Physics (English And Urdu) - Solved Model Paper 2022 - For Class IX (Science Group)

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Physics (English And Urdu)
Solved Model Paper 2022
For Class IX (Science Group)

Solution
Section 'B'

Q.2: Write down the names of any three renewable energy sources and any three nonrenewable energy sources?
Ans: Renewable Energy Sources:

1. Solar energy
2. Wind energy
3. Tidal energy and
4. Geothermal energy

Nonrenewable Energy Sources:

1. Coal
2. Petroleum and
3. Natural gases

Q.3: Write any three advantages and three disadvantages of friction.
Ans: ADVANTAGES OF FRICTION:

1. Friction enables us to walk on ground.
2. Friction protects from sliding, as sand is thrown to maintain friction on inclined railway tracks during rain.
3. The car brakes slow down the car to stop safely.
4. Threads and grooves are designed on tyres to increase the friction and improve grip between road and wheel.
5. Now vehicles are equipped with Anti-lock Braking System (ABS). ABS is designed to maintain steering stability, improve vehicle control, avoid skidding and decreases stopping distances on dry and slippery surfaces. The ABS maintains the static friction as the wheel starts slipping it releases the brake automatically for a fraction of a second then holds wheel again to create static friction between road and tyres.

DISADVANTAGES OF FRICTION:

1. A large amount of energy is wasted in the machines due to friction.
2. Friction leads to wear and tear of parts hence increases the service cost.
3. Failure of oil pump in car engine results contact between dry metals which yields high temperature hence the car engine is seized.
4. Due to friction, surface destroys.
5. Excess friction makes difficult to move object.

(Note: Write down any 3 for each as mentioned in question)

Q.4: Define equilibrium and State the two conditions for equilibrium?
Ans: EQUILIBRIUM:
When a body does not possess any acceleration neither linear nor angular it is said to be in equilibrium.
For example:

• A book lying on table in rest.
• A paratrooper moving downwards with terminal velocity.
• A chair lift hanging on supporting ropes.

CONDITIONS FOR EQUILIBRIUM:
A body must satisfy certain conditions to be in equilibrium. There are two conditions for equilibrium:

1. First Condition for Equilibrium
According to this condition for equilibrium:
"Sum of the all forces acting on a body must be equal to zero."

OR

"First condition for equilibrium is satisfied if net force acting on a body is zero."
Mathematically:

ΣF_y = 0

2. Second Condition For Equilibrium:
"Second condition for equilibrium is satisfied if sum of clockwise torques acting on a body is equal to the sum of the anticlockwise torques."
Mathematically:

Σ𝜏 = 0

Q.5: Define measurement, which is more accurate Vernier Caliper or a Screw Gauge and why?
Ans: MEASUREMENT:
It is the determination of the size or magnitude of something. It is explained with the help of describing the mathematical relations between various physical quantities.

Screw Gauge is more accurate than Vernier Caliper:
As compare to Vernier calipers, screw gauge is more accurate and precise because a screw gauge can even measure dimensions smaller (such as the diameter of a wire or sphere) than those measured by a Vernier Caliper. The least count of Vernier calipers is 0.1 mm and for screw gauge it is 0.01 mm. So, a screw gauge can measure accurately up to one hundredth part of a millimeter and more accurate than a Vernier caliper.

Q.6: Write three point of differences between:
Mass and Weight OR
Heat and temperature
Ans: Difference Between Mass and Weight

Comparison Chart	MASS	WEIGHT
Definition	Mass is the quantity of matter in a body regardless of its volume or of any forces acting on it.	Weight is a measurement of the gravitational force acting on an object.
Effect of gravity	Mass is always constant at any place.	The weight of an object depends on the gravity at that place.
Unit of Measurement	Mass is measured in kilogram (kg).	Weight is measured in Newton (N).
Nature	Mass has no direction so it is scalar quantity.	Weight always directed downward towards the center of the earth so it is vector quantity.
Formula	Its formula is m = F / a.	Its formula is W = mg.
Balance used for measurement	Mass is measured using a pan balance, a triple beam balance, lever balance or electronic balance.	Weight is measured using a spring balance.

OR

Difference Between Heat and Temperature

S.NO.	Heat	Temperature
1.	It is the form of energy.	It is a degree of hotness of a body.
2.	Heat transfers from hot body to cold body as a result of difference of temperature between them.	It determines the direction of flow of heat from one body to the other body.
3.	Heat of a body is the sum of all kinetic and potential energy of all the molecules constituting the body	Temperature of a body is the average kinetic energy of its molecules.
4.	Heat can be measures by a calorimeter.	Temperature of a body is measured by a thermometer.
5.	Its SI unit is joule. Its other unit is calorie.	SI unit of temperature is Kelvin. Its other units are Celsius (C) and Fahrenheit (F).

(Note: write down any 3 differences as mentioned in question)

Q.7: State Hooke's law and derive equation F = kx?
Ans: HOOKE'S LAW:
Robert Hooke, an English scientist first described the mathematical pattern of stretching a spring. He observed the dependence of displacement or size of the deformation upon the deforming force or load.
Hooke's law states that:

"Within elastic limit, the displacement produced in the spring is directly proportional to the force applied."

In other words,
Hooke’s Law state that the extension in spring is proportional to the load applied to it, provided the limit of elasticity is not exceeded.

Mathematically:
If ‘F’ is the applied force and ‘x’ is the displacement (extension) in the spring then the equation for Hooke's law may be written as:

F ∝ x

or F = kx ....... (i)

Where, k is spring constant (stiffness of spring).

Q.8: What is the mass of a solid iron wrecking ball of radius 18 cm. if the density of iron is 7.8 gm/cm³?
Data:
Radius of iron ball = 18 cm
Density of iron is 7.8 gm/cm³
Mass of a solid iron ball =?

Solution:
Step 1: write known physical quantities with units and point out the quantity to be found.

• Density of iron ball = ρ = 7.8 gm/cm³ = 7.8 x 1000 kg/ m³
• Radius of iron ball = r = 18 cm = 18 x 10^-2 m = 0.18 m
• Volume of the iron ball = V = (4/3) x π x r³ 
  V = (1.33) x 3.14 x (0.18 m)³ 
  V = 0.024m³

Step 2: write down the formula and rearrange if necessary

m = ρ x V

Step 3: put the values in formula and calculate
Since mass of iron ball is m = ρ x V = (7.8 x 10³) x (0.024)
m = 187.2 kg
Answer: The mass Of iron ball = 187.2 kg

Q.No.9:

Q.10: A ball of mass 400 gram, strikes the wall of velocity 4m/sec. How much is the kinetic energy of the ball at the time it strikes the wall?

Q.11: The "X" and "Y" components of a force F are 6N and 8N respectively. Find the magnitude and direction of F.
Solution:
Data:

• Horizontal component 'X' = F_x = 6 N
• Vertical component 'Y' = F_y = 8 N
• Magnitude or Resultant of force = F = ?
• Direction of Resultant of force = θ = ?

Working Formula:

Q.12: 2 kg of copper requires 2050 J of heat to raise its temperature through 10°C. Calculate the heat capacity of the sample.
Solution:
Data:

• mass of copper = m = 2 kg
• Temperature = ΔT = 10°C
• Specific heat capacity of copper= c = 385 J/K.°C
• Heat Capacity = Δ Q = ?

Working Formula:

Calculation:

Δ Q = cmΔT

Δ Q = 385 x 2 x 10 = 7700 Joules.

Answer: Hence the heat capacity of the sample is 7700 J.

Q.No.13

Section 'C'

Q.14: Define "g" & Derive the equation h= V_i + ¹/₂gt²?
Ans: Acceleration Due To Gravity "g":

“The acceleration produces in a free falling body due to force of gravity is called acceleration due to gravity.”

Acceleration due to gravity is a constant and it is denoted by "g". Its value near the surface of earth is found to be 9.81 ms^-2. However for ease of calculation value of ‘g’ is approximated to 10 ms^-2.

Negative Gravitational Acceleration:
Gravitational acceleration is taken negative for objects moving upward direction. (i.e. -9.8 m/sec²)

Positive Gravitational Acceleration
Gravitational acceleration is taken positive for objects moving downward direction. (i.e. 9.8 m/sec²)

Derivation Of The Equation Of Newton's Second Law Motion For Acceleration Due To gravity
For the motion of bodies under the influence of gravity the equation of motion are slightly modified.
Where,
distance is taken as (S = h) and
acceleration is taken as g (a = g)

Suppose a body is moving with uniform acceleration due to gravity “a =g” during some time interval “t”, its initial velocity “v_i” changes and denoted as final velocity “v_f ”. It covers a distance “s =h” in this duration of time.
The second equation of motion determines the distance covered during some time internal “t”, while a body is accelerating from a known initial velocity.
Let us take the

Q.15: State Newton's Law of gravitation. Determine the mass of the earth using Newton's Law of Gravitation?
Ans: NEWTON'S LAW OF GRAVITATION:
Sir Isaac Newton, was one of the greatest scientist of the world. He made fundamental contributions not only to several branches of Physics (like optics and mechanics) but also to Astronomy and Mathematics. He formulated the laws of motion and law of Universal gravitation.

Statement Of Newton’s Law Of Gravitation:
Newton's law of universal gravitation states that:

"Everybody in the 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 centres."

MASS OF THE EARTH:
Mass of Earth can not be measured directly by placing it on any weighing scale. But it can be measured by an indirect method. This method utilizes the Newton's law of universal gravitation.
Let us consider a small ball is placed on the surface of Earth.

• m ⟶ Mass of the ball.
• M_E ⟶ Mass of Earth.
• G ⟶ Universal gravitational constant.
• R_g ⟶ Radius of earth; which is also the distance between the ball and centre of earth.

Then according to Newton's law of universal gravitation, the gravitational force F of the Earth acts on the ball is:

Whereas the force with which Earth attracts the ball towards its centre is equal to the weight of the ball.
Therefore,

F = W = mg ............. (ii)

Comparing equation (i) and (ii); we get:

Numerical values of the constants at right hand side of equation (iii) are:

g = 10 Nkg^-1

R_E = 6.38 x 10⁶m

G = 6.673 x 10^-11Nm²kg^-2

Substituting these values in equation (iii), we get:

M_E = 6.0 x 10²⁴kg.

Thus, mass of Earth is 6.0 x 10²⁴kg.

Q.16: Define Co-efficient of Linear expansion. Also prove α = ¹/₃β.
Ans: Co-efficient of Linear expansion:
It is a constant and is denoted by 'α'. It depends upon the nature of the material. The Co-efficient linear expansion of is thus define as:
"Increase in length per unit length per degree rise in temperature" It has unit ¹/_k or K^-1

α = ¹/₃β:
We know that
β is the coefficient of volumetric expansion and
α is the coefficient of linear expansion.
Consider a block with dimensions l₁", "b₁" and "t₁ as the length breadth and thickness respectively at initial temperature T₁.

So the initial volume of the block will be:
Initial Volume = length x breadth x thicness

v₁ = l₁b₁t₁ ----- (i)

Let the temperature of the body is raised to T₂. The final volume of the block become v₂. So,

The increase in volume = ΔV = v₂ - v₁

and the increase in temperature = ΔT = T₂ - T₁

Due to increase in temperature the length, breadth and thickness of the body become "l₂", "b₂" and "t₂ respectively.through.
So,

Final Volume = v₂ = l₂ x b₂ x t₂ ----- (ii)

As we know that
l₂ = l₁ (1 + αΔT)
b₂ = b₁ (1 + αΔT)
t₂ = t₁ (1 + αΔT)

Putting the above values in equation (ii), we get

v₂ = l₁ (1 + αΔT) x b₁ (1 + αΔT) x t₁ (1 + αΔT)

v₂ = l₁b₁t₁(1 + αΔT)³

From equation (i) v₁ = l₁b₁t₁

v₂ = v₁ (1 + αΔT)³

By using Formula:

(a + b)³ = a³ + 3a²b + 3ab² + b³

v₂ = v₁ {(1)³ + 3(1)²αΔT + 3(1)(αΔT)² + (αΔT)³}

v₂ = v₁ (1 + 3αΔT + 3(αΔT)² + (αΔT)³)

Since α is small, higher ordered terms involving (α)² and (α)³are neglected. We have:

v₂ = v₁ (1 + 3αΔT)

v₂ = v₁ + 3v₁αΔT

v₂ - v₁ = 3αv₁ΔT

ΔV = 3αv₁ΔT

Since

ΔV = β v₁ΔT

∴ β v₁ΔT = 3αv₁ΔT

β v₁ΔT = 3αv₁ΔT

β = 3α

Or α = ¹/₃β

Prove that co-efficient of linear expansion is one third of co-efficient of volumetric expansion.

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