Physics 11th

1st Year Physics True and False

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1st Year Physics True and False

The Scope of Physics

1. Screw and liver was invented by Newton.
2. Pythagouras was famous in mathematics.
3. Logarathim was invented by Al-Beruni.
4. Omer Khyyam was mathematician and he was also a poet.
5. Pin hole camera was invented by Al-Razi.
6. Ibn-e-Sina was famous for his work in chemistry.
7. Al-Razi wrote about 200 books.
8. 20th century is the century of Physics.
9. Dimension of volume is L3.
10. Dimension of Linear momentum is MLT-1.
11. 9.8 contain two significant figures.
12. In C.G.S system the fundamental units of length, mass and time are kg, km and hour.
13. In British Engineering system the unit of force, length and time are chosen as the fundamental unit.
14. In M.K.S system unit of mass is pound.
15. Candela is the unit of luminous intensity.
16. The branch of physical science, which deal with interaction of matter and energy, is called physics.
17. The biological science deals with non-living things.
18. The class of science, which deals with the properties, and behaviour of living things is called physical science.
19. Ibn-e-Sina was famous in the field of mathematics.
20. In C.G.S system the unit of force is pound.
21. In the field of research the strong incentive comes from Bible.
22. The new era of modern physics began near the end of 19th century.
23. Al-Khwarizmi was the founder of analytical algebra.
24. Egyptian for the first time manufactured paper.
25. Chinese used to measure the flood level in the river Nile.
26. Ibn-e-Sina was a great astronomer.
27. The author of Alsh-Shifa was Al-Beruni.
28. The author of Al-Qannun-Fil-Tib was Ibn-e-Sina.
29. Yakoob- Bin Ishag Al-Kindi wrote many books on the mathematics, astronomy, medicine and other subjects.
30. In atomic clock the time be measured to an accuracy of part in 1012.
31. The unit of electric current is coulomb.
32. The SI unit of thermodynamic temperature is celsius.
33. The dimension of area is L-2.
34. The dimension of force is MLT-2.
35. 14.71 has four significant figures.

Work, Power and Energy

Work

Work is said to be done when a force causes a displacement to a body on which it acts.
Work is a scalar quantity. It is the dot product of applied force F and displacement d.


Diagram Coming Soon
W = F . d
W = F d cos θ .............................. (1)
Where θ is the angle between F and d.
Equation (1) can be written as
W = (F cos θ) d
i.e., work done is the product of the component of force (F cos θ) in the direction of displacement and the magnitude of displacement d.
equation (1) can also be written as
W = F (d cos θ)
i.e., work done is the product of magnitude of force F and the component of the displacement (d cos θ) in the direction of force.

Unit of Work
M.K.S system → Joule, BTU, eV
C.G.S system → Erg
F.P.S system → Foot Pound
1 BTU = 1055 joule
1 eV = 1.60 x 10(-19)

Important Cases
Work can be positive or negative depending upon the angle θ between F and d.

Case I
When θ = 0º i.e., when F and d have same direction.
W = F . d
W = F d cos 0º .............. {since θ = 0º}
W = F d .......................... {since cos 0º = 1}
Work is positive in this case.

Case II
When θ = 180º i.e., when F and d have opposite direction.

W = F . d
W = F d cos 180º ............................ {since θ = 180º}
W = - F d ......................................... {since cos 180º = -1}
Work is negative in this case

Case III
When θ = 90º i.e, when F and d are mutually perpendicular.
W = F . d
W = F d cos 90º ............................ {since θ = 90º}
W = 0 ........................................... {since cos 90º = 0}

Work Done Against Gravitational Force
Consider a body of mass 'm' placed initially at a height h(i), from the surface of the earth. We displaces it to a height h(f) from the surface of the earth. Here work is done on the body of mass 'm' by displacing it to a height 'h' against the gravitational force.
W = F . d = F d cos θ
Here,
F = W = m g
d = h(r) - h(i) = h
θ = 180º
{since mg and h are in opposite direction}
Since,
W = m g h cos 180º
W = m g h (-1)
W = - m g h
Since this work is done against gravitational force, therefore, it is stored in the body as its potential energy (F.E)
Therefore,
P . E = m g h


Power

Power is defined as the rate of doing work.
If work ΔW is done in time Δt by a body, then its average power is given by P(av) = ΔW / Δt
Power of an agency at a certain instant is called instantaneous power.

Relation Between Power and Velocity
Suppose a constant force F moves a body through a displacement Δd in time Δt, then
P = ΔW / Δt
P = F . Δd / Δt ..................... {since ΔW = F.Δd}
P = F . Δd / Δt
P = F . V ................................. {since Δd / Δt = V}
Thus power is the dot product of force and velocity.

Units of Power
The unit of power in S.I system is watt.
P = ΔW / Δt = joule / sec = watt
1 watt is defined as the power of an agency which does work at the rate of 1 joule per second.
Bigger Units → Mwatt = 10(6) watt
Gwatt = 10(9) watt
Kilowatt = 10(3) watt
In B.E.S system, the unit of power is horse-power (hp).
1 hp = 550 ft-lb/sec = 746 watt


Energy

The ability of a body to perform work is called its energy. The unit of energy in S.I system is joule.

Kinetic Energy
The energy possessed by a body by virtue of its motion is called it kinetic energy.
K.E = 1/2 mv2
m = mass,
v = velocity

Prove K.E = 1/2 mv2

Proof
Kinetic energy of a moving body is measured by the amount of work that a moving body can do against an unbalanced force before coming to rest.
Consider a body of mas 'm' thrown upward in the gravitational field with velocity v. It comes to rest after attaining height 'h'. We are interested in finding 'h'.
Therefore, we use
2 a S = vf2 - vi2 ............................ (1)
Here a = -g
S = h = ?
vi = v (magnitude of v)
vf = 0
Therefore,
(1) => 2(-g) = (0)2 - (v)2
-2 g h = -v2
2 g h = v2
h = v2/2g
Therefore, Work done by the body due to its motion = F . d
= F d cos θ
Here
F = m g
d = h = v2 / 2g
θ = 0º
Therefore, Work done by the body due to its motion = (mg) (v2/2g) cos0º
= mg x v2 / 2g
= 1/2 m v2
And we know that this work done by the body due to its motion.
Therefore,
K.E = 1/2 m v2

Potential Energy
When a body is moved against a field of force, the energy stored in it is called its potential energy.
If a body of mass 'm' is lifted to a height 'h' by applying a force equal to its weight then its potential energy is given by
P.E = m g h
Potential energy is possessed by
1. A spring when it is compressed
2. A charge when it is moved against electrostatic force.

Prove P.E = m g h OR Ug = m g h

Proof
Consider a ball of mass 'm' taken very slowly to the height 'h'. Therefore, work done by external force is
Wex = Fex . S = Fex S cos θ .................................. (1)
Since ball is lifted very slowly, therefore, external force in this case must be equal to the weight of the body i.e., mg.
Therefore,
Fex = m g
S = h
θ = 0º ......................... {since Fex and h have same direction}
Therefore,
(1) => Wex = m g h cos 0º
Wex = m g h .................................................. ................. (2)
Work done by the gravitational force is
Wg = Fg . S = Fg S cosθ ................................................. (3)
Since,
Fg = m g
S = h
θ = 180º ...................... {since Fg and h have opposite direction}
Therefore,
(3) => Wg = m g h cos 180º
Wg = m g h (-1)
Wg = - m g h .................................................. ................. (4)
Comparing (2) and (4), we get
Wg = -Wex
Or
Wex = - Wg
The work done on a body by an external force against the gravitational force is stored in it as its gravitational potential energy (Ug).
Therefore,
Ug = Wex
Ug = - Wg .............................. {since Wex = -Wg}
Ug = -(-m g h) ..................... {since Wg = - m g h}
Ug = m g h ............................................... Proved

Absolute Potential Energy
In gravitational field, absolute potential energy of a body at a point is defined as the amount of work done in moving it from that point to a point where the gravitational field is zero.

Determination of Absolute Potential Energy
Consider a body of mass 'm' which is lifted from point 1 to point N in the gravitational field. The distance between 1 and N is so large that the value of g is not constant between the two points. Hence to calculate the work done against the force of gravity, the simple formula W = F .d cannot be applied.
Therefore, in order to calculate work done from 1 to N, we divide the entire displacement into a large number of small displacement intervals of equal length Δr. The interval Δr is taken so small that the value of g remains constant during this interval.
Diagram Coming Soon Now we calculate the work done in moving the body from point 1 to point 2. For this work the value of constant force F may be taken as the average of the forces acting at the ends of interval Δr. At point 1 force is F1 and at point 2, force is F2.


Law of Conservation of Energy

Statement
Energy can neither be created nor be destroyed, however, it can be transformed from one form to another.

Explanation
According to this law energy may change its form within the system but we cannot get one form of energy without spending some other form of energy. A loss in one form of energy is accompanied by an increase in other forms of energy. The total energy remains constant.

Proof
For the verification of this law in case of mechanical energy (Kinetic and potential energy). Let us consider a body of mass 'm' placed at a point P which is at a height 'h' from the surface of the earth. We find total energy at point P, point O and point Q. Point Q is at a distance of (h-x) from the surface of earth.

Gravitation

The property of all objects in the universe which carry mass, by virtue of which they attract one another, is called Gravitation.


Centripetal Acceleration of the Moon

Newton, after determining the centripetal acceleration of the moon, formulated the law of universal gravitation.
Suppose that the moon is orbiting around the earth in a circular orbit.
If V = velocity of the moon in its orbit,
Rm = distance between the centres of earth and moon,
T = time taken by the moon to complete one revolution around the earth.
For determining the centripetal acceleration of the moon,. Newton applied Huygen's formula which is
a(c) = v2 / r
For moon, am = v2 / Rm ..................... (1)
But v = s/t = circumference / time period = 2πRm/T
Therefore,
v2 = 4π2Rm2 / T2
Therefore,
=> a(m) = (4π2Rm2/T2) x (1/Rm)
a(m) = 4π2Rm / T2
Put Rm = 3.84 x 10(8) m
T = 2.36 x 10(6) sec
Therefore,
a(m) = 2.72 x 10(-3) m/s2

Comparison Between 'am' AND 'g'
Newton compared the centripetal acceleration of the moon 'am' with the gravitational acceleration 'g'.
i.e., am / g = 1 / (60)2 ................. (1)
If Re = radius of the earth, he found that
Re2 / Rm2 - 1 / (60)2 ......................... (2)
Comparing (1) and (2),
am / g = Re2 / Rm2 ..................................... (3)
From equation (3), Newton concluded that at any point gravitational acceleration is inversely to the square of the distance of that point from the centre of the earth. It is true of all bodies in the universe. This conclusion provided the basis for the Newton' Law of Universal Gravitation.
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Newton's Law of Universal Gravitation

Consider tow bodies A and B having masses mA and mB respectively.


Diagram Coming Soon
Let,
F(AB) = Force on A by B
F(BA) = Force on B by A
r(AB) = displacement from A to B
r(BA) = displacement from B to A
r(AB) = unit vector in the direction of r(AB).
r(BA) = unit vector in the direction of r(BA).

From a(m) / g = Re2 / Rm2, we have
F(AB) ∞ 1 / r(BA)2 ......................... (1)
Also,
F(AB) ∞ m(A) ............................... (2)
F(BA) ∞ m(B)
According to the Newton's third law of motion
F(AB) = F(BA) .................... (for magnitudes)
Therefore,
F(AB) ∞ m(B) ................................ (3)
Combining (1), (2) and (3), we get
F(AB) ∞ m(A)m(B) / r(BA)2
F(AB) = G m(A)m(B) / r(BA)2 ........................... (G = 6.67 x 10(-11) N - m2 / kg2)

Vector Form
F(AB) = - (G m(A)m(B) / r(BA)2) r(BA)
F(BA) = - (G m(B)m(A) / r(AB)22) r(AB)
Negative sign indicates that gravitational force is attractive.

Statement of the Law
"Every body in the universe attracts every other body with a force which is directly proportional to the products of their masses and inversely proportional to the square of the distance between their centres."


Mass and Average Density of Earth

Let,
M = Mass of an object placed near the surface of earth
M(e) = Mass of earth
R(e) = Radius of earth
G = Acceleration due to gravity
According to the Newton's Law of Universal Gravitation.
F = G M Me / Re2 ............................. (1)
But the force with which earth attracts a body towards its centre is called weight of that body.


Diagram Coming Soon
Therefore,
F = W = Mg
(1) => M g = G M Me / Re2
g = G Me / Re2
Me = g Re2 / G .................................. (2)
Put
g = 9.8 m/sec2,
Re = 6.38 x 10(6) m,
G = 6.67 x 10(-11) N-m2/kg2, in equation (2)
(1) => Me = 5.98 x 10(24) kg. ................... (In S.I system)
Me = 5.98 x 10(27) gm .................................... (In C.G.S system)
Me = 6.6 x 10(21) tons
For determining the average density of earth (ρ),
Let Ve be the volume of the earth.
We know that
Density = mass / volume
Therefore,
ρ = Me / Ve ........................... [Ve = volume of earth]
ρ = Me / (4/3 Π Re3) .............. [since Ve = 4/3 Π Re3]
ρ = 3 Me / 4 Π Re3
Put,
Me = 5.98 x 10(24) kg
& Re = 6.38 x 10(6) m
Therefore,
ρ = 5.52 x 10(3) kg / m3


Mass of Sun

Let earth is orbiting round the sun in a circular orbit with velocity V.
Me = Mass of earth
Ms = Mass of the sun
R = Distance between the centres of the sun and the earth
T = Period of revolution of earth around sun
G = Gravitational constant
According to the Law of Universal Gravitation
F = G Ms Me / R2 .................................... (1)
This force 'F' provides the earth the necessary centripetal force
F = Me V2 / R ............................................ (2)
(1) & (2) => Me V2 / R = G Ms Me / R2
=> Ms = V2 R / G ........................................ (3)
V = s / Π = 2Π R / T
=> V2 = 4Π2 R2 / T2
Therefore,
(3) => Ms = (4Π R2 / T2) x (R / G)
Ms = 4Π2 R3 / GT2 ........................................ (4)
Substituting the value of
R = 1.49 x 10(11),
G = 6.67 x 10(-11) N-m2 / kg2,
T = 365.3 x 24 x 60 x 60 seconds, in equation (4)
We get
Ms = 1.99 x 10(30) kg

Variation of 'g' with Altitude
Suppose earth is perfectly spherical in shape with uniform density ρ. We know that at the surface of earth
g = G Me / Re2
where
G = Gravitational constant
Me = Mass of earth
Re = Radius of earth
At a height 'h' above the surface of earth, gravitational acceleration is
g = G Me / (Re + h)2
Dividing (1) by (2)
g / g = [G Me / Re2] x [(Re + h)2 / G Me)
g / g = (Re + h)2 / Re2
g / g = [Re + h) / Re]2
g / g = [1 + h/Re]2
g / g = [1 + h/Re]-2
We expand R.H.S using Binomial Formula,
(1 + x)n = 1 + nx + n(n-1) x2 / 1.2 + n(n + 1)(n-2)x3 / 1.2.3 + ...
If h / Re < 1, then we can neglect higher powers of h / Re.
Therefore
g / g = 1 - 2 h / Re
g = g (1 - 2h / Re) ................................. (3)
Equation (3) gives the value of acceleration due to gravity at a height 'h' above the surface of earth.
From (3), we can conclude that as the value of 'h' increases, the value of 'g' decreases.

Variation of 'g' with Depth
Suppose earth is perfectly spherical in shape with uniform density ρ.
Let
Re = Radius of earth
Me = Mass of earth
d = Depth (between P and Q)
Me = Mass of earth at a depth 'd'
At the surface of earth,
g = G Me / Re2 ...................................... (1)
At a depth 'd', acceleration due to gravity is
g = G Me / (Re - d)2 ........................... (2)
Me = ρ x Ve = ρ x (4/3) π Re3 = 4/3 π Re3 ρ
Me = ρ x Ve = ρ x (4/3) π (Re - d)3 = 4/3 π (Re - d)3 ρ
Ve = Volume of earth
Substitute the value of Me in (1),
(1) => g = (G / Re2) x (4/3) π Re3 ρ
g = 4/3 π Re ρ G ................................. (3)
Substitute the value of Me in (2)
g = [G / Re - d)2] x (4/3) π (Re - d)3 ρ
g = 4/3 π (Re - d) ρ G
Dividing (4) by (3)
g / g = [4/3 π (Re - d) ρ G] / [4/3 π Re ρ G]
g / g = (Re - d) / Re
g / g = 1 - d/Re
g / g = g (1 - d / Re) ........................... (5)
Equation (5) gives the value of acceleration due to gravity at a depth 'd' below the surface of earth
From (5), we can conclude that as the value of 'd' increases, value of 'g' decreases.
At the centre of the earth.
d = Re => Re / d = 1
Therefore,
(5) => g = g (1-1)
g = 0
Thus at the centre of the earth, the value of gravitational acceleration is zero.


Weightlessness in Satellites

An apparent loss of weight experienced by a body in a spacecraft in orbit is called weightlessness.
To discuss weightlessness in artificial satellites, let us take the example of an elevator having a block of mass ;m; suspended by a spring balance attached to the coiling of the elevator. The tension in the thread indicates the weight of the block.
Consider following cases.

1. When Elevator is at Rest
T = m g

2. When Elevator is Ascending with an Acceleration 'a'
In this case
T > m g
Therefore, Net force = T - mg
m a = T - m g
T = m g + m a
In this case of the block appears "heavier".

3. When Elevator is Descending with an Acceleration 'a'
In this case
m g > T
Therefore
Net force = m g - T
m a = m g - T
T = m g - m a
In this case, the body appears lighter

4. When the Elevator is Falling Freely Under the Action of Gravity
If the cable supporting the elevator breaks, the elevator will fall down with an acceleration equal to 'g'
From (3)
T = m g - m a
But a = g
Therefore
T = m g - m g
T = 0
In this case, spring balance will read zero. This is the state of "weightlessness".
In this case gravitation force still acts on the block due to the reason that elevator block, spring balance and string all have same acceleration when they fall freely, the weight of the block appears zero.
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Artificial Gravity

In artificial satellites, artificial gravity can be created by spinning the space craft about its own axis.
Now we calculate frequency of revolution (v) of a space craft of length 2R to produce artificial gravity in it. Its time period be 'T' and velocity is V.