Physics 9 Solved Paper 2018 Federal Board

Class 9 Physics Solved Paper 2018

Class 9 Physics Solved Paper 2018 for Federal Board is available in this post. You can see more FBISE solved past papers of class 9 on our website.

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Class 9 Physics Solved Paper 2018

Section A

Q1. MCQs

Class 9 Physics Solved Paper 2018
Section B


(i) On closing the stud and spindle of a screw gauge, the zero of the circular scale is behind the index line and the 8th division of the circular scale coincides with the index line. There are 50 divisions on the circular scale and the distance between two consecutive threads on the spindle is 0.5mm. Find its zero error and zero correction.

Sol: Number of divisions on the circular scale = 50
Pitch of the screw gauge = 0.5 mm

On closing the stud and spindle of a screw gauge:

Main scale reading             =          0 mm
Circular scale reading        =          8 x 0.01 mm
                                              =          0.08 mm

Zero error of the screw gauge  =          (0+ 0.08) mm
                                                       =          + 0.08 mm

Zero Correction of the screw gauge   = – 0.08 mm

(ii)  Define Atomic Physics and Nuclear Physics.

Ans. Atomic Physics: It is the study of the structure and properties of atoms.

Nuclear Physics: It deals with the properties and behaviour of nuclei and the particles within the nuclei.

(iii)    A cyclist completes half round of a circular track of diameter 636 m in 1.5 minutes. Find his velocity.

Sol.   Diameter of circular track = D = 636 m
Radius of circular track = r = D/2 = 636 / 2 = 318 m
Time taken   t = 1.5 min = 1.5 × 60 = 90 s
Distance covered      S = π × radius
                                        = 3.14 × 318 m = 999 m
Displacement            d = 2 r
                                        = 2 × 318 m = 636 m

(iv)   Define Inertia. Describe the factor on which it depends.

It is a property of a body due to which it resists any change in its state of rest or motion.
Factors on which Inertia Depends
Inertia depends on the mass of a body. Greater is the mass of a body greater is its inertia.

(v)   How much time is required to change 22 Ns momentum by a force of 20 N?

Sol.     Change in momentum = Pf – Pi = 22 Ns
Force = F = 20 N
Time = t =?
                                                                        We know that

(vi)   Can a small child play with a fat child on the sea-saw? Briefly explain now?

Ans:  Yes, a small child can play with a fat child on a see-saw. Since the fat child has a larger weight, therefore, he should be at a smaller distance from the center of the see-saw while the small child should be at a larger distance from the center of the see-saw.

(vii) On the surface of the earth, the weight of a boy is 400N but on a mountain peak, his weight is 360N. Calculate the value of ‘g’ on the mountain peak.

Sol.     Weight of boy on Earth = we = 400 N
Weight of boy on mountain peak = wm = 360 N
Value of g on Earth = ge = 10 ms-1
Value of g on mountain peak = gm =?

(viii)  State the law of gravitation.

Ans.  Law of Gravitation: This law states that “everybody in this universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers”. 

(ix) Make a flow diagram of energy conversion for a hydroelectric power station.

Ans. Energy Conversion Diagram for a Hydroelectric Power Station

(x) Describe Hooke’s Law.

Ans. Hooke’s Law
This law states that within the elastic limit, the strain produced in a body is directly proportional to the stress applied on it.

 (xi) Describe the factors on which the upthrust of a liquid depends.

Ans.    Upthrust is given by the formula
                                                            F = V ρ g
V = Volume of liquid displaced by the object
ρ = Density of liquid
Hence upthrust of a liquid depends upon the density of a liquid and the volume of liquid displaced
by an object.

(xii)  Why are a large number of slots made in radiators?

Ans. The larger the area, the greater will be the transfer of heat. Therefore, a large number of slots are made in radiators to increase the surface area to allow more radiation of heat in less time.

(xiii)  Convert 1 gcm-3 into kgm-3.

(xiv)  a.  Why does the temperature of a solid substance not increase during its fusion?

Ans. The temperature of a solid substance does not increase during its fusion because all the heat given to the solid is utilized in changing the state of a solid into a liquid.

b. Differentiate between boiling and condensation.

Ans. In boiling, we heat a substance present in a liquid state to change it into a gaseous state while in condensation we cool down a substance present in a gaseous state to change it into a liquid state.

(xv) a.  How does evaporation differ from vaporization?

Ans:  Vaporization: Vaporization is a transitional phase of an element or compound from a solid phase or liquid phase to a gas phase. It changes matter from one state or phase into another without changing its chemical composition.

Evaporation: Evaporation, wherein the transition from a liquid phase to a gas phase takes place below the boiling temperature at a given pressure, and it occurs on the surface.

Class 9 Physics Solved Paper 2018
Section C

Attempt any TWO questions. All questions carry equal marks.

Q3.  a.   i. Draw a distance-time graph for an object moving with variable speed. (2.5)
              ii. How can the slope of the graph be determined? (1.5)
             iii. What information does the slope provide? (2)

Ans. i. Distance-time graph for an object moving with variable speed

ii. The slope of the above graph at any point can be found from the slope of the tangent at that point.
For example                                               
The slope of the tangent at P = RS / QS
The slope of the tangent at P = 30m / 10 s = 3 ms-1        

iii. The slope provides the speed of the object. The speed is higher at instants when the slope is greater; speed is zero at instants when the slope is horizontal.

Q3. b. State and explain Newton’s first law of motion. (4)

Ans.   First Law of Motion
This law states that a body at rest remains at rest and a body in motion continues to move in a straight line, provided that no net force acts on it.

The first part of this law states that a body at rest remains at rest when no net force acts on it which is true. Because a book lying on a table remains at rest as long as no net force is applied on it. However, the second part of this law seems to be against our observation, because a ball rolling on the ground stops after covering some distance. On close observation, we come to know that the ball stops because of some external forces like the force of friction between the ball and the ground. The air resistance also opposes the motion.

Q4. a. Define three states of equilibrium. Explain these states with reference to the center of gravity. (2+2+2)

Ans. Stable Equilibrium: A body is said to be in stable equilibrium if after a slight 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 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.

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.
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.

Neutral Equilibrium: If a body remains in its new position when disturbed from its previous position, it is said to be in a state of neutral equilibrium.
In neutral equilibrium, all the new states in which a body is moved, are the stable states, and the body remains in its new state. In neutral equilibrium, the center of gravity of the body remains at the same height, irrespective of its new position.

Q4. b. A motorboat moves at a steady speed of 4 ms – 1. Water-resistance acting on it is 4000 N. Calculate the power of its engine. (4)

Solution:  Speed of the boat = v = 4ms – 1
Force = F = 4000 N
Power = P = ?
P = W / t
P = F.S / t
P = F v (S/t = v)
P = 4000 x 4
P = 16000 W
P = 16 x 103 W
P = 16 kW

Q5. a. Define Pascal’s Law and name any four working systems (machines) of its application. (1+2)

Ans:   Pascal’s Law: This law states that pressure exerted from outside on a liquid is transmitted equally in all directions. This law has some useful applications.

1. Hydraulic press is used to compress raw cotton and cloth into bails.
2. Hydraulic brake system is used in motor vehicles.
3. This law is also used at the service stations to lift the cars for washing.
4. Hydraulic Pumps also work on Pascal’s law.

Q5. b. What is a bimetal strip? Name any two instruments where the bimetal strip is used.  (1+1)

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:
1. Thermostats as an automatic switch to control the temperature of any device e.g. electric iron.
2. Thermometers to measure temperature in furnaces and ovens.

Q5. c.  How much heat is lost in an hour through a glass window measuring 2.0 m by 2.5 m when the inside temperature is 25 °C and that of the outside is 5°C, the thickness of the glass is 0.8 cm and the value of k for glass is  0.8 W m – 1 K – 1? (3.6×10 7 J) (5)

Solution:  Time = t = 1 hour = 3600 s
Thickness of glass = L = 0.8 cm = 0.8/100 = 0.008 m                      
Area of a glass window = A = 2.0 m x 2.5 m = 5 m2
Temperature outside the house = T1 = 25°C = 25 + 273 = 298 K
Temperature inside the house = T2 = 5°C = 5 + 273 = 278 K
Change in temperature = DT = T1 – T2 = 298 – 278 = 20 K
Value of conductivity for concrete = k = 0.8 W m – 1 – 1
Rate of conduction of thermal energy = Q/t = ?
Q/t = kA (T1 – T2) / L
Q =  kAt (T1 – T2) / L
Q / t = 0.8 x 5 x 3600 x 20 / 0.008 = 36,000,000 J = 3.6 x 10 7 J

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