Superior Lalazar Public School and College Thana
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Solved physics 9th 2013 paper Swat Board.
Section B (Marks 32)
Q. 2 Answer any eight parts. Each part carries 4 marks
(i) write different methods of reducing friction
Ans. METHODS OF REDUCING FRICTION
There are a number of methods to reduce friction in which some are
discussed here.
USE OF LUBRICANTS:
The parts of machines which are moving over one another must be
properly lubricated by using oils and lubricants of suitable viscosity.
USE OF GREASE:
Proper greasing between the sliding parts of machine reduces the
friction.
USE OF BALL BEARING:
In machines where possible, sliding friction can be replaced by rolling
friction by using ball bearings.
DESIGN MODIFICATION:
Friction can be reduced by changing the design of fast moving
objects. The front of vehicles and airplanes made oblong to minimize
friction.
(ii) Write note on motion and its types.
Motion.
Definition.
The state of a body in which it changes its position with respect to an
observer or surrounding is called motion.
Example. A person walking on the road is in state of motion because
he is changing his position with respect to an observer.
Types of motion. There are three types of motion.
1. Translatory motion.
Definition.
The Motion of a body in which all particles of the body move parallel
to each other along any path (straight or curved) is called
Translatory Motion.
Examples.
Motion of a person on a road, motion of bicycle on a road, a falling
objects etc
2. Rotatory motion.
Definition.
The type of motion in which each point of a body rotates around
a fixed point or axis is called Rotational Motion.
Examples
 motion of a wheel,
 motion of the blades of fan
 motion of the hands of clock.
3. Vibratory motion.
Definition.
The to and fro motion of a body about a fixed point is called
vibratory motion OR the back and forth motion of a body over
the same path about its position of equilibrium is called
vibratory motion.
Examples.
 Motion of simple pendulum
 Motion of the wire of guitar
 the pendulum of a wall clock
 oscillation of a mass attached to elastic spring
(iii) Name the seven SI base units of measurement
Base quantities
SI unit
Symbol
1. length
1. Metre
1. m
2. mass
2. Kilogram
2. Kg
3. time
3. Second
3. S
4. electric current
4. Ampere
4. A
5. temperature
5. Kelvin
5. K
6. luminous
6. Candela
6. Cd
intensity
7. Mole
7. mol
7. amount of
substance
(iv) State the two condition of equilibrium.
Conditions of equilibrium
There are two conditions of equilibrium given below.
First condition of equilibrium
Definition: A body will be in equilibrium if the sum of all the forces acting on
the body is zero
Mathematically:
where ƩF= ƩFx + ƩFy +ƩFz
For example, stationary person
Second condition of equilibrium
Definition: A body will be in equilibrium if the sum of all the torques acting on
the body is zero
Mathematically:
where ƩƬ = sum of clockwise torque + sum of anticlockwise torque
For example. Helicopter has two rotors to satisfy second condition of
equilibrium
(v) Determine the mass of earth by applying the law of gravitation
MASS OF EARTH:
The mass of earth was determined by Newton with the help of law of
gravitation. Consider a body of mass m which is lying on the surface
of earth. Let the mass of earth is Me and radius is Re.
The gravitational force between earth and body
mass m will be
𝐺𝑀𝑒𝑚
F=
------(1)
𝑅𝑒2
But the force with which the earth attracts a body
towards its centre is called weight so
F= W= mg
Putting the value of F in above equation 1
mg =
𝐺𝑀𝑒𝑚
𝑅𝑒2
𝐺𝑀𝑒
g=
𝑅𝑒2
OR
𝑔𝑅𝑒2
Me=
---(2)
𝐺
By putting the value of g(10m/sec2), G (6.67 x 10-11 Nm2/kg2) and
Re (6.4x106m) we will get the value of Me
Me=
9.8 𝑥(6.4x106m)2
6.67 x 10−11
Me = 6x 1024Kg
(vi) A mason of 600 weight is climbing a ladder of 10 m high. Find his
potential energy at the middle of the ladder.
Ans See videos lecture chap 6 Q 6
Solution:
Given Data : W = 600N
h= 10 m
Required data : P.E at middle of ladder (h/2)
Formula used : P.E = mgh = Wh
as w= mg
Solution:
ℎ
10
2
2
height at middle of ladder = =
ℎ
= 5m
P.E = W = 600X 5 = 3000J so P.E at middle of ladder (h/2) =
2
3000J
(vii) What is meant by evaporation. On what factor evaporation
depends?
Evaporation of liquid:
Definition: The spontaneous change of a liquid state into gaseous
state is known as evaporation. OR the process by which liquid slowly
changes into its vapours at any temperature below its boiling point
without the aid of external source of heat is called evaporation of
liquid.
Factors affecting evaporation:
a) Nature of liquid:
b) Temperature..
c) Area of the exposed surface of liquid.
d) Water vapours in air:
e) Movement of air:
f) Dryness of air:
g) External air pressure on liquid surface:
(viii) The temperature of normal body is 37 0C find the temperature on
the Fahrenheit scale.
Given data
Temperature = 37 0C
Required data :
Fahrenheit = ?
Solution
F = 1.8C + 32 = 1.8 x 37 + 32 = 66.6+ 32= 98.60F
(ix) explain Archimedes’s principle
Archimedes Principle:
Statement: When a body is immersed partially or fully in a liquid,
it experienced upthrust force which is equal to the weight of liquid
displaced by the object during immersing. OR
Buoyant Force or upthrust force acting on body = Weight of liquid
displaced by that object.
Mathematically:
Apparent Weight = Actual Weight - Weight of the liquid displaced
by the object
Explanation: consider a cylindrical object of cross sectional area “A”
and height “h” immersed in a fluid of density “ρ”. From above figure
it is clear that on the vertical sides, the liquid exert horizontal forces,
which balance each other. On the top face, the liquid exerts a
downward force and on the bottom face an upward force.
The force acting on upper end a by liquid will be
𝐹1
P1 =
OR F1 = P1A
𝐴
But the pressure due to liquid is P= ρgh so P1 = ρgh1 put value in
above equation
F1 = ρgh1A
Similarly the upward force on bottom b of cylinder will be
F2 = ρgh2A
The net upward thrust force will be
Upthrust = F2 – F1 = ρgh2A - ρgh1A = ρgA (h2 – h1)
Upthrust = ρgA (h2 – h1)
But h2- h1 = h (height of cylinder)
So
Upthrust = ρgAh (i)
As Ah = V
Upthrust = g ρV (ii)
But mass = density x volume or m= ρV put value of ρV = m in
equ ii
Upthrust = mg = W
Upthrust = W (iii)
Above equation shows that upthrust is equal to the weight of the
liquid displaced which is knows as Archimedes principle.
(x) State the law of conservation of energy
This law states “energy can neither be created nor destroyed
but may be transformed from one form to another and total
amount of energy remains constant.
Examples:
1. A diver on a spring board. Stored chemical energy in the
body of the diver allows him to bend the diving board.
This causes the bent diving board to store elastic potential
energy which is then converted into kinetic energy for
the diver by giving him an upward push
2. Potential energy of water: the potential energy of water
stored in the dam is converted into kinetic energy to
produce electric energy due to falling of water on turbine
3. Connecting battery to filament lamp. Stored chemical
energy in battery is first converted into electrical energy
and then to light and heat enegy
4. Einstein mass energy relation: According to this relation
mass can be converted into energy and energy into mass
E= mc2
(xi) Describe three application of conduction of heat
a. Cooking pods and pan are made of metals which are good
thermal conductor. They conduct heat to the food inside and
cook the food
b. Woollen cloths have fines pores filled with air. As wool and air
are bad conductor of heat so conduction of heat does not take
place and keep our body warm in winter season
c. Ice box has double wall made of thin or iron. The space between
walls is filled with cork which is bad conductor so prevent heat to
flow inside the box
Q. 3 (a) State Newton third law of motion with examples
NEWTON'S 3RD LAW OF MOTION
STATEMENT:
"To every action there is a reaction equal in magnitude but opposite in
direction"
OR
"When a body exerts a force on another body, the second body also
exerts a force on the first body of same magnitude but in the opposite
direction"
Examples: i) When we hit a ball on the wall, the ball exerts a force on
the wall which is known as action. While the wall also exerts an equal
and opposite force (reaction) on the ball and as a result the ball comes
back
(ii) Motion of rocket: fuel burns rapidly, exerts force in downward
direction and rocket moves upward
as a reaction.
(iii) Book lying on a table: weight of the book on the surface of table
is action and the force exerted by the surface of table (R) is the
reaction.
R = -W
(iv) When we walk on the ground, we push the ground in the
backward direction. This is action. While in reaction, the earth pushes
us in the forward direction.
Q 3 (b) ) Calculate the mass of a body when a force of 700 N
produces and acceleration of 12.5 m/s2
Ans. see videos Q 2 chap 3
Given data: Force F = 700 N
acceleration a = 12.5 m/s2
Required data: Mass m = ?
Formula used: F = ma ( Newton 2nd law)
Solution :
𝑭
𝟕𝟎𝟎
𝒂
𝟏𝟐.𝟓
m= =
F = ma
= 56 Kg
so mass is 56 Kg
Q 4 (a) prove that S = vit + ½ at2
(i) S = vit + ½ at2
Ans. Suppose a body has initial velocity ‘Vi’ at point A.
let the body accelerates uniformly and reaches to final
velocity’ Vf’ at point B and covers distance ‘S’ in time ‘t’.
Now consider velocity- time graph shown in figure 1
The total distance covered by a body will be equal to areal of
trapezium
So S = area of trapezium OABD
But area of trapezium =
( 𝒔𝒖𝒎 𝒐𝒇 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒔𝒊𝒅𝒆𝒔)𝑿 𝒉𝒆𝒊𝒈𝒉𝒕
𝟐
( 𝒔𝒖𝒎 𝒐𝒇 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒔𝒊𝒅𝒆𝒔)𝑿 𝒉𝒆𝒊𝒈𝒉𝒕
S=
𝟐
In above figure parallel sides are OA and BD and height is
OD or AC ( as OD = AC)
So
S=
( 𝑶𝑨+𝑩𝑫)𝑿 𝑶𝑫
(a)
𝟐
Now we know that OA = vi BD = BC + CD = at + vi and
OD = t put all values in eq a
( 𝑶𝑨+𝑩𝑫)𝑿 𝑶𝑫
S=
=
𝟐
S = vit + ½ at2
( 𝑽𝒊+𝒂𝒕+𝑽𝒊)𝑿 𝒕
𝟐
= vit + ½ at2
Q. 4 (b) A brick fall from the tower 15m high. How much time it will
take to reach the ground.
Ans See example 2.10 page 49 text book
Given data;
h= 15m
Vi = 0
g= 10m/s2
Required data.
Time t = ?
Formula used
h= vit + ½ gt2
Solution: h= vit + ½ gt2
Putting values
15 = 0xt + ½ 10t2
15 = 5t2
√𝒕𝟐 = √𝟑
t2 = 15/5 = 3
t = 1.73s
Q. 5 (a) Define the Kinetic energy. Derive its expression.
1
K.E = 2 mv2
KINETIC ENERGY:
Definition: Energy possessed by a body due to its motion is
called Kinetic Energy.
Mathematically: It is denoted by K.E and obtained when the
product of mass and square velocity is divided by 2.
𝟏
K.E. = 𝟐 mv2
Where K.E is Kinetic energy, m is mass and v is velocity.
Derivation of equation of K.E: Consider a body of mass “m”
is initially at rest. When a force of “F” is applied to the body,
it is accelerated and covered a distance “S”. its final velocity
is “V”. During this process the work done is given by:
W = FS -------------- (i)
But according to Newton second law of motion
F = ma put value of F in above equation(i)
W = FS = maS OR
W = maS ------------------(ii)
Now the acceleration produced by force can be calculated from
third equation of motion
2aS = Vf2 – Vi2 where Vf= V and Vi = 0
So 2aS = V2 –( 0)2
𝑽𝟐
a = 𝟐𝑺
Put value of a in above equation ii
𝑽𝟐
𝟏
W = maS = m 𝟐𝑺 S = = 𝟐 mv2
As work done by force is due to kinetic energy therefore
𝟏
K.E = 𝟐 mv2
The above equation shows relation between K.E of moving
object with its mass and velocity.
Q 5 (b) A ball of weight 100N is moving on a frictionless
surface with a velocity of 10 m/s, calculate the kinetic
energy.
Ans. See videos lecture chap 6 Q-7
Given data;
W= 100N V= 10 m/s g= 10 m/s2
Required data;
K.E = ?
𝟏
Formula used K.E = 𝟐 mv2
Solution:
As mass in unknown in above equation so first find out
mass
W = mg so m= W/g = 100/10 = 10 Kg
So m= 10 Kg
𝟏
𝟏
K.E = 𝟐 mv2 = = 𝟐 10X (10)2 =
500 J
Q 6. (a) State and explain Pascal principles.
Pascal Principles:
Statement: This law states that
“ All liquids transmit pressure equally
in all direction”
In other words
"When pressure on any portion of a confined liquid is changed, the
pressure on every other part of the liquid is also changed by the same
amount."
Explanation: This principle was put forward by French Scientist
Pascal. He observed that when pressure is applied to a liquid in a
container, then it is transmitted equally in all direction & acts
perpendicularly to the walls of the container. To understand the
principle consider hydraulic lift which is given in figure 1
In hydraulic lift we consider two interconnecting cylinder having
area of different dimeter and piston attached to both cylinder. The
chambers are filled with liquid. Suppose a force F1 is applied to area
𝐹1
A1 of piston so pressure P1 =
𝐴1
Similarly the pressure P2 on the other side will be
𝐹2
P2 =
𝐴2
According to Pascal principle the pressure on liquid transmitted
equally so
P1 = P2
𝐹1
𝐴1
=
F2 =
𝐹2
𝐴2
𝐹1 𝐴2
𝐴1
-----(1)
From above equation it is clear that depending on ratio of A2/A1 force
F2 can be greater as possible so small efforts can be used to overcome a
much larger load.
Q 6 (b) Calculate the pressure at the depth of 100 m water take g=
10m/s2
Ans.
Given data : h= 100m g= 10m/s2 water density ρ = 1000Kg/m3
Required data :
Formula used: P= ρh g
Solution:
Pressure p = ?
P = ρh g = 1000 x 100x 10 = 1000000 = 106 pa
P= 106 pa