Class 9 Physics Solved Paper 2019

Class 9 Physics 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


(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´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 ´ 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.

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 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?

                                                Excluded from Syllabus for Annual Exam 2021

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

                                                Excluded from Syllabus for Annual Exam 2021

(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.

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(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 = h = 20´18 = 360 cm = 360/100 m = 3.6 m
Time = t = 20 s
Power = P =?                            

                                                                      Using formula

(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 upthrust of the liquid. It is given by
                                                                                    F = V p g
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 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?

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

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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)

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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 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.     
When a body is in stable equilibrium, its center of gravity is at the lowest position. When it is tilted, its centre of gravity rises. It returns to its stable state by lowering its centre of gravity. A body remains in stable equilibrium as long as the centre of gravity acts through the base of the body.
A book lying on the table.

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

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

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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)

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b.   A container has 2.5 litres 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)

                                                Excluded from Syllabus for Annual Exam 2021

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