments – Chlorophylls a and b, and phycoerythrin

Stored food – Floridean starch

Cell wall – Cellulose, pectin, and polysulphate esters

Flagella number – Absent

Differentiate between the following:-

(i)red algae and brown algae

(ii) liverworts and moss

(iii) homosporous and heterosporous pteridophyte

(iv) syngamy and triple fusion

Answer

(i) Red algae and brown algae

Red algae		Brown algae
1.	Red algae are grouped under the class Rhodophyceae.	1.	Brown algae are grouped under the class Phaeophyceae.
2.	They contain floridean starch as stored food.	2.	They contain mannitol or laminarin as stored food.
3.	They contain the photosynthetic pigments chlorophylls a and d, and phycoerythrin.	3.	They contain the photosynthetic pigments chlorophylls a and c, and fucoxanthin.
4.	Their cell walls are composed of cellulose, pectin, and phycocolloids.	4.	Their cell walls are composed of cellulose and algin.
5.	Flagella are absent	5.	Two flagella are present

(ii) Liverworts and moss

Liverworts		Moss
1.	They have unicellular rhizoids.	1.	They have multicellular rhizoids.
2.	Scales are present very often	2.	Scales are absent
3.	They are generally thalloid, with dichotomous branching.	3.	They are foliage, with lateral branching.
4.	Gemma cups are present	4.	Gemma cups are absent
5.	Sporophyte has very little photosynthetic tissue	5.	Sporophyte has abundant photosynthetic tissue

(iii) Homosporous and heterosporous pteridophyte

Homosporous pteridophytes		Heterosporous pteridophytes
1.	They bear spores that are of the same type.	1.	They bear two kinds of spores – microspores and megaspores.
2.	They produce bisexual gametophytes.	2.	They produce unisexual gametophytes.

(iv) Syngamy and triple fusion

Syngamy		Triple fusion
1.	It is the process of fusion of the male gamete with the egg in an angiosperm.	1.	It is the process of fusion of the male gamete with the diploid secondary nucleus in an angiosperm.
2.	A diploid zygote is formed as a result of syngamy.	2.	A triploid primary endosperm is formed as a result of triple fusion.

How would you distinguish monocots from dicots?

Answer

Monocots and dicots can be differentiated through their morphological and anatomical characteristics.

Characteristic			Monocot	Dicot
Morphology
Roots			Fibrous roots	Tap roots
Venation			Generally parallel venation	Generally reticulate venation
Flowers			Trimerous flowers	Pentamerous flowers
Cotyledons in seeds			One	Two
Anatomy
No. of vascular bundles in stem			Numerous	Generally 2 – 6
Cambium			Absent	Present
Leaves			Isobilateral	Dorsiventral
	Column I		Column II
(a)	Chlamydomonas	(i)	Moss
(b)	Cycas	(ii)	Pteridophyte
(c)	Selaginella	(iii)	Algae
(d)	Sphagnum	(iv)	Gymnosperm

Answer

	Column I		Column II
(a)	Chlamydomonas	(iii)	Algae
(b)	Cycas	(iv)	Gymnosperm
(c)	Selaginella	(ii)	Pteridophyte
(d)	Sphagnum	(i)	Moss

Describe the important characteristics of gymnosperms.

Answer

Important features of gymnosperms:

1. The term gymnosperm refers to plants with naked seeds (gymnos – naked, sperma – seeds), i.e., the seeds of these plants are not enclosed in fruits.

2. The plant-body ranges from medium to tall trees and shrubs. The giant redwood tree Sequoia is one of the tallest trees in the world.

3. The root system consists of tap roots. The coralloid roots present in Cycas are associated with nitrogen-fixing cyanobacteria.

4. The stem can be branched (as in Pinus and Cedrus) or un-branched (as in Cycas).

5. The leaves can be simple (as in Pinus)or compound (pinnate in Cycas). The leaves are needle-like, with a thick cuticle and sunken stomata. These help in preventing water loss.

6. Gymnosperms are heterosporous. They bear two kinds of spores – microspores and megaspores.

7. Flowers are absent. The microsporophylls and megasporophylls are arranged to form compact male and female cones.

8. Pollination occurs mostly through wind and pollen grains reach the pollen chamber of the ovule through the micropyle.

9. The male and female gametophytes are dependent on the sporophyte.

10. The seeds contain haploid endosperms and remain uncovered.

———————————————————–BEST OF LUCK—————————————————————————-

IMPORTANT QUESTION FOR SA-II EXAM

GIVE REASON:—

1.A CAMEL WALKS EASILY ON A SANDY SURFACE THAN A MAN INSPITE OF THE FACT THAT A CAMEL IS MUCH HAVIOUR THAN A MAN.

2.RAILWAY TRACKS ARE LAID ON A LARGE SIZED WOODEN OR IRON OR CONCRETE SLEEPER.

3.IT IS COMFORTABLE TO WEAR THE FLAT- BOTTOMED SHOES THAN TO WEAR THE HIGH HEELED SHOES.

4.IT IS COMFORTABLE TO WALK ON THS SOFT GRASS OF LAWN THAN TO WALK ON STONE CHIPS.

5.THE TTIP OF THE NIDDLE IS MADE POINTED.

6.THE SCHOOL BAG HAS WIDE STRAP .

7.WIDE STEEL BELTS ARE PROVIDED OVER THE WHEELS OF ARMY TANK.

8. A SHARPY KNIFE CUTS OBJECTS MORE EFFECTIVELY THAN A BLUNT KNIFE.

9. THE BUILDINGS OR DAMS HAVE A WIDE FOUNDATIONS .

10.WHY DOES AN IRON SHIP FLOAT ON WATER AND AN IRON NAIL SINK IN WATER.

11.GLASS OR COPPER SINK IN WATER BUT FLOAT ON MURCURY.

12.A BLOCK OF PLASTIC RELEASED UNDER WATER COME UP TO THE SURFACE OF WATER .

13.A MUG FULL OF WATER FEELS LIGHTER INSIDE WATER.

14.WE CAN NOT TALK WITH OUR FRIEND ON MOON ,LIKE EARTH.

15. BATS ARE ABLE TO FLY AT NIGHT WITHOUT COLLIDING WITH OTHER OBJECTS.

16. LUNGS BECOME INFECTED MOST WHEN AN AIR BORNE INFECTION OCCURS.

17.YOU ARE MORE LIKELY TO SUFFER FROM INFECTIONS WHEN YOU TRAVEL BY BUS OR TRAIN FOR LONG HOURS.

FILL IN THE BLANKS:———————

1. SI UNIT OF FORCE IS ———-
2. 2 SI UNIT OF PRESSURE IS———————-
3. 3 SI UNIT OF THRUST IS ————————
4. 1 PASCAL IS ————————–.
5. A FORCE OF 100 N IS APPLIED ON AN OBJECT OF AREA 4 SQUARE METRE. THE PRESSURE WILL BE ——–Pa.
6. DUE TO —— FORCE ,A SOLID APPEARS LIGHTER IN WATER.
7. MASS OF A BODY IS EQUAL TO THE PRODUCT OF ITS VOLUME AND ———– OF THE MATERIAL OF THE BODY.
8. MASS PER UNIT VOLUME IS CALLED ————-.
9. RELATIVE DENSITY OF THE MATERIAL OF A BODY =DENSITY OF THE MATERIAL OF THE BODY

——————————————————–

DENSITY OF THE———————————-

10.BUOYANT FORCE IS EQUAL IN MAGNITUDE TO THE WEIGHT OF THE———————————————LIQUID.

11.THE APPARENT WEIGHT OF AFLOATING BODY IS———————————————.

12.THE SI UNIT OF WORK IS ————-.

13.THE SI UNIT OF ENERGY IS ——————–.

14. THE SI UNIT OF POWER IS ————————.

15. 1 HP = ———————WATTS.

16.THE RATE TO DO WORK IS CALLED …………………

17.THE SUM OF KINETIC AND POTENTIAL ENERGY IS CALLED ———————–.

18.DURING FREE FALL THE POTENTIAL ENERGY OF A BODY ——————– AND KINETIC ENERGY ——————–TILL IT REACHES THE EARTH SURFACES.

19.A COMPRESSED SPRING HAS ———– POTENTIAL ENERGY.

20.POWER = FORCE * (——————).

21. POTENTIAL ENERGY =m*(______)*h.

22.THE COMMERCIAL UNIT OF ENERGY IS —————————–.

23.IN SOLAR CELL, SOLAR ENERGY IS DIRECTLY CONVERTED INTO ———————— ENERGY.

24. IN AN ELECTRIC BULB, ELECTRICAL ENERGY IS CONVERTED INTO ———————– ENERGY.

25.SOUND IS A FORM OF ——————- WAVE MOTION.

26. SOUND IS PRODUCED BY A ————————BODY.

27. THE SPEED OF SOUND IS ——————– AT ZERO DEGREE CELCIOUS.

28.THE RANGE OF AUDIBILITY OF HUMAN IS 20 Hz TO ———————-.

29.HIGH FREQUENCY SOUND WAVES HAVE ——————— WAVELENGTH.

30. REFLECTED SOUND IS CALLED————————–.

31.A WAVE OF VERY SHORT DURATION IS CALLED ———————-.

32.THE METHOD USED FOR DIAGONOSTING THE DIFFERENT INNER PART OF A HUMAN BODY WITH THE HELP OF ULTRASOUND IS KNOWN AS —————————————.

———————————————————————————————————————————————

33.THE GEOGRAPHICAL DISTRIBUTION OF LIVING ORGANISMS IS KNOWN AS ——————————.

34.PLANTS ARE AUTOTROPHIC ,WHEREAS ANIMALS ARE————————————.

35.PLANTS BEARING FLOWERS AND SEEDS ARE GROUPED UNDER SUB-KINGDOM ———————————-.

36.ALGAE ARE ———————- BEARING LOWER PLANT.

37. YEAST IS AN WXAMPLE OF ———————– FUNGI.

38. ————————- ARE REGARDED AS AMPHIBIANS OF THE PLANT WORLD.

39.CYCAS IS ALSO CALLED AS —————————-.

40. THE LOCOMOTION OF AMOEBA IS WITH THE HELP OF —————————-.

41. SPONGES HAVE A SINGLE BODY CAVITY CALLED ——————————.

42.CORALS ARE TINY ——————–.

43.THE ————————- INCLUDE THE COMMON EARTHWORM.

44.VERTEBRATES ARE ALSO CALLED AS ———————.

45.ROHU IS A ——————— FISH.

46. THE EARLIEST KNOWN BIRD IS ——————.

47. IN MAMMALS, THE RED BLOOD CELLS ARE WITHOUT ———————.

48. LSD IA S ——————-DRUG.

49. NIGHT BLINDNESS DISEASE IS CAUSED DUE TO DEFICIENCY OF ———————.

50. THE FULL FORM OF AIDS IS —————————————————————————–.

51. MALERIA IS CAUSED BY A PROTOZOAN PARASITE, NAMED AS ———————————–.

52. BCG VACCINE IS GIVEN FOR THE TREATMENT OF —————————.

53. RABIES DISEASE IS ALSO KNOWN AS —————————–.

54.RATIO BETWEEN MASSES OF CARBON AND OXYGEN IN CO2 is ——————————-.

55.——————–and ————————are the example of polyatomic ions.

56. —————————————-is avogadro’s constant.

57.mass of 1 mole of oxygen atom is —————————–.

58. MOLAR MASS OF CO2 IS ————————————–.

59.FORMULA OF ALUMINIUM CHLORIDE IS —————————.

60. FORMULA OF MAGNESIUM HYDROXIDE IS —————————.

61.NAME O0F THE COMPOUND Al2(SO4)3 is —————-

64.THE ELECTRON HAS ————-CHARGE EQUAL TO———————.

65.THE MASS OF AN ELECTRON IS —————gm.

66.ELECTRON WAS DISCOVERED BY ——————————AND PROTON WAS BY ——————————.

67.THE ELEMENTARY PARTICLE NOT PRESENT IN H-ATOM IS —————————–.

68.THE MASS OF AN ELECTRON IS ——————————– OF THE MASS OF 1 H-ATOM.

69.MOST OF THE MASS OF AN ATOM IS CONCENTRATED IN A SMALL REGION CALLED————————–.

70.NUCLEUS OF AN ATOM WAS DISCOVERED BY ———————————-.

71.IN A NEUTRAL ATOM ———————–ARE EQUAL TO THE NUMBER OF ——————-.

72.ATOMIC NUMBER IS EQUAL TO THE NUMBER OF —————————.

73.AN ELEMENT HAS MASS 23 ;THE NUMBER OF ELECTRON IN IT IS 11 ,THE NUMBER OF NEUTRON WILL BE EQUAL TO THE ————————.

74.THE ELECTRONIC CONFUGERATION OF Ca (20) IS ——————,———————-,——————–,—————-.

75.NUMBER OF VALENCE ELECTRON IN ion are ————————.

76.RUTHERFORD’S ALPHA-PARTICLE SCATTERING EXPERIMENT WAS RESPONSIBLE FOR THE DISCOVERY OF ———–77.ISOTOPES OF AN ELEMENT HAVE SAME NUMBER OF ——————.

78.THE AVERAGE ATOMIC MASS OF A SAMPLE OF AN ELEMENT X IS 16.2 u. THE PERCENTAGE OF ISOTOPE IN THE SAMPLE ARE ————————- AND —————————-.

79.IF Br ATOM IS IN THE FORM OF SAY TWO ISOTOPES THEN ATOMIC MASS OF Br-ATOM IS —————————————————-.

80.IF K-SHELL AND L-SHELLS OF AN ATOM ARE COMPLETELY FILLED THEN NAME OF THE ATOM IS ——————-.

81.IF ATOMIC NUMBER OF Na IS 11 THEN ATOMIC NUMBER AND ATOMIC MASS OF IS——————-AND————–.

82.LIMITATIONS OF J.J. THOMSON MODEL OF ATOM ARE ———————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————.

NUMERICAL PROBLEM

1THE VOLUME OF A SUBSTANCE IS 20 .IF THE DENSITY OF WATER IS 1 . the substance will ————-2. THE VOLUME OF A 500 gm SEALED PACKET IS .WILL THE PACKET FLOAT OR SINK IN WATER IF THE DENSITY OF WATER IS ?WHAT WILL BE THE MASS OF WATER DISPLACED BY THE PACKET?

3. FIND THE energy possessed by an object of mass 10 kG when it is at a height of 6m above the ground .Given , .

4.Define 1 Watt of Power.

5.A lamp consumes 1000J of electrical energy in 10 sec.What is its power?

6.Certain force acting on a body of mass 20 Kg changes its velocity from . Calculate the Work done by the force.

7.A certain household has consumed 250 units of energy during a month. How much energy is this in a joules?

8.An electric heater is rated 1500W . How much energy does it use in 10 hours?

9.Calculate the work required to be done to stop a car of 1500 Kg moving at a velocity of 60 Km/h ?

10.How much work is done to lift a 2 Kg masses to a height of 10 m?(g= )

11.What is the work done when you apply a 5 N forece on a wall?

12.Two masses in the ratio 1:10 are dropped from the same height . Which will lose more potential Energy?

13.When the velocity becomes 3 times the original velocity , What is the new K.E.?

14.Which will bring more changes — Doubling the speed or doubling the mass ?

15.How much joules of energy are consumed by a 1000W heater used for 2 hours?

16.Calculate the power of an engine which can lift 300 Kg of water through a vertical height of 15 m in 15 sec.

17.An electric motor used 40 j of energy in 5 sec.What is its power?

18.Relate 1 kWh with joule.

19.A force of 10 N moves a body with a constant speed of 2 m/s .What is the power spent?

20.A man carries a 10 Kg mass without altering its height .What is the value of work done against gravity?

21.Is work is a scalar or vector?

22.If 1 joule of work is done in lifting a 0.5 kg of mass .How much high will it rise?

23.A car and a truck have the same speed of 30 m/s .If their masses are in the rario 1:3 ,find the ratio of K.E.

24.Two masses in the ratio 1:4 have their speed in the ratio 4:5 ,find the kinetic Energy?

25.A man of 50 kg jumps to a height of 3 m.What is the maximum P.E.that he will have?

26.When the speed of a body is tripled ,What is the change in its kinetic energy?

27.A pair of bullock exerts a force of 140N on a plough. The field ploughed is 15m long. How much work is done in ploughing the length of the field?

28.An object of mass 15 kg is moving with a uniform velocity of 4 m/s .What is the K.E possessed by the object?

29.What is the work to be done to increase the velocity of a car from 30 km/hr to 60 km/hr, if the mass of the car is 1500 kg?

30. The K.E.of an object of mass m,moving with velocity of 5 m/s is 25 J.What will be its K.E. when its velocity is doubled? What will be its K.E. when its velocity is increased three times?

31.An object of mass 12 kg is at a certain height above the ground. If the P.E. of the object is 480J ,find the height at which the object is with respect to the ground.Given

32.a BOY OF MASS 50 KG RUNS UP a staircase of 45 steps in 9 sec.If the height of each step is 15 cm,find his power. Take .

33.An electric bulb of 60W is used for 6h per dayu. Calculate the ‘units ‘ of energy conpsumed in one day by the bulb.

34.A mass of 10 kg is at a point A on a table.It is moved to a point B. If the line joining A and B is horizontal.WhAT IS THE WORK DONE ON THE OBJECT BY THE GAVITATIONAL FORCE?explain your answer.

35.A machine does 1960 J of work in 4 minutes. What is its power?

36.Calculate the energy possessed by a stone of mass 10 g kept at a height of 5m.

37.The heart does 1.5J of work in every beat.How many times per minute doesit beat if its power is 2 watt?

38.An engine supplies 196 J of energy . If it is used to raise ½ kg mass, how high can it be lifted?

39.A 2 kg mass is thrown up with a velocity of 30 m/s .After 2 sec. ,what is ita (I)P.E. (II)Mechanical enrgy

40.How much 0/0 variation in K.E.is seen when the momentum of a body is doubled?

41.How much should be the mass of a man who has K.E. of 625 j while moving with a speed of 5m/s

42.A man drops a10kg rock from the top of a 5m ladder. What is its K.E on reaching the ground? What is its speed just before it hits the ground?

43.A man whose mass is 50 kg climbs up 30 steps of the stairs in 30 sec.If each step is 20 cm high, calculate the power used in climbing the stairs.

44.If an electric bulb of 100W is lighted for 20 hrs everyday.How much is the electrical energy consumed?Find the cost if 1 unit is costing Rs 2.

45.100J of work was done to shift a body by 10m using a force of 20N. Find the angle of its application with the horizontal.

46.A man holding 20 kg masses remains at rest. Even though he is not doing any work, why does he feel tiered after sometimes?

47.The height of Qutab Minar is 75 m.How much work is done when a 50 kg man reach the top?

48.A man pulls a bucket of water of mass 5 kg from a well 10m deep in 20 s.Calculate the power used?

49.A woman pulls a bucket of water of total mass 5 kg from a well which is 10m deep in 10 sec.Calculate the power used by her.()

50.A boy pushes the cart through a distance of 20m by applying a force of 50N .How much work is done by the boy?

51.Two girls, each of weight 400N climbs up a rope through a height of 8m.We name one of the girlA and the other B.Girl A takes 20s while B takes 50 sec to accomplish this task.What is the power expended by each girl?

52.An object of mass 40 kg is raised to a height of 5 m above the ground. What is the P.E.? If the object is allowed to fall , find its K.E. when it is half-way down.

53.IN each of the following ,a force F is acting on an object of mass m.The direction of displacement is from west to east shown by the longer arrow.Observe the diagram carefully and state whether the work done by the force is negative,positive or zero. West————————– east

54.A horse exerts a pull on a cart of 300N. This makes the horse –cart system moves with a uniform speed of 18km/hr on a level road.Calculate the power developed by the horse in watt and also find its equivalent in horse power.

55.Calculate the electricity bill amount for a month of february 2012 ,if the following devices are used as specified:

(I) 4 bulbs of 50W for 6 hrs

(II)3 tube lights of 40 W for 8 hrs

(III)A T.V.of 100 W for 6 hrs

(IV)A refrigerator of 300W for 24 hrs.

The cost per unit is Rs.4.75.

56.The figure below shows two paths to shift a 5 kg body from point A to point B.IN which case work done is more?Give reason for your answer.

57. If a 60 W bulb,a1500W A.C. and a 15W table lamp are used in a house hold for 6 hrs a day. Find the energy consumed in (I)Joule (II)KWh

58.A human heart beats 75 times a minute. Calculate its frequency.

59.A source of wave produces 40 compressions and 40 rarefactions in 0.4 sec.Find its frequencies.

60.A child hear an echo from a cliff 4 sec after the sound from a powerful cracker is produced. How far is the child from the cliff?(velocity-330m/sec)

61.If the speed of small ripples on w nater is 0.1 sec and their wavelength is 5 cm, what is the speed of the wave?

62.Sound travels with a speed of 330m/s.What is the wavelength of sound whose frequency is 550 Hz.

63.Calculate the wavelength of sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.

64.A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound.What is the time interval between successive compressions from the source?(speed of sound in air =330m/s)

65.A person clapped his hands near a cliff and heard the echo after 5 s.What is the distance of the cliff from the person if the speed of the sound ,v is taken as 346m/s?

66.An echo returned in 3 s.What is the distance of the reflecting surface from the source,(V=342m/s)

67.A ship sends out ultrasound that returns from the sea bed and is detected after 3.42 s. If the speed of ultrasound through sea water is 1531m/s,what is the distance of the sea bed from the ship?

68.A person has a hearing range from 20 Hz to 20KHz.What are the typical wavelengths of sound waves in air corresponding to these two frequencies?(V=344m/s)

69.Two children are at opposite ends of an aluminium rod..One strikes the end of the rod with a stone.Find the ratio of time taken by the sound wave in air and in aluminium to reach the second child.

70.A stone is dropped from the top of a tower 500m high into a pond of water at the base of tower. When is the splash heared at the top ? and V=340m/s

71.A source produces sound waves of wavelength 1.2m in air. If the same is produced in water, the wave length is 0.8 m .If the velocity of sound in air is 330m/s, find its velocity in water.

72.If an observer standing between two cliffs receives echo at 1.5 s and 4 s after clapping,find the distance between the cliffs if the velocity of sound is 320 m/s.

73.A stone is dropped into a well whose water layer is at a depth of 80 m.What will be the time interval from the time of dropping the stone, when the sound of splash is heard? (V in air =320 m/s)

74.A bat can hear sound at frequency upto 120kHz.Determine the wavelength of sound in air at this frequency. (v=344 m/s)

——————————————————————————————————————————————————

Human Health And Disease

1.list two disease caused due to deficiencies of nutrients.

2. mention any two disease caused by hormonal deiciency.

3. Define Vector of a disease. Give suitable example.

4. define (I)an antigen (b) an antibody (III) immunization

5 . Name three disease that can be prevented by using safe drinking water.

6.DPT acts against which disease.

7. What are differences between communicable and non-communicable disease?

8.What is vaccination?Mention names of any two vaccines and their use.

9.MATCH

(I) Iodine dificiency	(A)diabetes
(II) insulin	(B)sezually transmitted
(III)mantoux test	© goiter
(IV)syphilis	(D)tuber culosis
(V)plasmodium	(E)jaundice
(VI)hepatitis virus	(F)malaria
(VII)vitamin C	(g) EDWARD JENNER
(viii)FATHER OF IMMUNIOLOGY	(H) scurvy

Natural resources

1 what is soil erosion?

2.what is the green house effect

3.How is ozone layer formed?

4.which component of air is essential for animals?

5.which radiation is absorbed by ozone?

6.Name any two g ases produced by the combustion of fossil fuels.

7.Name few alternative sources of energy other than conventional fossil fuels.

8.Which radiation is absorbed by carbon di oxide?

9.How does the ozone layer protect us from harmful effects of the environment?

10.What destructive effects do the chloro fluorocarbons brings about in the atmosphere?

11.what is global warming?

12.Name some abiotic factors of biosphere.

13.what causes acid rain?

14.what is the role of nitrogen fixing bacteria in the atmosphere?

15.draw a schematic labeled diagram of carbon cycle. State the importance of carbon cycle in nature.

16. Draw a labeled diagram of nitrogen cycle in nature.

17.What is Biogeochemical Cycle?Explain the pathway of carbon on Nitrogen cycle with the help of a labeled diagram.

18.How does energy enter the biosphere?

19.Name one natural and one man – made process by which RETURNS TO THE atmosphere.

20 In the following bio- geochemical cycle , name and define the processes marked as (X),(Y)AND (z).

HOW DO WE FALL ILL

1. WHAT IS IMMUNISATION?
2. NAME A SEXUALLY TRANSMITTED DISEASE CAUSED BY VIRUS.

3 . N AME TWO DISEASE SPREAD BY MOSQUITO.

4 . GIVE THE CAusitive agents of tb and typhoid.

5.name two disease caused by—-(I) virus (II)bacteria (III)protozoa

6. what disease is AIDS? How does it spread from one person to another?

7.State difference between acute and chronic disease.

8.what are the different means by which infectious diseases are spread?

ANIMAL KINGDOM/ QUESTION- ANSWER/

THE ANIMAL KINGDOM

SKELTON SYSTEM

NERVOUS SYSTEM

MUSCULAR SYSTEM

DIGESTIVE SYSTEM

CARDIO VASCULAR SYSTEM

VERTEBERATES

INVERTEBERATES

ANIMAL KINGDOM

http://www.excellup.com/InterBiology/animalkingdom.aspx CLASSIFICATION OF ANIMAL KINGDOM

http://kinzgirl588.webs.com/animalclassification.htm/ANIMAL CLASSIFICATION

Animal Classification

Animal Classification

The Animal Kingdom is made up of two groups known as Vertebrates, (animals with backbones) and Invertebrates, (animals without backbones). The Vertebrate group includes fish, amphibians, reptiles, birds, and mammals. The Invertabrates group include creatures such as worms, sponges, mollusks, and insects.

DIVERSITY IN THE LIVING WORLD

DIVERSITY IN THE LIVING WORLD II (PLANT)

http://kinzgirl588.webs.com/animalclassification.htm )

MOLE-CONCEPT/ATOMIC STRUCTURE/THRUST/ALGAE

Niels Bohr Atomic TheorY

An atom is made up of three particles, electrons protons and neutrons. Electrons have a negative charge and protons have a positive charge whereas neutrons have no charge. They are neutral. Due to the presence of equal number of negative electrons and positive protons, the atom as a whole is electrically neutral.
The protons and electrons are located in a small nucleus at the centre of the atom. Due to the presence of protons, the nucleus is positively charged.
The electrons revolve rapidly around the nucleus in fixed circular paths called energy levels or shells. The ‘energy levels’ or ‘shells’ or ‘orbits’ are represented in two ways: either by the numbers 1, 2, 3, 4, 5 and 6 or by letters K, L, M, N, O and P. The energy levels are counted from centre outwards.
Each energy level is associated with a fixed amount of energy. The shell nearest to the nucleus have minimum energy and the shell farthest from the nucleus have maximum energy.
There is no change in the energy of electrons as long as they keep revolving in the same energy level. But, when an electron jumps from a lower energy level to a higher one, some energy is absorbed while some energy is emitted. When an electron jumps from a higher energy level to a lower one, the amount of energy absorbed or emitted is given by the difference of energies associated with the two levels. Thus, if an electron jumps from orbit 1 (energy E₁) to orbit 2 (energy E₂), the change in energy is given by E₂ – E₁. The energy change is accompanied by absorption of radiation energy of E = E₂ E₁ = h where, h is a constant called ‘Planck’s constant’ and is the frequency of radiation absorbed or emitted. The value of h is 6.626 x 10^–³⁴ J^–^s. The absorption and emission of light due to electron jumps are measured by use of spectrometers.

This model of the atom was able to explain the stability of the atom. It also explained the phenomenon of atomic spectra and ionization of gases

Sub Topics

Niels Bohr Atomic Theory

Rutherford’s model had a major drawback. It could not explain why ultimately electrons did not fall into the nucleus by taking a spiral path. This was in concurrence with the electromagnetic theory that states “if a charged particle undergoes accelerated motion, then it must radiate energy (lose) continuously”.In order to explain the stability of an atom, Neils Bohr gave a new arrangement of electrons in the atomin 1913. According to Neils Bohr, the electrons could revolve around the nucleus in only ‘certain orbits’ (energy levels), each orbit having a different radius. When an electron is revolving in a particular orbit or particular energy level around the nucleus, the electron does not radiate energy (lose energy) even though it has accelerated motion around the nucleus.

Arrangement of energy levels around the nucleus

Top

Structure of an atom

In this page we are going to discuss about explain the structure of an atom concept .Electrons revolve around the nucleus in different energy levels or shells and each shell is associated with definite energy. The energy of the K shell is the least while those of L, M, N and O shells increases progressively. We also know that any system that has least energy is the most stable.1^st energy level is K shell2^nd energy level is L shell3^rd energy level is M shell4^th energy level is N shell and so on

Bohr and Bury Scheme – Important Rules

Maximum number of electrons that can be accommodated in a shell is given by 2n² where n = shell number

For 1^st energy level, n = 1Maximum number of electrons in 1^st energy level = 2n²2 x (1) 2 = 2

For 2^nd energy level n = 2Maximum number of electrons in the 2^nd energy level = 2n²2 x 2² = 2 x 4 = 8

For 3^rd energy level n = 3Maximum number of electrons in the 3^rd energy level = 2n²= 2x(3) ²= 2 x 9 = 18

For 4^th energy level n = 4Maximum number of electrons in the 4^th energy level = 2n²= 2x(4) ²= 2×16 = 32

Sl.No Electron Shell Maximum Capacity

1 K Shell 2 electrons

2 L Shell 8 electrons

3 M shell 18 electrons

4 N shell 32 electrons

The outermost shell of an atom cannot accommodate more than 8 electrons, even if it has a capacity to accommodate more electrons. This is a very important rule and is also called the Octet rule. The presence of 8 electrons in the outermost shell makes the atom very stable.

Electronic Configuration of an Element

The arrangement of electrons in the various shells/orbits/energy levels of an atom of the element is known as electronic

configuration. Keeping the Bohr and Bury rules in mind let us write the electronic configuration of elements.

Electronic Configurations of Some Important Elements

Element	Symbol	Atomic number	Electronic configuration (or Electron arrangement) K L M N
Hydrogen	H	1	1
Helium	He	2	2
Lithium	Li	3	2,1
Beryllium	Be	4	2,2
Boron	B	5	2,3
Carbon	C	6	2,4
Nitrogen	N	7	2,5
Oxygen	O	8	2,6
Fluorine	F	9	2,7
Neon	Ne	10	2,8
Sodium	Na	11	2,8,1
Magnesium	Mg	12	2,8,2
Aluminium	Al	13	2,8,3
Silicon	Si	14	2,8,4
Phosphorus	P	15	2,8,5
Sulphur	S	16	2,8,6
Chlorine	Cl	17	2,8,7
Argon	Ar	19	2,8,8
Potassium	K	19	2,8,8,1
Calcium	Ca	20	2,8,8,2

Geometric Representation of Atomic Structure

Below are the examples on Geometric Representation of Atomic Structure

Example: 1

24 Mg 12

Steps:

The first 2 electrons will go to the 1^st shell = K Shell (2n²)
The next shell L takes a maximum of 8 electrons (2n²)
In this way 2 + 8 = 10 electrons have been accommodated. The next 2 electrons go to the M Shell.

K	L	M
2	8	2

Atomic parameters of the first twenty elements

Name and Symbol	Atomic no.(Z)	Mass no.(A)	Sub Atomic Particals	Electronic configuration				Valency	Metal/Non-metal or Noble gases and Nature of element
				K	L	M	N
Hydrogen (H)	1	1	p = 1 e = 1n = 0	1				±1	Metal as well as Non-Metal (Gas)
Helium(He)	2	4	p=2e =2n =2	2				2	Noblegas (Gas)
Lithium (Li)	3	7	p = 3e =3n =4	2	1			+1	Metal (Solid)
Beryllium (Be)	4	9	p = 4e=4n=5	2	2			+2	Metal (Solid)
Boron(B)	5	11	p=5e =5n = 6	2	3			+3	Metal (Solid)
Carbon (C)	6	12	p = 6e =6n =6	2	4			4	Non-metal (Solid)
Nitrogen (N)	7	14	p = 7e =7n =7	2	5			-3	Non-metal (Gas)
Oxygen (0)	8	16	p = 8e =8n =8	2	6			-2	Non-metal (Gas)
Fluorine (F)	9	19	p=9e=9n=10	2	7			-1	Non-Metal(Gas)
Neon(Ne)	10	20	p=10e=10n=10	2	8			0	Noble Gas(Gas)
Sodium (Na)	11	23	p = 11e =11n = 12	2	8	1		+1	Metal (Solid)
Magnesium (Mg)	12	24	p = 12e =12n = 12	2	8	2		+2	Metal (Solid)
Aluminium (Al)	13	27	p = 13e =13n = 14	2	8	3		+3	Metal (Solid)
. Silicon (Si)	14	28	p = 14e =14n = 14	2	8	4		4	Non- Metal (Solid)
Phosphorus (P)	15	31	p = 15e =15n = 16	2	8	5		-3	Non- Metal (Solid)
Sulphur (S)	16	32	p = 16e = 16n = 16	2	8	6		-2	Non- Metal (Solid)
Chlorine (Cl)	17	35	p = 17e = 17n = 18	2	8	7		-1	Non- Metal (Gas)
Argon(Ar)	18	40	p=18e = 18n = 22	2	8	8		0	Noblegas (Gas)
Potassium (K)	19	39	p = 19e = 19n = 20	2	8	8	1	+1	Metal (Solid)
Calcium (Ca)	20	40	p = 20e = 20n = 20	2	8	8	2	+2	Metal (Solid)

Atomic structures of the First Twenty Elements

Early Theories of Atomic Structure

Thomson’s Plum Pudding & Rutherford’s Nuclear Models of the Atom

Feb 24, 2008
Paul A. Heckert

J. J. Thomson and Ernest Rutherford pioneered studies of atomic structure, but their early models of the structure of the atom had serious flaws.

Thomson’s Plum Pudding Model of the Atom

In 1897, J. J. Thomson announced his discovery of the electron and the fact that atoms must therefore have some structure. At the time not all scientists were convinced that atoms even existed, but Einstein‘s 1905 explanation of Brownian motion provided convincing evidence that atoms were real.

Thomson’s model of atomic structure is often called the plum pudding model. He postulated that the negatively charged electrons, he had recently discovered, were scattered throughout a cloud of positive charge, like the plums in plum pudding.

The plum pudding model could not however predict why atoms absorbed and emitted spectral lines. The real deathblow to the plum pudding model came from experiments showing that atoms have nuclei.

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Rutherford’s Scattering Experiments

In 1909 Ernest Rutherford, one of Thomson’s students, conducted the first nuclear scattering experiments. Rutherford and his students scattered alpha particles, now known to be helium nuclei, off the atoms in a sheet of gold foil. Rutherford’s scattering experiments showed that atoms consist of a positively charged nucleus surrounded by negatively charged electrons.

Rutherford therefore envisioned the electrons orbiting the nucleus in a matter analogous to the planets orbiting the Sun. The electrical forces between protons and electrons follow Coulomb’s law, which has the same mathematical form as Newton’s law of gravity. Hence negatively charged electrons orbiting a positively charged nucleus in a manner analogous to the solar system is a very reasonable idea.

Problems with Rutherford’s Model of Atomic Structure

Despite Rutherford’s indisputable evidence for a very small atomic nucleus, physicists of the time had problems with the solar system analogy of atomic structure.

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The density of the nucleus would have to be unimaginably high. It in fact is.

The positively charged nucleus should fly apart because of the electromagnetic repulsive forces between positive charges, whereas in the plum pudding model the negative charges mixed in with the positive charges could hold the atom together. Physicist later discovered two types of nuclear forces: strong and weak. The strong nuclear force overwhelms the electromagnetic force within atomic nuclei and holds the nuclei together.

There were however two much more serious problems with Rutherford’s solar system analogy of atomic structure.

The orbiting electrons should emit electromagnetic waves as predicted by Maxwell’s equations for electromagnetic phenomena. This loss of energy should cause the electrons’ orbits to decay until they crashed into the nucleus. That returns us to the plum pudding model.
Rutherford’s model was unable to explain atomic emission and absorption line spectra. Why do atoms produce spectra at specific discrete wavelengths? Rutherford was unable to explain why spectral emissions and absorptions were at discrete wavelengths rather than continuous wavelength bands.

These problems were finally resolved by Niels Bohr and the Bohr model of the hydrogen atom.

THRUST AND PRESSURE

The perpendicular force acting per unit area on a surface is called pressure.

The total force acting on a body perpendicular to its surface is called thrust.

SI UNIT OF PRESSURE IS N/m 2 or pascal( Pa). Other unit of pressure is bar and Torr.

1 bar=10 5Pa

1 Torr = 1 mm Hg
1Atm=760 mm Hg at sea level

1.A concrete block measuring 2m*1m*o.5m has mass of 3000 kg.Calculate the pressure exerted when the block is placed with different faces different faces on a horizontal plane ground.Find the thrust in each case. (g=10 m 2) 2.The relative density of silver is 10.5. Find its density.

3.The weight of solid in air ,water, and liquid are 20N,16N,and 17N respectively . Find the r elative density of the liquid. Ans-0.75

4. A solid body of mass 100gm. And volume 140 cubic Cm is placed in water.Will it float or sink? Ans- Float

5. How much will a body of weight 60N weigh in water if it displaces 1 litre of water? (g=9.8m/s 2) Ans—50.2 N

6.The crown of Hiero weighted 9.072 kg.Archimedes found that weight of crown in water was 8.505 kg.The crown was made of gold and silver .What was the mass of gold and silver in the crown? (Relative density of gold=19.3 and relative density of silver =10.5) Ans—-6839.4375g and 2232.5625g

7.The pressure of earth on a person is minimum when he is (a) sitting (b) lying (c) standing on the leg (d) standing on the both legs

8.When the base area of the applied force is doubled then the pressure will be—(a)double (b) halved (c) one quarter (d) four times

9.A body floats in a liquid if the possible buoyant force is (a) zero (b) less than its weight (c) greater than its weight (d) none

10.The foundation of building or dam is increased to decrease ———————-

11Due to the ————–force , a solid appears lighter in water.

12.Why have school bags wide stripe?

13. State Archemedis principle. Give its application

14.state the meaning of 1 pascal.

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	kvlibrary on BIRBHUM PHOTO
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	ANIRUDDHA BHAWSINKA on BIRBHUM PHOTO
	ANIRUDDHA BHAWSINKA on BIRBHUM PHOTO