Class 9 Physics Solved Paper 2019 Overseas

Class 9 Physics Solved Paper 2019 Overseas 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 Overseas

Section A

Q1. MCQs

Class 9 Physics Solved Paper 2019 Overseas
Section B


(i) Define the term prefixes and write four prefixes commonly used.

Ans. Prefixes:
The words or letters added before a unit and stand for the multiples or submultiples of that unit are known as prefixes. 
For example, kilo (k), mega (M), milli (m), and micro, etc.

(ii) Write the following in standard form:   
a. 11.68 x 10-27 m

b. 32 x 105

c. 725 x 10-5 kg

(a) 1168 x 10-27

= 1168 x 10-27

= 1.168 x 10-24

(b) 32 x 10-5 s

= 32 x 10-5

= 3.2 x 10-4

(c) 725 x 10-5 kg

= 725 x 10-5 kg

= 725 x 10-5 x 103 g    (k = 103)

= 725 x 10-2 g

7.25 g

(iii) Define gravitational acceleration. Give its SI unit and value.

Ans. Gravitational Acceleration
The acceleration produced in a freely falling body due to the attraction of the earth is called gravitational acceleration or acceleration due to gravity. It is denoted by ‘g’. 

Its SI Unit is ms-2 and its value is 9.8 ms-2 on the surface of the earth.

(iv) How are vector quantities important to us in our daily life?

Ans.    Vector quantities have direction. Without direction, we cannot find the location of an object. The whole GPS system is based on location for which we need to know the direction from a particular place. Therefore, vector quantities are very important in our daily life.

(v) A cyclist of mass 40 kg exerts a force of 200N to move his bicycle with an acceleration of   3ms-2. How much is the force of friction between the road and the tyres?

Sol. Mass = m = 40 kg
Force = F = 200 N
Acceleration = a = 3 ms-2 
Force of friction = f =?
We know that
Net force = Applied Force – Force of friction
ma = F – f
or  f = F – ma
f = 200 – (40) (3)
f = 200 – 120
f = 80 N

(vi) How does oiling the moving parts of a machine reduce friction?

Ans. Friction exists due to the interlocking of ups and downs on the surfaces. While oiling various parts of the machine, downs are filled by oil which makes them smooth and slippery. They can easily move against each other, and friction between them is reduced.

(vii) Name three objects that work by the turning effect of forces.

Ans. Following are the three objects that work by the turning effect of forces:
(i) Pencil Sharpener 
(ii) Water tap
(iii) Doorknob
(iv) Steering wheel

(viii) Why is there a need of second condition for equilibrium if a body satisfies first condition for equilibrium?

Ans. The first condition does not ensure that a body is in equilibrium.
Example: Consider a body is acted upon by two equal and opposite forces. Although the first condition of equilibrium is satisfied, still the body is not in equilibrium and it will rotate in the anticlockwise direction. Thus, there is a need for the second condition of equilibrium in addition to the first condition of equilibrium. 

(ix) Can you determine the mass of moon? If yes, then what do you need to know?

Ans. Like the mass of earth, we can also find the mass of the moon by the following equation:

                                                                        M= g R2 / G
For this purpose, we need to know:
i) The value of ‘g’ on the moon
ii) The radius of the moon

(x) On reaching top of a slope 6m high from its bottom, a cyclist has speed of 1.5 ms-1. Find the kinetic energy and the potential energy of the cyclist. The mass of the cyclist and his bicycle is 40kg.

Sol. Height of slope = h = 6 m
Speed = v = 1.5 ms-1
Mass = m = 40 kg
Kinetic Energy = K.E =?
Potential Energy = P.E =?
Using formula for Kinetic Energy
K.E = ½ m v2
K.E = ½ (40) (1.5)2
K.E = (20) (2.25)
K.E = 45 J

Now using the formula for Potential Energy
P.E = m g h
P.E = (40) (10) (6)
P.E = 2400 J

(xI) What is meant by efficiency of a system?

Ans. The efficiency of a system is the ratio of the required form of energy obtained from a system as output to the total energy given to it as input.
The efficiency of a machine is given by:

It is observed that the efficiency of a working system is always less than 100% because every system meets energy losses due to friction that causes heat and noise etc. However, an ideal system is that in which output energy is equal to the total energy given to it. In other words, its efficiency is 100%.

(xiI) Write three features of kinetic molecular model of matter.

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(xiiI) Why is water NOT suitable to be used in barometer?

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(xiv) Convert 100°F into the temperature on Celsius scale.

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(xv) Briefly explain why does land breeze blow from land towards sea?

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Class 9 Physics Solved Paper 2019 Overseas
Section C

Q3. a. On what factors does the accuracy in measuring a physical quantity depend?    (3)

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Q3. b. Derive S=Vit +½ at2    (4)

Ans.    Second Equation
S = vit + ½ at2
Consider the speed-time graph in which the initial velocity vi=OA, final velocity vf=BD, and BC=at. Also, AC=OD=t

Total distance covered       
S= (Area of rectangle OACD + Area of triangle ABC)
As area of rectangle OACD = OA x OD
= vi x t = vi t
Area of triangle ABC = ½ (AC x BC)
= ½ (t x at) = ½ at2
As S = Area of rectangle OACD + Area of Δ ABC
Therefore, S = Vit + ½ at2

Q3. a. A bullet of mass 20g is fired from a gun with a muzzle velocity 100 . Find the recoil of the gun if its mass is 5kg.   (3)

Sol.     Mass of the bullet = m = 20 g = 20/1000 kg = 0.02 kg
Velocity of the bullet = v = 100 ms-1
Mass of rifle = M = 5 kg
Velocity of rifle = V =?

According to the law of conservation of momentum:

MV + mv = 0
(5) V + (0.02) (100) = 0
5 V + 2 = 0
5 V = – 2
V = – 2 / 5
V = – 0.4 ms-1

Thus, the recoil of the gun will be – 0.4 ms-1.

Q4. a.  Differentiate between centre of mass and centre of gravity.  (2)

Ans. Centre of mass
Centre of mass of a system is such a point where an applied force causes the system to move without rotation.

Centre of gravity
A point where the whole weight of the body appears to act vertically downward is called centre of gravity of a body.

Q4. b. Calculate the value of g at an altitude of 1000 km. The mass of the earth is  6.0 x 1024 kg. The radius of the earth is 6400km.   (4)

Sol.  Height above the Earth = h =1000 km =1000 x 1000 m = 1 x 106 m

Mass of earth = Me = 6.0 x 1024 kg

Radius of earth = R = 6.4 x 106 m

Gravitational constant = G = 6.673 x 10-11 Nm2 kg-2

Acceleration due to gravity =gh =?

Q4. c. What are fossil fuels? Explain.   (4)

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Q5. a. Explain construction and working of hydraulic press.   (4)

Ans: Hydraulic Press
Hydraulic Press is a machine which works on the principle of Pascal’s Law. It consists of two cylinders of different cross-sectional areas. Each cylinder is fitted with water tight pistons joined with each other by means of pipe. When a force ‘F1’ is applied on a small piston of area ‘a’, the pressure produced by it is transmitted equally on the large piston. Therefore a large force ‘F2’ acts on the larger piston of area ‘A’. Thus

The pressure on small piston        P = F1 / a
The pressure on large piston         P = F2 / A
Comparing the two equations
F2 / A = F1 / a
Or F2 = (F1 x A) / a

As the force F2 is much greater than the force F1. Therefore, the Hydraulic Press systems working in this way are known as “Force multipliers”.

Q5. b. An electric heater supplies heat at the rate of  1000 Js-1. How much time is required to raise the temperature of 200g of water from 20°C to 90°C?   (3)

Sol. Rate of heat supplied by heater = P = 1000 Js-1
Mass of water = m = 200 g = 0.2 kg
Initial temperature = T1 = 20 °C = 20+273 = 293 K
Final temperature = T2 = 90 °C = 90+273=363 K
Change in temperature = ΔT = T2 – T1= 363 – 293 = 70 K
Specific heat of water = 4200 J kg-1 K-1
Time = t =?
Using formula
ΔQ = m c ΔT
ΔQ = 0.2 x 4200 x 70
ΔQ = 58800 J
Now we know that
Total Heat Supplied = Rate of Heat supplied x time
ΔQ = P x t
Or t = ΔQ / P
t = 58800 / 1000
t = 58.8 s

Q5. c. Which measures can be taken to save energy in houses?   (3)

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