Saturday, 13 October 2018
Wednesday, 10 October 2018
CLASS 7 / MATHEMATICS / holiday homework
ARMY PUBLIC SCHOOL, GOLCONDA
DUSHERRA BREAK HOMEWORK
CLASS 7 / MATHEMATICS
Q.1: Find sum and
difference of 312.456 and 213.546
Q.2: How much is
the sum of 32.33 and 33.32 is more than the sum of 21.12 and 12.21.
Q.3: The
complement and supplement angle of the given angles 45˚, 790, 350
AND 180.
Q.4: Find the
reciprocal of 3/5, 9/10, 3/9, 7/4 and 9/2. Also classify them as proper,
improper fractions.
Q.5: Find mean,
mode and median of the data 1,1,2,4,3,2,1,2,2,4.
Q.6: Write down a
pair of integers whose: a:
Sum is – 7 b: Difference is -10
Q.7: Find:- i) ½ of 24 ii) 2/3 of 18
Q.8: Find:- i) 1.3 X 10 ii) 153.7 X100
Q.9: Find the
mean of first five whole numbers.
Q.10: Find the
median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14
Q.11: Find: i) 0.35 ÷ 5 ii) 52.5 ÷ 10
iii) 7 ÷ 3.5
Q.12: The marks
(out of 100) obtained by a group of 10 students in a science test are 85, 76,
90, 85, 39, 48, 56, 95, 81 and 75.
Find the: a:
Highest and the lowest marks obtained by the students.
b:
Range of the marks obtained.
c.
mean
d.
median
Q.13.Find the
values of the angles x and y in the following figure.
Q.14. In the
figure given below, identify
(a) a pair of corresponding angles
(b) a pair of alternate interior angles
(c) a pair of interior angles on the same side of the
transversal
Q.15.There are 6
marbles in a box with numbers 1 to 6 marked on each of them.
(a) What is the
probability of drawing a marble with number 2 ?
(b) What is the
probability of drawing a marble with number 5 ?
Q.16. The runs
scored in a cricket match by 11 players is as follows:
6,15,120,49,93,80,11,10,15,8,10,15
Find the mean, median and mode of this data.
Q.17. Evaluate
(a) (-30) ÷ 10 (b) (50) ÷ (-5)
Q.18. Find the
product (a) (-1) x (-225) (b) (-15) x ( -10) x 0
Q.19. Express the
following as rupees using decimals (a) 7 paise (b) 230 paise
Q.20 . Multiply
and reduce to the simplest form
Q.21 Find
Q.22 Find (-1)x(-1)x(-1)x(-1) is equal to____________
Q.23 In the given
figure
The line l ǁ m and p
is a Transversal, then find the x Values of x, y, z( Give reasons)
Q.24 Find The
value of ( 30 + 20 ) x 50
Q.25 Express 1000
as product of powers of its prime factors.
Q.26 Find the
value of

Q.27 Draw 2 plane
figures with more than one line of symmetry.
Q. 28 solve the given expression

Q.29 Write the
Oder of Rotation and angle of symmetry of
(a) Square (b)
Rectangle (c) Equilateral Triangle
Q.30 Which of the
following English alphabet has reflectional symmetry about a vertical mirror?
(a) B (b) D (c) V (d)
E
Q.31 Find the value of angle x
Q.32 An exterior
angle of a triangle is 1200and interior opposite angles are in ratio
7:5. Find all the three angles of a triangle.
Q.33 Find the
value of 1.8 + 16.03 + 2.985
Q.34 Solve the
given expression using BODMAS

Q.35 In the given
figure PQ // RT. Find the value of angle x
and y
Q.36 Represent
the given data on double bar graph
Q.37 Weight of 73
bags of cement is 4106.25 kg. Find the weight of 100 such bags.
Q.38 There are 45
people in a building. Out of them ½ are men and 1/3 are women and rest are
children. Find the number of men, women and children in the building.
Q.39 The cost of
a pair of shoes is Rs.346.72. How much will it cost for 17 such pair of shoes?
Q.40 In a
triangle PQR, angle P = 800 and angle Q = 650. Find the
measure of the third angle of triangle and also find the measure of exterior
angle at vertex R.
**********
Monday, 24 September 2018
CLASS 9 / LAB ACTIVITY 3
EXPERIMENT - 3
AIM
To observe and compare the pressure exerted by a solid iron cuboid on sand while resting on its three different faces and to calculate the pressure exerted in the three cases.
THEORY
Simply stated, weight is the force acting vertically downward. The weight of an object is the force with which it is attracted towards the earth, that is:
For an object in free fall, when gravity is the only force acting on it, the expression for weight follows Newton's Second Law.
W = F, thus:
Here ‘g’ is the Earth's gravitational field strength, equal to about 9.81 m s−2.
An object's weight depends on its environment, while its mass does not. The SI unit of weight is the same as that of force, that is, Newton (N).The force acting on an object perpendicular to the surface is called thrust. The effect of thrust depends on the area on which it acts. Thus:
The thrust on unit area is called pressure. Thus:
SI unit of pressure is N/m2 or Nm-2 (Newton per square metre).
In honour of scientistBlaise Pascal, the SI unit of pressure is called pascal, denoted by Pa.
MATERIAL REQUIRED
PROCEDURE
Fill ¾ ths of a tray with dry sand and level it.
- Measure the dimensions of a solid iron cuboid accurately using a scale. Mark the three faces of the cuboid as A, B and C.
- Place the solid iron cuboid by the surface A on the plane levelled sand in the tray.
- After a few minutes, remove the Iron cuboid and you will see that it has made a depression in the sand.
- Measure the depth (depression) it has made in the sand using the scale.
- Repeat the same procedure for the other two surfaces.
OBSERVATIONS
Gravitational force on the environment = …….. 
- Calculate the area occupied by each surface of the solid iron cuboid.
- Area occupied by surface A in the sand = .............
- Area occupied by surface B in the sand = .............
- Area occupied by surface C in the sand = .............
- Area occupied by surface A in the sand = .............
- Calculate the pressure made by each surface of the solid iron cuboid.
- Pressure made by the surface A in the sand = ............. N
- Pressure made by the surface B in the sand = ............. N
- Pressure made by the surface C in the sand = ............. N
- Calculate the Depression.
- Depression made by the surface A in the sand = ............. cm
- Depression made by the surface B in the sand = ............. cm
- Depression made by the surface C in the sand = ............. cm
RESULT
PRESSURE EXERTED BY THREE DIFFERENT SURFACES IS DIFFERENT ALTHOUGH THE WEIGHT IS SAME.
PRECAUTIONS
- Dried sand must be used.
- The tray must have significant length and width.
- Appropriate cuboid of dimension must be used.
ANIMATION
CLASS 9 / LAB ACTIVITY - 2
EXPERIMENT - 2
AIM
To study the Third Law of Motion using two spring balances.
THEORY
Newton’s Third Law of Motion states: ‘To every action there is an equal and opposite reaction’.
It must be remembered that action and reaction always act on different objects. The Third Law of Motion indicates that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first object. These two forces are always equal in magnitude, but opposite in direction.
These forces act on different objects and so they do not cancel each other. Thus, Newton’s Third Law of Motion describes the relationship between the forces of interaction between two objects.
Whenever two bodies interact with each other, the force exerted by the first body on the second is called action. The force exerted by the second body on the first body is called reaction. The action and reaction are equal and opposite.
For example, when we placed a wooden block on the ground, this block exerts a force equal to its weight, W = mg acting downwards to the ground. This is the action force. The ground exerts an equal and opposite force W’ = mg on the block in the upward direction. This is the reaction force.
MATERIAL REQUIRED
Two spring balances, hook
PROCEDURE
- Take two similar spring balances of different ranges, say A and B.
- Note the least count of the spring balances.
- Attach the ring of spring balance A on a hook fixed in the wall and the spring balance B is attached to the hook of spring balance A.
- Hold the spring balances exactly horizontal to the table.
- Pull the ring of spring balance B gently.
- Observe and note the reading of both the spring balances.Repeat the experiment by applying different forces.
OBSERVATIONS
- Least count of the spring balance = value of 1 small division
- Least count of the spring balance A=………N
- Least count of the spring balance B =…………………N
No. of observations
|
Reading of B (Second spring balance) when force applied (N)
|
Reading of A (First spring balance) when force applied (N)
|
Difference in reading of A and B (N)
|
RESULT
The readings on both the balances are the same in each case. Thus action and reaction forces are equal and opposite and act on two different bodies.
PRECAUTIONS
- Spring balance of different least count should be taken.
- The spring balance should be brought in elastic mode before doing the experiment.
- The second spring balance should not be pulled beyond its elastic limit.
- The reading of the spring balance should be taken without any parallax error.
ANIMATION
CLASS 9 / LAB ACTIVITY - 1
EXPERIMENT - 1
AIM
To establish relationship between weight of a rectangular wooden block lying on a horizontal table and the minimum force required to just move it using a spring balance.
THEORY
According to Newton’s Second Law of Motion, the force acting on a body is directly proportional to the product of the mass of the body and the acceleration produced in the body by the application of the force. The acceleration takes place in the direction in which the force acts.
Newton’s Second Law of Motion precisely explains the relationship between force and acceleration.
Suppose a body of mass ‘m’ is moving by the application of a force ‘F’, then the produced acceleration ‘a’ will be directly proportional to the applied force.
Acceleration ∝ Force
F ∝ a
F ∝ m
Therefore, F = kma ………… (1)
Where k is constant of proportionality
In SI unit, k =1
Therefore, F = ma
Importance of the Newton’s Second Law of Motion
- Newton’s Second Law of Motion gives a quantitative measure of force.
F = ma
- The Second Law is a basic law of motion because both First and Third Law can be derived from this law.
MATERIAL REQUIRED
Wooden block and spring balance
PROCEDURE
- A wooden block with a hook is placed on a table.
- Note down the least count of the spring balance.
- The spring balance is attached to the hook.
- Hold the spring in such a way that the reading displayed in the spring balance can be noted down.
- The applied force on the spring balance is gradually increased until it just starts to move.
- The reading of the spring balance is noted.
OBSERVATION
Least count of the spring balance = value of 1 small division = …….g.wt
| Trial No; | Mass of the block, M (kg) | Force required, F (N) |
RESULT
Force applied to just move the block is directly proportional to the mass of the block.
ie, F ∝ M
PRECAUTIONS
- A suitable spring balance to pull the block must be used.
- The spring balance should be brought in elastic mode before doing the experiment.
- The reading of the spring balance should be taken without any parallax error.
- A smooth table or surface must be used.
https://www.youtube.com/watch?time_continue=71&v=1srdDxmWe34
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