Science Formulae for Class 9

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Science formula for class 9 is considered to be the foundation for the next three years. For the students who wish to proceed into the field of science, class 9 physics holds utmost importance because its syllabus is carried forward into the coming years. It is of utmost importance to have its basics cleared to build a robust base for further advancements and topics including electric current, optics, magnetism, laws of motion, etc. 

Keyterms: Electric current, Optics, Magnetism, Laws of motion, Length conversion, Time Unit conversion, Energy Unit conversion, Force Unit conversion, Viscosity Unit conversion


Conversion of Units

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Length Conversion

1 mm  =  0.001 m

1 cm = 0.01 m

1 dm (1 decimeter) = 0.1 m

1 dcm (1 decameter) = 10 m

1 hm = 100 m

1 km = 1000 m

1 mile = 1.609 km

1 inch = 2.54 * 10?2 m

1 foot = 3.05 * 10?2 m

1 fermi = 10?¹? m

1 angstrom = 10?10 m

1 light year = 9.46 * 1015 m

Time Unit Conversion

1 min = 60 secs

1 hour = 3600 secs

1 day = 86400 secs

1 week = 604800 secs

1 year = 31536000 secs (365 days)

Energy Unit Conversion

1 erg = 10?? Joules

1 foot-pound = 1.356 Joules

1 kilowatt-hr = 3.6 * 106 Joules

1 eV = 1.602 * 1019 Joules

1 calorie = 4.186 Joules

1 litre atmosphere = 101.13 Joules

Force Unit Conversion

1 dyne = 10-5 Newton

1 pound = 4.448 Newton

Viscosity Unit Conversion

1 poise = 0.1 kg/m s

1 slug/ft = 47.9 kg/m s

Also Read: 


Force and Laws of Motion

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Speed = total distance travelled / time taken [S = d/t]

Velocity = displacement / time [v = x/t]

acceleration = change in velocity / time [a = Δv/Δt]

acceleration = a = final velocity - initial velocity / time

[a = (v-u) / t]

The three equations of motion: 

(i) v = u + at  Relation between velocity and time

(ii) s = ut + 1/2at2  Relation between displacement and time

(iii) v2 - u2 = 2as  Relation between displacement and velocity

Rate of change of momentum = Δp/t

where, Δp = m(v-u)

In the absence of an external force, an object will remain in its state of motion/rest i.e., maintain inertia [First law of Motion]

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object i.e., F = ma. [Second law of Motion]

For every action, there is an equal and opposite reaction [Third law of Motion]

Derivations of the three equations of motion:

v = u + at

Here, v is the final velocity, u is the initial velocity, a is acceleration and t is the time taken. 

Acceleration = change in the velocity / time taken

= (final velocity – initial velocity) / time taken

a = (v - u) / t

at = v – u

v = u + at

s = ut + ½ at2

We know that displacement = average velocity * time

Displacement = {(initial velocity + final velocity) / 2 } * time

s = {(u + v) / 2} * t

From the first equation of motion, we know that v = u + at

Substituting this value of v into this equation, we get

s = [ {u + (u + at)} / 2 ] * t

s = {(2u + at) / 2 } * t

s = {(2u / 2) + (at / 2)} * t

s = (u + ½ at) * t

s = at + ½ at2

v2 – u2 = 2as

From the first equation of motion, we know that v = u + at

v-u = at

t = (v - u) / a

displacement = average velocity * time

s = {(u + v) / 2} * t

s = {(v  + u) / 2 } * {(v  - u) / a }

s = (v² – u²) / 2a

2as = v2 – u2

v2 – u2 = 2as

Derivation of the Second Law of Motion:

The statement of this law states that:

F ∝ change in momentum / time taken

We know that change in momentum = mv – mu

Hence, the law becomes

F ∝ (mv – mu) / t

F ∝ m (v – u) / t

But (v -u) / t = acceleration

Hence,

F ∝ ma

F = kma [k is the constant of proportionality whose value is taken to be 1]

F = ma

This is the second law of motion.


Gravitation

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The force exerted by an object on another object is proportional to the product of their masses and is inversely proportional to the square of the distance between them.

F = G (M1*M2/d2)

Where G = 6.673 * 10-11 is Universal Gravitation Constant [Nm2Kg?2]

Derivation: 

From the statement, we can write that

F ∝ (M1 * M2) / d2

F = G (M1*M2/d2)  [Here, G is the constant of proportionality whose value is 6.673 * 10-11 Nm2kg?2]


Work and Energy 

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The work done is defined as the product of the force in the direction of the displacement and the magnitude of the object’s displacement.

W = (F cosθ) *s Joules 

When angle is 0, 

W = F * s

Energy is the ability of an object to do work. It has two primary forms – Kinetic and Potential. Energy stored is referred to as Potential Energy while that which we observe while in motion is called Kinetic energy.

K.E. = ½*( mv2 )

P.E. = mgh 

Also Read: 

Work – Energy Theorem

This theorem states that the total work done by all the forces combined is equal to the change in kinetic energy of the object. 

W = ΔK = K Final  - K Initial

Derivation of Kinetic Energy Formula:

We know that W = F Δs = ma Δs

Now, from the third equation of motion,

v2 = u2 + 2aΔs

aΔs = (v2 – u2) / 2

Combining these, we get

ΔK = m * ((v2 – u2) / 2)

ΔK = ½ mv2 – ½ mu2

We know that the kinetic energy of a body at rest is zero.

Hence, 

K = ½ mv2

Derivation of Potential Energy Formula

We know that W = Force * Displacement

Therefore, for a body of mass m that is being raised to a height h with g being the gravitational force acting on it,

W = mg * h

W = mgh

When the body is at rest, its Kinetic Energy is zero.

Hence, according to the work energy theorem, 

Potential Energy = mgh

P.E. = mgh


Sound

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A form of mechanical energy that is created by the vibration of objects via propagation through a medium

The number of oscillations performed per second are measured using frequency ν (Hz)

The time taken per oscillation is called Time Period (T)

The distance between two consecutive crests or falls is called the Wavelength (λ)

A few noteworthy relations are:

ν = 1 / T

Sound of Speed Wave v = ν λ 


Things to Remember

  • v = u + at
  • s = at + ½ at2
  • v2 – u2 = 2as
  • F = ma
  • The kinetic energy of a body at rest is zero.
  • All objects have a tendency to maintain inertia.
  • Gravitational force is the force between two bodies while the force of gravity is the force between the earth’s surface and any object.
  • Sound comes from vibrations which create sound waves.
  • Sound requires a medium to travel. Hence, it cannot propagate through vacuum. 

Also Read: 


Sample Questions

Ques 1. Find the force that is required to accelerate a car with a mass of 2000 kg at 4 m/s2. (3 mark)

Ans. With the usage of the Second law of motion i.e., F = ma

Given, a = 4 m/s2

m = 2000 kg

Therefore, F = m* a

= 2000 * 4

= 8000 N

Ques 2. A body that has a mass of 1 kg undergoes a change in velocity of 4 m/s in a time of 4 seconds. Find the force that is acting on the body. (4 mark)

Ans. Given,

Δv = 4 m/s, t = 4 s, m = 1 kg

Therefore, acceleration a = Δv / t = 4/4 = 1 m/s²

F = ma

= 1 * 1

= 1 N

Ques 3. Why is ‘G’ known as the Universal Gravitational Constant? (1 mark)

Ans. G is called the Universal Gravitational Constant because it is independent of the shape, size etc. of the bodies involved. Its value is constant and remains the same for all bodies.

Ques 4. How does the force of gravitation between two objects change when the distance between the two objects is brought to half of its original value? (2 marks)

Ans. F = G (M1*M2/d2)

Now that d = ½ d

F = G (M1*M2/d2)

F = G (M1*M2/(d2)2)

F = 4G (M1*M2/d2)

Therefore, F= 4F

Hence, if the distance is reduced to half, the force of gravitation will become 4 times the original force.

Ques 5. An object with a mass of 2 kg is thrown up at a speed of 10 m/s. Find the kinetic energy of the object and also its potential energy when it is at the highest point. (2 mark)

Ans. Given, m = 2 kg, v = 10m/s

K = ½ mv2

= ½ * 2 * 10 * 10

= 100 J

The potential energy of the object at the highest point is equal to the initial kinetic energy of the object = 100 J

Ques 6. Find the power of an electric motor which can lift 1000kg of water to store in a tank at a height of 15 m in 20 seconds. (2 mark)

Ans. Taking g = 10m/s2

Weight = 1000 * 10 = 10000 N

T= 20 s

Power = Work done / time taken

= mgh/t

= 10000 * 15 / 20 

= 7500 W

Ques 7. The frequency of a source of sound is found to be 100 Hz. Find the number of times it vibrates in a second. (1 mark)

Ans. Number of vibrations produced in 1 second = 100

Hence, number of vibrations in 1 minute = 100 * 60 = 6000

Ques 8. The pulse rate of a person is found to be 80 beats per minute. Find its frequency. (2 marks)

Ans. Given, number of beats per minute = 80

        Number of beats per seconds = 80/60 = 1.3

        The frequency is equal to 1.3 Hz.

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