CBSE Class IX Chapters for Mathematics

Test Paper And Study Materials

Tuesday, 29 July 2014
Geetha told her classmate radha that (root2  1 upon root root2 + 1) is an irrational number. Radha told Geetha that she was wrong and further claimed that if there is a number x such that x3 is irrational then x5 is also irrational. Geetha said that Radha was wrong. Radha accepted her mistake.justify both these statements.
Thursday, 20 February 2014
PRACTICE QUESTIONS CLASS VIII: CHAPTER – 15_ INTRODUCTIONS TO GRAPH
QUESTIONS CLASS VIII
PRACTICE CHAPTER – 15_ INTRODUCTIONS TO
GRAPH
1. If
y – coordinate of a point is zero, then this point always lies:
(a) I quadrant (b) II quadrant (c) x – axis (d) y – axis
2. If
x – coordinate of a point is zero, then this point always lies:
(a) I quadrant (b) II quadrant (c) x – axis (d) y – axis
3. Point
(–6, 4) lies in the quadrant:
(a) I (b) II (c) III (d) IV
4. The
point (–4, –3) means:
(a) x = –4, y = –3 (b) x = –3, y = –4 (c) x = 4, y = 3 (d)
None of these
5. Point
(0, 4) lies on the:
(a) I quadrant (b) II quadrant (c) x – axis (d) y – axis
6. Point
(5, 0) lies on the:
(a) I quadrant (b) II quadrant (c) x – axis (d) y – axis
7. On
joining points (0, 0), (0, 2), (2,2) and (2, 0) we obtain a:
(a) Square (b) Rectangle (c) Rhombus (d) Parallelogram
8. Point
(–2, 3) lies in the:
(a) I quadrant (b) II quadrant (c) III quadrant (d) IV
quadrant
9. Point
(0, –2) lies:
(a) on the xaxis (b) in the II quadrant (c) on the yaxis
(d) in the IV quadrant
10. Abscissa
of the all the points on x – axis is:
(a) 0 (b) 1 (c) –1 (d) any number
11. Ordinate
of the all the points on x – axis is:
(a) 0 (b) 1 (c) –1 (d) any number
12. Abscissa
of the all the points on y – axis is:
(a) 0 (b) 1 (c) –1 (d) any number
13. Ordinate
of the all the points on y – axis is:
(a) 0 (b) 1 (c) –1 (d) any number
14. The
point whose ordinate is 4 and which lies on y – axis is:
(a) (4, 0) (b) (0, 4) (c) (1, 4) (d) (4, 2)
15. The
perpendicular distance of the point P(3,4) from the y – axis is:
(a) 3 (b) 4 (c) 5 (d) 7
2 or 3 marks
11. Draw
the graph of y = 3x. From the graph, find the value of y when (i) x = 4 and
(ii) x =5.
12. Consider
the relation between the perimeter and the side of a square, given by P = 4a.
Draw a graph to show this relation. From the graph, find the value of P when
(i) a = 4 and (ii) a =5.
13. Consider
the relation between the area and the side of a square, given by A = x^{2}. Draw a
graph to show this relation. From the graph, find the value of P when x = 4.
14. Simple
interest on a certain sum is Rs. 40 per year then S = 40x, where x is the
number of years. Draw a graph of this relation. From the graph, find the value
of S when (i) x = 5 and (ii) x =6.
15. Plot the points
(0, 2), (3, 0), (–3, 0) and (0, –2) in the graph sheet. Join these points. Name
the figure obtained and find the area of the figure so obtained.
1.(c)

2.(d)

3. (b)

4. (a)

5. (d)

6. (c)

7. (a)

8. (b)

9. (c)

15. (a)

10. (a)

11.

12. (d)

13. (a)

14. (d)

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Saturday, 8 February 2014
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Thursday, 16 January 2014
9th SA2 Solved Practice paper [Physics and Chemistry] 2014
1.
Name the anion and cation which constitute the molecule of magnesium
oxide [1]
 
2. A cube of side 5cm is
immersed in water and then in saturated salt solution. In which case will it
experience a greater buoyant force. If each side of the cube is reduced to
4cm and then immersed in water compare the force experienced by the cube, as
compared to the first case. Give reasons for each case. 2
 
4. The percentage of
three elements calcium, carbon and oxygen in a sample of calcium carbonate is
given as : [3]
 
Calcium =40% ; Carbon
=12.0% ; Oxygen = 48%
 
If the law of constant
proportion is true, what weights of these elements will be present in 1.5 g
of another sample of calcium. Carbonate ? (Atomic mass of Ca = 40 u, C
= 12 u, O = 16 u)
 
5. The description of atomic particles of two
elements X and Y is given below[3]
 
 
(i) What is the atomic
number of Y?
 
(ii) What is the mass
number of X?
 
(iii) What is the
relation between X and Y?
 
(iv) Which
element/elements do they represent?
 
(v) Write the electronic
configuration of X?
 
(vi) Write the
cation/anion formed by the element
 
6. Which of the
following are isotopes and which are isobars?
 
Argon, Protium, Calcium,
Deuterium. Explain why the isotopes have similar chemical properties but they
differ in physical properties? [3]
 
7.What do you mean by
buoyancy? Why does an object float or sink when placed on the surface of
water?[3]
 
8. When is the work done
by a force said to be negative? Give one situation in which one of the forces
acting on the object is doing positive work and the other is doing negative
work. [3]
 
9. Define one Joule of
work. Calculate the work done in lifting a box weighting 150 kg through a
vertical height of 7 meters (take g=10 ms2) [3]
 
10. A construction
worker’s helmet slips and falls when he is 78.4 m above the ground. He hears
the sound of the helmet hitting the ground 4.23 seconds after it slipped.
Find the speed of sound on air. [3]
 
11. Answer these questions[5]
 
(i) Define mole
 
(ii) How is it related
to Avogadro constant, relative mass and molecular mass?
 
(iii) What is the number
of molecules in 0.25 moles of oxygen ? Avogadro’s no. = 6 .22 x10^{23}.
 
OR
 
(i) Define atomicity
 
(ii) Give an example
each of a polyatomic element and a polyatomic ion.
 
(iii) How many atoms are
present in CaCl2 molecule and ion? SO_{4}^{2}
 
(iv) Write down the
formulae of
 
(a) sodium carbonate
 
(b) Ammonium chloride
 
(c) Zinc oxide
 
(d) Aluminium hydroxide
 
12. (i) An object thrown
at a certain angle to the ground moves in a curved path and falls back to the
ground. The initial and the final points of the path of the object lie on the
same horizontal line. What is the total work done against the force of gravity
and by the force of gravity on the object? Explain with proper mathematical
expressions.
 
(ii) Certain force
acting on 20kg mass changes its velocity from 5ms1 to 2ms1 calculates the
work done by the force. [5]
 
OR,
 
(i) Define kinetic
energy. Derive an expression for the kinetic energy of an object.
 
(ii) The power of a
motor pump is 5kW. How much water per minute the pump can raise to height of
20m? Take g=10ms^{2}

Solution:
1.

Magnesium oxide – MgO
Anion – O2
Cation – Mg+2

½
½


2

We know buyant force =
V.P g
V = Volume of body
P = density of liquid
g = acceleration
Since density of
saturated salt solutions is more than water the cube will experience greater
upthrust.
In second case as volume
of cube is less then first so in 2nd case it will experience less upthrust

½
½
½
½


3

Initial K.E = 25J
Since K.E
=
½ mv2 D as velocity doubles K.S
becomes 4 times
K.E = 4x25=100J

1
1


4

Mass of calcium in 1.5 g of sample = 40 x [1.5/100] =
9.6 g
Mass of carbon in 1.5 g of sample = 12x [1.5/100] =
0.18 g
Mass of oxygen in 1.5 g of sample = 48x[1.5/100] =
0.72 g

1
1
1


5

(i) Atomic number of y –
8
(ii) Mass number of x –
16
(iii) x and y are
isotopes
(iv) x and y represent –
oxygen
(v) ^{16}_{8}O
 2, 6
(vi) It will form an
anion – O^{}^{2}

1
1
1


6

Isotopes – Protium,
Diuterium
Isobars – Argon and
calcium
Since isotopes have
identical electronic configuration containing same number of valence
electrons they have similar chemical properties.
Since the masses are
slightly different the physical properties (density, melting pt, boiling pt,
etc) are different

1
1


7

When a body is immersed
in a fluid, it experiences an upward force. This upward force is the force of
buoyancy.
Forces acting an the
body inside water are buoyant force weight of body and if the buoyant force
is greater than the gravitational force it floats, otherwise it sinks.

1
2


8

Negative work – When the
force is acting opposite to the direction of the displacement, the work done
by the force is taken as negative.
When we lift an object
two forces act on the object.
Muscular force – Doing
positive work in the direction of the displacement.
Gravitational force –
doing negative work opposite to the direction of the displacement.

1
1
1


9

One Joule is the amount
of work done on an object when a force of 1N displaces it by 1 m along the
line of action of force
m = 150kg ; G = 10m/s^{2 } ;^{
}h = 7m
P.E = mgh
P.E = 150 x 10 x7 =
10500J

1
2


10

We know, h = ut + 1/2 gt^{2}
u =
0, h =
78.4, g =
9.8 m/s^{2}
using these
t^{2} = 16 Ãž
t = 4 s
time taken by the helmet reach the ground = 4 s
Ãž
time taken by sound to reach the height =
4.23 – 4 =
0.23s
ÃžSpeed
of sound = 78.4 /0.23 = 340.87 m/s

1
1
1


11

(i) Mole – one mole of
any species (atoms, molecules, ions or particles) is that quantity in number
having a mass equal to its atomic or molecular mass in grams
(ii) 1 mole = 6.022 x
1023 in number – Relative mass in g.
(iii) 1 mole of oxygen
contain 6.022 x 1023 molecules
0.25 mole of oxygen
contain 6.022 x1023 x 0.25
= 1.505 x 1023 molecules

1
2
2


OR

(i) Atomicity – Number
of atoms constituting a molecule
(ii) Polyatomic element
– sulphur
Polyatomic ion – NH4^{}, OH^{}, SO_{4}^{2}
(any one)
(iii) Cacl2 – atomicity – 3
SO42 – atomicity  5
(iv) (a) Na2CO3 Sodium carbonate
(b) NH4Cl Ammonium chloride
(c) ZnO zinc oxide
(d) Al(OH)3 Aluminium hydroxide

1
½+½
½+½
½+½+
½+½


12.

(i) Work done =
mgh
Difference in height of initial and final position is
zero.
Work done =
mg ( h2 
h1)
= mg(o) = O
(ii) Work done = change in K.E
Work done = 1/2 m
(v_{1}^{2 }– v_{2}^{2}) = ½ x 20(5^{2 } ^{ }2^{2}) = 210 J

1
½
1
1
½


OR


(i) Energy posesed by a
body by virtue of motion.
F = ma
W =
F.S =
mas
But u^{2} 
v^{2} =
2as Ãž as =
½ (v2u2)
w = energy Ãž
S = v^{2}/2a
m. a x x v^{2} x 2a = 1/2mv^{2}

½
½
½
½


(ii) Energy = power x
time =
5kw x 1 min = 3 x 10^{5} J
mgh = E /gh = (3
x 10^{5})/ 10x 20 = 1.5 x 10^{3}
kg
volume of water
= mass/ density = (1.5 x 10^{3}
)/10^{3} = 1.5 m3

½
½
1


13

(i)


(ii)

Eardrum moves inwards
when compression reaches it and moves outwards when rarefaction reaches it
thus the ear drum vibrates.
(ii) Audible rang 20Hz
to 20,000 Hz
OR


(i)

Ceiling and walls are
made curved so that sound after reflection reaches the target audience.
(ii) Reflection
The sound goes to the
obstacle and reaches back the ear of the listeners on reflection after 0.1s.
If speed of sound in air is 344 m/s
then the minimum
distance of obstacle from the source of sound must be 17.2 cm
(iii) Repeated
reflection of sound that results in persistence of sound Reverberation.
sound is blurred,
distorted, confusing
sound absorbent material
on roof, walls, seats

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