Physics 9 Solved Paper 2019 Federal Board

Class 9 Physics Solved Paper 2019


Physics 9 Solved Paper 2019 for Federal Board is available in this post. You can see more FBISE solved past papers of class 9 on our website.


Class 9 Physics Solved Paper 2019

Section A

Q1. MCQs


Class 9 Physics Solved Paper 2019
Section B

Q2.

(i) Define ‘base quantities’ and ‘derived quantities’. Also, give one example of each.

Ans. Base Quantities
Base quantities are the quantities on the basis of which other quantities are expressed.     
Examples: length, mass, time

Derived Quantities
The quantities that are expressed in terms of base quantities are called derived quantities.
Examples: area, volume, speed


(ii)   Express the following quantities using prefixes:
a.  5000g
b. 2000000W
c. 52 x 10-10 kg

(a) 5000 g
Sol.  5000 g = 5 x 1000 g
= 5 x 103 g
= 5 kg   (∵ 103 = k)

(b) 2000 000 W
Sol. 2000 000 W = 2 x 1000 000
= 2 x 106 W
2 MW  (∵ 106 = M)

(c) 52 x 10-10 kg 
Sol. 52 x 10-10 kg = 52 x 10-10 x 103 g     (∵ k = 103)
= 52 x 10-7 g
= 5.2 x 10-6 g
= 5.2 mg   (∵ 10-6 = m)


(iii) Draw distance-time graphs for an object
a. At rest
b. Moving with constant speed
c. Moving with variable speed

Ans.  a. Distance time graph for an object at rest

b. Distance time graph for an object moving with constant speed

c.  Distance time graph for an object moving with variable speed 


(iv)   A body of mass 5 kg is moving with a velocity of 10 ms-1. Find the force required to stop it in 2 seconds.

Sol.    
Mass = m = 5 kg
Initial velocity = vi = 10 ms-1
Final velocity = vf = 0 ms-1
Time = t = 2 s
Force required to stop = F =?
                                                                        We know that


(v)  Friction is a necessary evil. Why?

Ans. Friction is a necessary evil because it is not desirable due to its so many disadvantages but at the same time it is necessary because without friction our lives will become miserable as we will not be able to write on the paper or walk on the ground etc.


(vi)   Two children are sitting on the see-saw such that they cannot swing. What is the net torque in this situation?

Ans. The net torque in this situation is zero because their weights are acting downwards due to which two opposite and equal torques are produced. SInce the net torque is zero therefore it satisfies second condition of equilibrium.


(vii)   What is the relation between stability and position of the center of mass?

Ans. The position of centre of mass of an object plays an important role in its stability. To make them stable, their centre of mass must be kept as low as possible. It is due to this reason, racing cars are made heavy at the bottom and their height is kept to be minimum.


(viii)  How can you say that gravitational force is a field force?

Ans. Gravitational force exists all around the earth, while field force exists all around anybody including earth. Therefore we can say that gravitational force is also a kind of field force.


(ix)  What is chemical energy? Explain briefly.

Ans. Chemical energy is present in food, fuels, and in other substances. We get other forms of energy from these substances during chemical reactions. The burning of wood, coal, or natural gas in the air is a chemical reaction that releases energy as heat and light. Electric energy is obtained from electric cells and batteries as a result of a chemical reaction between various substances present in them.

(x)   A girl carries a 10 kg bag upstairs to a height of 18 steps, each 20 cm high. Calculate the amount of work she has done to carry the bag. (g=10ms-2)

Sol.  Mass of bag = m = 10 kg
Distance = S = 20 x 18 = 360 cm = 360/100 m = 3.6 m
Time = t = 20 s
Work = W =?    
Using formula
Work = Force x Distance
W = F x S
W = mg x S (F=w=mg)
W = 10 x 10 x 3.6
W = 360 J


(xi)  The weight of a metal spoon in air is 0.48 N, its weight in water is 0.42 N. Find its density.

Sol.  Weight of the spoon = w1 = 0.48 N
Weight of spoon in water = w2 = 0.42 N
Density of water = r = 1000 kg m-3
Density of spoon = D =?


(xii)  What is up thrust? Explain the principle of floatation.

Ans: Upthrust
When an object is totally or partially immersed in a liquid, the net force acting on the object is called the upthrust of the liquid. It is given by
F = V p g
Where
V = Volume of liquid displaced by the object
P = Density of liquid

Principal of Floatation
This law states that when an object floats, it displaces a fluid having its weight equal to the weight of the object. According to this principle, the conditions of the floating bodies can be given as:
1. An object sinks if its weight is greater than the upthrust of liquid acting on it.
2. An object floats, if its weight is less than or equal to the upthrust. In this case, the object may be partially immersed in the liquid.


(xiii)  Why is water used in the cooling system of automobiles?

Ans.  Water has a large specific heat capacity. For this reason, it is used in the cooling system of automobiles. In an automobile, a large amount of heat is produced by its engine due to which its temperature goes on increasing. The engine would cease unless it is not cooled down. Water circulating around the engine maintains its temperature. Water absorbs unwanted thermal energy of the engine and dissipates heat through its radiator.


(xiv)  How does bimetallic strip work?

Ans. Bimetal Strip:
A bimetal strip consists of two thin strips of different metals such as brass and iron joined together. On heating the strip, brass expands more than iron. This unequal expansion causes bending of the strip in a curve.

The bimetal strip is used in the thermostat as an automatic switch to control the temperature of any device e.g. electric iron. Bimetal thermometers are used to measure temperatures, especially in furnaces and ovens.


(xv)  Deserts soon get hot during the day and soon get cold after sunset. Why?

Ans. Since the specific heat of sand is low therefore it gets hot or cold soon by taking the normal amount of heat provided by the sun during the day. In the daytime, it gets hot due to the heat of sunlight and after sunset when the temperature comes down then it gets cold due to a decrease in temperature.


Class 9 Physics Solved Paper 2019
Section C


Attempt any TWO questions. All questions carry equal marks.

Q3. a.  What is a physical balance? Explain. (03)

Ans: Physical balance: A physical balance is used in the laboratory to measure the mass of various objects by comparison.
It consists of a beam resting at the Centre on a fulcrum. The beam carries scale pans over the hooks on either side. An unknown mass is placed on the left pan. Find some suitable standard masses that cause the pointer to remain at zero on raising the beam.


b.  Derive 2aS = vf2 – vi2                                                                                               (04)

c.   A body has a weight of 20 N. How much force is required to move it vertically upwards with an acceleration of  2ms-2 ?                  (03)    

Sol.  Weight = w = 20 N
Acceleration = a = 2 ms-2
Force = F =?
First of all, we need to find mass m.
So using the formula for mass
m = w/g = 20/10 = 2 kg
Now we know that
Net force = Applied Force – Weight
ma = F – w
(2)(2) = F – 20
4 + 20 = F
F = 24 N


Q4.  a.   What is meant by stable and unstable equilibrium? Explain.                                (02+02)

Ans. Stable Equilibrium     
A body is said to be in stable equilibrium if after a light tilt it returns to its previous position.     
Explanation     
When a body is in stable equilibrium, its center of gravity is at the lowest position. When it is tilted, its center of gravity rises. It returns to its stable state by lowering its center of gravity. A body remains in stable equilibrium as long as the center of gravity acts through the base of the body.
Example
A book lying on the table.

Unstable Equilibrium
If a body does not return to its previous position when set free after the slightest tilt is said to be in unstable equilibrium.
Explanation
The center of gravity of the body is at its highest position in a state of unstable equilibrium. As the body topples over about its base (tip), its center of gravity moves towards its lower position and does not return to its previous position.
Example
A pencil balanced at its tip.


b.   Derive an expression for the orbital speed of an artificial satellite.                                        (04)

Ans. Consider a satellite of mass m revolving around the Earth at an altitude h in an orbit of radius ro with orbital velocity vo. The necessary centripetal force required is given by the equation:
Fc = mvo2 / ro
This force is provided by the gravitational force of attraction between the Earth and the satellite and is equal to the weight of the satellite w (mg.). Thus
Fc = w’ = mgh
or mgh= mvo2 / ro
or vo2 = gh ro
or vo = √ gh ro
as ro= R+h
vo = √ gh (R+h)
The above equation gives the velocity, that a satellite must possess when launched in an orbit of radius ro = (R + h) around the Earth. An approximation can be made for a satellite revolving close to the Earth such that R >> h.
R+h = R
and gh=g
∴ vo =√g R
A satellite revolving around very close to the Earth has a speed vo nearly 8 kms -1 or 29000 kmh-1 .


c.  A motorboat moves at a steady speed of 4ms-1. Water resistance acting on it is 4000N. Calculate the power of its engine.                                                                (02)

Sol. Speed of boat = 4 ms-1
Water-resistance = F = 4000 N
Power of engine = P =?
Using formula


Q5.a. Write a note on hydraulic braking system in vehicles. (03)

Ans. The braking systems of cars, buses, etc. also work on Pascal’s law. The hydraulic brakes allow equal pressure to be transmitted throughout the liquid. When the brake pedal is pushed, it exerts a force on the master cylinder, which increases the liquid pressure in it. The liquid pressure is transmitted equally through the liquid in the metal pipes to all the pistons of other cylinders. Due to the increase in liquid pressure, the pistons in the cylinders move outward pressing the brake pads with the brake drums. The force of friction between the brake pads and the brake drums stops the wheels.


b.   A container has 2.5 liters of water at 20°C. How much heat is required to boil the water?          (03)

Sol.  Volume of water = V = 2.5 litres
Mass of water = m = 2.5 kg
(Since density of water is 1000 kgm-3 or 1kgL-1 )
Specific heat of water = c = 4200 Jkg-1K-1
Initial temperature = t1 = 20 °C
Final temperature = t2 = 100 °C
Temperature Increase = Δt = t2 – t1= 100 – 20 = 80 °C  or  80 K
Since  Q = c  m  Δt
Q = 4200 x  2.5 x 80
or  Q  = 840, 000 J
Thus, the required amount of heat is 840 000 J or 840 kJ


c.  Write a note on application and consequences of radiation.  (03)

Ans. Different objects absorb different amounts of heat radiations falling upon them reflecting the remaining part. The amount of heat absorbed by a body depends upon the colour and nature of its surface. A black and rough surface absorbs more heat than a white or polished surface. Since good absorbers are also good radiators of heat. Thus, a black coloured body gets hot quickly absorbing heat reaching it during a sunny day and also cools down quickly by giving out its heat to its surroundings. The bottoms of cooking pots are made black to increase the absorption of heat from the fire.

Like light rays, heat radiations also obey laws of reflection. The amount of heat reflected from an object depends upon its colour and the nature of the surface. White surfaces reflect more than coloured or black surfaces. Similarly, polished surfaces are good reflectors than rough surfaces and the reflection of heat radiations is greater from polished surfaces. Hence, we wear white or light coloured clothes in summer which reflect most of the heat radiation reaching us during hot day. We polish the interior of the cooking and hot pots for reflecting back most of the heat radiation within them.

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