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Class IX Unit Test on Polynomials

2013-06-02T22:33:00.001-07:00

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Unit Test For Class 8 (Exponents And Powers, Polynomials, Compound Interest)

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Sample Questions on Work, Energy And Power

2012-12-05T19:51:00.000-08:00

Sample Questions On Work, Energy And Power

Q:1- An engine of 2 × 10⁶ kW power is drawing a train of total mass of 3.0 × 10⁶ kg up an inclined plane, inclined at an angle of 60°. The frictional force is 8 × 10³ N. What is the maximum speed of the train?(Ans: 78.7 m/s.)

Q:2- A bullet of mass 15 grams moving with a velocity of 350 m/s enters a block of length 1.2 metres and goes out of the other side with a velocity of 290 m/s. What is the work done by the bullet in passing through the block?(Ans: 288 J.)

Q:3- A block of mass of 500 grams starts from the top of a frictionless curved hill top with an initial speed of 2 m/s, it arrives at the bottom of the hill with a speed of 3.5 m/s.What is the height of the hill?(Ans:0.42 m. )

Q:4- An iron ball of 0.2 kg, which is moving with a speed of 12 m/s, collides with an identical steel ball at rest. After collision the direction of each ball makes an angle of 45° with the original direction. What is the speed of the iron balls after the collision?(Ans: 8.48 m/s.)

Q:5- A motor of 2 kW power is able to pump out water from a well 12 m deep. What quantity of water is being pumped out per second?(Ans: 17.01 kg.)

Q:6- The bob of a pendulum is able to rise to vertical height of 15 cm. What is the velocity of the bob at its mean position?(Ans: 1.71 m/s.)

Q:7- A drop of water of radius 1.5 mm falls from a height of 60 cm. What is the work done by the gravitational force on the drop?(Ans: 8.3 × 10⁻⁵ J.)

Q:8- A body of mass 3.2 kg, which was at rest initially, is subjected to a force of 12 N for 10 seconds. What is the kinetic energy acquired by the body?(Ans: 2250 J.)

Q:9- A bullet of mass 30 grams passes between the two ends of a 40 metres long road in 0.3 seconds. The bullet travels with a constant speed. What is the kinetic energy of the bullet?(Ans: 266.66 J.)

Q:10-Work done by the boy in the given figure is 250 J per second. The speed of the boy is 1.2 m/s.

What is the force applied by the boy?(Ans: 294.6 N.)

Q:11- The linear momentum of a body is increased by 21%. What is the percentage increase in kinetic energy?(Ans: 46.41 %.)

Q:12-A motor is used to pump water from a well, and rise to a height of 12 m. The power of the motor 3 kW. The amount of water raised per minute by the motor is (Ans: 1530.61 kg.)

Q:13- A body of mass 2 kg falls from a height of 8 m and collides with another body of the same mass. After collision, the two bodies move through a distance of 2 m and ultimately come to rest. What is the work done against the resistive force?(Ans: 156.8 J.)

Q:14- A ball of mass 350 grams is dropped from a height of 12 metres; it rebounds from the hard floor. It loses 30 % of the energy in impact. The height to which the ball would rise after the impact is (Ans: 3.6 m.)

Q:15- Momentum for a body is twice the magnitude of kinetic energy of the body. What is the velocity of the body?(Ans: 1 unit. )

SOLUTIONS

Answer:1-

Power of the engine,

P = 2 × 10⁶ kW = 2 × 10⁶ kW
Mass of the train, m = 3 × 10⁶ kg
Frictional force = 8 × 10³ N
Angle of the inclined plane,
θ = 60°
Acceleration due to gravity, g = 9.8 m/s²
Force, F = mg sin θ + f

Therefore, the velocity of the train is 78.7 m/s.

Answer:2-

Mass of the bullet, m = 15 g = 0.015 kg
Initial velocity, v₀ = 350 m/s
Final velocity, v = 290 m/s
Work done by the bullet = loss in kinetic energy

Therefore, the work done by the bullet is 288 J.

Answer:3-

Mass of the block, m = 500 g = 0.5 kg
Initial speed of the block, u = 2 m/s
Final speed of the block v = 3.5 m/s
Acceleration due to gravity, g = 9.8 m/s²
Let h be the height of the hill.
The track is frictionless, so the mechanical energy will be conserved
Energy at the top of the hill
Energy at the base of the hill
Equating (1) and (2),

Hence, the height of the toy hill is 0.42 m.

Answer:4-

Mass of the iron ball, m₁ = 0.2 kg
Initial velocity of iron ball, v₁ = 12 m/s
Mass of the other iron ball, m₂ = 0.2 kg
The second ball is at rest.
Initial velocity of the second ball, v₂ = 0
Along x-axis,
Applying the law of conservation of momentum,

Where,
is the final velocity of the first ball.
is the final velocity of the second ball.
is the angle of deflection of the ball = 45°

Along y-axis,
Applying law of conservation of momentum,

From equation (1) and (2),

The final velocity of both the balls is 8.48 m/s.

Answer:5-

Power of the motor, P = 2 kW = 2 × 10³ W
Depth of the well, h = 12 m
Acceleration due to gravity = g = 9.8 m/s²
Energy = mgh
Where m is the mass of water.
Time, t = 1s
Power ,

Therefore, the mass of water pumped out per second is 17.01 kg.

Answer:6-

According to the principle of conservation of energy, the kinetic energy the bob at the mean position is equal to the potential energy at the maximum vertical height.
Kinetic energy of the bob at mean position =
Potential energy of the bob at maximum vertical height = mgh
Where,
m is the mass of the bob.
v is the velocity of the bob at the mean position.
g is the acceleration due to gravity = 9.8 m/s²
h is the height reached by the bob = 15 cm = 0.15

Therefore, the velocity of the bob in the mean position is 1.71 m/s.

Answer:7-

Radius of the water drop, r = 1.5 mm = 1.5 × 10⁻³ m
Height of the drop from the ground, h = 60 cm = 0.6 m
Volume of the drop,
Density of water, ρ = 1000 kg/m³
Mass of water, m = Vρ
Acceleration work done by the gravitational force = mgh

Therefore, the work done by the gravitational force on the drop of water is 8.3 × 10⁻⁵ J.

Answer:8-

Mass of the body, m = 3.2 kg
Force applied, F = 12 N
Time for which force is applied, t = 10 s.
The body is at rest initially.
∴Initial velocity, u = 0
Let, a be the acceleration of the body, and v be the final velocity of the body.

Kinetic energy =
Therefore, kinetic energy possessed by the body is 2250 J.

Answer:9-

Mass of a bullet, m = 30 g = 0.0 3 kg
Distance covered by the bullet, d = 40 cm
Time taken to cover the distance, t = 0.3s.
Speed,
Kinetic energy,
Therefore, the kinetic energy of the bullet is 266.66 J.

Answer:10-

Work done by the man, W = 250 J
Speed of the man, v = 1.2 m/s
Angle made by the direction of force to the horizontal θ = 45°
Let F be the force applied by the boy.

S is the distance traveled in 1s = 1.2 m/s × 1s
∴S = 1.2 m

Therefore, the force applied by the boy is 294.6 N.

Answer:11-

Let m be the mass of the body.
v₁, be the initial velocity of the body.
v₂ be the final velocity of the body.
The linear momentum increased by 21%.

Initial kinetic energy,

Final kinetic energy,

Percentage increase in kinetic energy

Therefore, the percentage increase in kinetic energy is 46.41 %.

Answer:12-

ower of the motor, P = 3 kW = 3 × 10³ W
Height to which the water is raised, h = 12 m
Acceleration due to gravity, g = 9.8 m/s².
Time for which the water is been lifted, t = 1 min = 60 s
Let m be the mass water been raised.

Therefore, the amount of water raised per minute is 1530.61 kg.

Answer:13-

Mass of both the body, m = 2 kg
Height from which the body falls, h = 8m
Distance covered by both the bodies together, d = 2 m
Acceleration due to gravity, g = 9.8 m/s²
According to the principle of conservation of energy,
mgh
Where, v is the velocity of the first body.

Let v′ be velocity of both bodies together.

Work done against the resistive force

Therefore, the work done against the resistive is 156.8 J.

Answer:14-

Mass of the ball, m = 350 g = 0.35 kg.
Height from which the ball is dropped, h = 12 m
Acceleration due to gravity, g = 9.8 m/s².
Energy loss after impact = 30 %
Let h₁ be the height to which the ball would rise after the impact.

Therefore, the height to which ball would rise after the impact is 3.6 m.

Answer:15-

Let, m be the mass of the body.
v be the velocity of the body.
Momentum, p = mv
Kinetic energy,
Given: P = 2 E_x

Hence, the velocity of the body is 1 unit.

Sample Question on Qudrilaterals

2012-12-05T19:48:00.001-08:00

Practice Question Sheet

Q:1: The diagonals AC and BD of parallelogram ABCD intersect at O. What is the ratio of the area of parallelogram ABCD and ΔAOB?

Ans:13

Q:2: If the length of the diagonal of a square is cm, then what is the perimeter of the square?

Q:3: The three angles of a quadrilateral are in the ratio 2:5:7. The sum of the greatest and the smallest among these angles is 180°. What is the measure of the fourth angle?

Q: 4: In a rhombus of side 5 cm, if the length of one of the diagonals is 8 cm, then what is the length of the other diagonal?

Ans: 5000

Q:5: In a quadrilateral PQRS, if the diagonals PR and QS are perpendicular to each other and PR bisects QS, but QS does not bisect PR, then the quadrilateral PQRS is a

Q:6: If the angles of a quadrilateral are in the ratio 3:3:7:7, then the quadrilateral is a/an

Q:7: In the given figure, ABCD is a square and PQRS is a rhombus.

The relation in which alternative is correct with respect to the given figure?

Ans:5

Q:8: The given figure shows a parallelogram ABCD where AD = BD and ∠ABD = 50°.

What are the measures of the adjacent angles of the parallelogram?

Q:9: The width and length of a rectangle are in the ratio 3:4. If its perimeter is 28 cm, then what is the area of the rectangle?

Q: 10: If one of the angles of an isosceles trapezium is 125°, then what are the measures of the other angles?

Ans:8400$

Q:11: The length and width of a rectangle ABCD are respectively a and 3a. The bisector of ∠A meets CD at P. What is the value of CP:DP?

Q:12: If in quadrilateral PQRS (vertices being taken in order), PS = QR and PQ||RS, then PQRS is necessarily a/an

Q:13: A quadrilateral in which opposite angles are equal and diagonals are bisecting each other is always a

Ans: 3, 24, 6, and 4

Q:14: The given figure shows a parallelogram ABCD. M is a point on CD such that AM = BM.

If ∠DAM = 25° and ∠ADC = 100º, then what is the measure of ∠AMB?

Ans:14

Q:15: The given figure shows a rhombus ABCD.

Which of the following relations is correct with respect to the given figure?

A) x < y B) x > y C) x = y D) y=180 - x

Solutions:

Answer:1:

Parallelogram ABCD can be drawn as:

It is known that diagonals of a parallelogram divide it into two triangles of equal area.
∴ Area of ΔABD = Area of ΔBCD
∴ Area of ΔABD =× Area of ABCD
In a parallelogram, diagonals bisect each other.
∴ AO is the median of ΔABD.
and Area of ΔAOB = Area of ΔAOD
⇒ Area of ΔAOB = × Area of ΔABD = ×× Area of ABCD

Thus, the required ratio is 4: 1.

Answer:2:

Let the length of each side of the square be a cm.
It is given that the length of its diagonal is cm.
Therefore, applying Pythagoras theorem in the triangle formed by a pair of adjacent sides and the diagonal:
a² + a² =
2a² =
a² = 16²
a = 16
Thus, perimeter of the square = 4a cm = 4 × 16 cm = 64 cm

Answer:3:

Let the three angles of the quadrilateral be 2x, 5x, and 7x.
Here, the greatest angle is 7x and smallest angle is 2x.
It is given that the sum of the greatest and the smallest angles is 180°.
∴ 7x + 2x = 180°
⇒ 9x = 180°
⇒ x = 20°
∴ Measure of fourth angle = 360° − (2x + 5x + 7x)
= 360° − 14x
= 360° − 14 × 20° [Putting x = 20°]
= 360° − 280°
= 80°
Thus, the measure of the fourth angle is 80°.

Answer:4:

Let ABCD be the rhombus such that AB = BC = CD = AD = 5 cm and AC = 8 cm

In a rhombus, diagonals are perpendicular bisectors of each other.
∴AO = OC = cm = 4 cm
In ΔAOB, ∠AOB = 90°
∴ AB² = AO² + OB² [Pythagoras theorem]
⇒ (5 cm)² = (4 cm)² + OB²
⇒ OB² = (25 − 16) cm² = 9 cm²
⇒ OB = 3 cm
∴BD = 2OB = 2 × 3 cm = 6 cm
Thus, the length of the other diagonal is 6 cm.

Answer:5:

A quadrilateral whose diagonals are perpendicular to each other and one of the diagonals bisect the other is a kite.

Answer:6:

It can be observed that in the given quadrilateral, there are exactly two pairs of adjacent equal angles.
Thus, the given quadrilateral is an isosceles trapezium.

Answer:7:

In a square, all the angles are 90°.
s + q = 90° + 90° = 180°
In a rhombus (which is also a parallelogram), adjacent angles are supplementary.
∴a + b = 180°
Thus, s + q = a + b

Answer:8:

Since AD = BD and ∠ABD = 50°, ΔABD is isosceles with ∠DAB = ∠ABD = 50°.
In a parallelogram, adjacent angles are supplementary.
∴∠ABC = 180° − ∠A = 180° − 50° = 130°
Thus, the measures of adjacent angles of the parallelogram are 50° and 130°.

Answer:9:

Since the width and length of the rectangle are in the ratio 3:4, let the width be 3x cm and the length be 4x cm.
∴ 2(3x + 4x) = 28 [Perimeter = 28 cm]
⇒ 14x = 28
⇒ x = 2
∴Length = 3 × 2 cm = 6 cm and width = 4 × 2 cm = 8 cm
Thus, area of the rectangle = 6 cm × 8 cm = 48 cm²

Answer:10:

Let PQRS be an isosceles trapezium where ∠S = 125°

Since, SRPQ and ∠P = ∠Q.
So, ∠R =∠S= 125°

Also the sum of all angles of trapezium is 360°.
So, ∠P + ∠Q + ∠R + ∠S = 360°
⇒ 2∠P + 250° = 360°
⇒ 2∠P = 110°
⇒ ∠P = 55°
Thus, the measures of the other angles are 55°, 55°, and 125°.

Answer:11:

The given rectangle ABCD can be drawn as:

Here, AD = a, DC = 3a
∠A = 90°
∴∠DAP = ∠A = × 90° = 45°
In ΔADP:
∠DAP + ∠ADP + ∠DPA = 180°
⇒ ∠DPA = 180° − 90° − 45° = 45°
⇒∠DAP = ∠DPA
∴ ΔADP is isosceles where ∠DAP = ∠DPA
Therefore, AD = DP = a
CP = DC − DP = 3a − a = 2a
Thus, CP: DP = 2a: a = 2:1

Answer:12:

The given quadrilateral PQRS can be drawn as:

It is known that a quadrilateral with exactly one pair of parallel sides where the non-parallel sides are of equal lengths is called an isosceles trapezium.
Thus, the given quadrilateral PQRS is an isosceles trapezium.

Answer:13:

A quadrilateral in which opposite angles are equal and diagonals are bisecting each other is always a parallelogram.

Answer:14:

In ΔAMD, ∠DMA + ∠MDA + ∠DAM = 180º
⇒ ∠DMA + 100º + 25º = 180º
⇒ ∠DMA + 125º = 180º
⇒ ∠DMA = 180º − 125º
∴∠DMA = 55º
Since ABCD is a parallelogram, AB||CD
∴∠MAB = ∠DMA = 55º [Alternate interior angles]
It is also given that AM = BM
∴∠MAB = ∠MBA = 55º
In ΔMAB:
∠MAB + ∠AMB + ∠MBA = 180º
⇒ ∠AMB + 55º + 55º = 180º
⇒ ∠AMB + 110º = 180º
⇒ ∠AMB = 180º − 110º = 70º
Thus, the measure of ∠AMB is 70°.

Answer:15:

Since ABCD is a rhombus, all the sides are equal.
∴ΔABD is isosceles.
⇒ ∠ADB = ∠ABD
∴ x = y

Class -8 Sample Questions

2012-11-21T02:25:00.001-08:00

Hello,

This is for the class -8th (VIII) student to attempt the following sample papers of mathematics.

Sample papers for class -VIII : Mathematics

Class-8

Class-8

Class-8

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SCDL PAPERS AND ASSIGNMENT

2012-09-24T04:06:00.000-07:00

Get SCDL(SYMBIOSIS CENTER FOR DISTANCE LEARNING) SOLVED SAMPLE SCDL PAPERS AND ASSIGNMENT

General Knowlege

2012-09-21T04:30:00.001-07:00

Question of the Day

Question 1: What is the name of Indian Milk Man ?

Question 2: Which type of cancer is found in indian cricketer yuvraj singh ?

Question 3: What is name of God Particle discoverd by scientsts ?

Assignment On real Numbers For Class X Free download

2012-09-05T02:57:00.001-07:00

Assignment on Real numbers

SAMPLE QUESTION PAPER SA-I for CBSC Class IX(Mathematics)-2012

2012-08-30T04:31:00.000-07:00

Hello,

This is for the class -9th (IX) student to attempt the following sample papers of mathematics.

Sample papers for class -IX : Mathematics

SA-I chapters covered: Number Systems, Polynomials, Coordinate geometry, Euclid Geometry, Heron's Formula, Lines and Angles and Triangles.

https://docs.google.com/document/d/1xQRfC7KNaUqz-kBPkiHASxP7z9o4zw6XPrBNRBIBHcw/edit

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SAMPLE QUESTION PAPER for CBSC Class IX(Mathematics)-2012

2012-08-28T04:21:00.000-07:00

Hello,

This is for the class -9th (IX) student to attempt the following sample papers of mathematics.

Sample papers for class -IX : Mathematics

https://docs.google.com/document/d/11DrynAaH70YCUr8Lkz0TZsksZp1c8I-XWfeF757TpDQ

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Sample Question on FORCE, LAW OF MOTION and Linear Momentum Conservation for CLASS IX(9th) CBSC Board

2012-08-07T23:41:00.002-07:00

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Free Download of Assignment and Notes on Triangles for Class IX of CBSC Board

2012-08-01T22:58:00.001-07:00

Hello,

Assignment on Triangles chapter 7 for class IX of CBSC Board is uploaded. you see some new question that will help you in brain storming.

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Assignment on Polynomials is uploaded for class(9th)

2012-07-19T23:31:00.004-07:00

Hello,

Assignment on polynomials for class IX of CBSC Board is uploaded.

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New assignment on Number System is uploaded.

2012-07-17T01:51:00.001-07:00

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Free Download CBSC Maths for Class 9(ix) Chapter-08 Test-Motion(Level-Medium)

2012-07-06T00:45:00.001-07:00

Chapter 08-Test-01

Total Marks:30 Time:45 Min

Note: All question of 2 marks each. Solve all question

Q.1: A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

(A)10m

(B)10

(D)None

Q.2: An Object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of object?

(A)5 m/s

(B)5.33 m/s

(D)None

Q.3: A man swim in 90 m long pool. He covers 180 m in one minute from one end to other and back on the same straight path. What is the average speed and velocity?

(A)3 m/s and 0 m/s

(B)0 m/s and 3 m/s

(D)None

Q.4: which is not true ?

(A) The acceleration is taken to be positive if it is in the direction of velocity

(B)The acceleration is taken to be negative if it is opposite to the direction of velocity

(C)SI unit of acceleration is m/

(D) Acceleration is negative for a accelerated motion of car.

Q.5: Which is not true?

(A)The motion of a freely falling object is an example of uniform accelerated motion.

(B) If a car travelling along straight road increases its speed by unequal amounts in equal intervals of time, then car is said to be moving with non-uniform motion.

(C)If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, Then the acceleration of the object is said to be uniform.

(D)All above are false statement.

Q.6: Starting from a stationary position, a boy start to run attains velocity of 6 m/s in 30 s. Calculate the acceleration ?

(A)0.2 m/

(B) 0.1 m/

(D)None

Q.7: A bus decreases its speed from 80 km/h to 62 km/h in 5 s, what will be the acceleration of the bus ?

(A)1 m/

(B) 0.1 m/

(D) None

Q.8:The breaks applied to a car produce an acceleration 6 m/in opposite to the direction of motion. If the car takes 2 s to stop after the application of breaks. Calculate the distance travelled during this time.

Q.9: A boat starting from rest on a lake accelerates in straight line at constant rate of 3.0 m/for 4.0 s. find the distance covered by boat during this time.

Q.10: An artificial satellite is moving in a circular orbit of radius 7/22 km. Calculate its speed if it takes 2000 s to revolve around the earth.

Q.11:A car moves with a speed of 160 km/h for 30 minutes and then with a speed of 240 km/h for the next 30 minutes. Calculate the total distance(km) covered by the car in 60 minutes.

Q.12: An airplane accelerates down a runway at 3.20 m/s² for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

Q.13:A feather is dropped on the moon from a height of 0.835 meters. The acceleration of gravity on the moon is 1.67 m/s². Determine the time for the feather to fall to the surface of the moon.

Q.14: A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well. (hint: assume acceleration 9.8 m/)

Q.15: A driver starts her car and steps on the gas pedal. The car gradually accelerates to 50 km/hr. A few minutes later, the driver suddenly slams on the brakes to avoid hitting a box in the road. As the car comes to a stop, the driver’s body appears to lurch forward in the seat until it is restrained by the seat-belt.

Use the following graphs to answer the next two questions.

1. Which graph best matches the motion of the car described in the paragraph above?

a. A
b. B
c. C
d. D

2. A student rides her bicycle from her home to the library. She stays there for a while and then goes to a friend’s house. Which graph best matches this situation?

a. A
b. B
c. C
d. D

Answer:

12 m
24 m
1m/s
200 km
1720 m
1 s
57 m
C,A

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Free Download CBSC Maths for Class 7(vii) Chapter13 Test-Motion(Level-Medium)

2012-07-06T00:30:00.002-07:00

Chapter 13-Test-01

Total Marks:20 Time:45 Min

Note: Q1-Q5 of one marks each and Q6-Q12 of 2 marks each. One marks for presenting the solutions.

Q.1: What is the basic unit of time ?

(A)Second

(B)Minute

(C)Hour

(D)None

Q.2: How many seconds in a year ?

(A)365*24*60*60

(B)24*60*60

(D)24*60*60*300

Q.3: Which of the following is not the straight line motion ?

(A)Motion of a horse pulling a cart on a straight road.

(B)Motion of car on a horizontal road

(C)Motion of your hands while running.

(D)Motion of wooden block on a horizontal path.

Q.4: The distance between the home and school is 20km. A car takes 1000 seconds. Calculate the speed of car ?

(A)200000 m/s

(B)20 m/s

(D)None

Q.5: A boy daily runs 10 km in 5 hours. Calculate the distance covered in 15 days ?

(A)75 km

(B)150 km

(D)None

Q.6: The distance traveled by a car and the time taken by it to cover the distance is given in below table: find out the speed(km/h) of the car.

S.No.	Time(hours)	Distance(km)
1	0	0
2	5h	10 km
3	10 h	20km
4	15 h	30km
5	20 h	40km
6	25h	50 km

Q.7: Draw and explain in one line speed- time(V-t) graph for following scenarios:

A car is moving with a constant speed.
A car parked on high way.

Q.8: Tanya takes 30 minutes from her house to reach her school by car. If the car has a speed of 2 m/s, calculate the distance between her house and the school.

Q.9: The distance time graph for a car is given below. Describe the behavior of graph for AB and BC.

Q.10:The odometer of a car reads 3.5 km when the clock shows the time 08:40:10 AM. What is the distance(m) moved by the car, if at 08:40:20 AM, the odometer reading has changed to 3.6 km? Calculate the speed of the car in m/s during this time.

Q.11:Write down the example of circular motion and oscillatory motion and their definition.

Q.12:A car moves with a speed of 80 km/h for 30 minutes and then with a speed of 120 km/h for the next 30 minutes. Calculate the total distance(km) covered by the car in 60 minutes.

Answer:

A
A
C
B
B
2 km/h

3600 m
AB-- Straight line motion and BC-- rest
100m, 10m/s
Circular Motion : pedal of bicycle in motion, Oscillatory Motion: motion of pendulum.
100 km

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Free Download CBSC Maths for Class 7(vii) Chapter13 Test-Motion(Level-Easy)

2012-07-06T00:25:00.000-07:00

Chapter 13

Q.1: What is the full form of SI ?

Q.2: What is the basic unit of speed ?

Q.3: How many seconds in a day ?

Q.4: The distance traveled by a car and the time taken by it to cover the distance is given in below table: find out the speed of the car.

S.No.	Time(hours)	Distance(km)
1	0	0
2	1 h	2 km
3	2 h	4 km
4	3 h	6 km
5	4 h	8 km
6	5 h	10 km

Q.5: A simple pendulum takes 32 s to complete 16 oscillations. What is the time period of the pendulum?

Q.6: Tanya takes 30 minutes from her house to reach her school by car. If the car has a speed of 2 m/s, calculate the distance between her house and the school.

Q.7: The distance time graph for three car A, B and C is given below. Which one of them is moving faster?

Q.8: Convert the speed of car 18 km/h into basic unit.

Answer:

SI Stands for "Système international". its a french word. It arrives from Système international d'unités i.e. International System of Units.
m/s
24*60*60 second
2 km/h
2 second
3600 m
B>C>A
5 m/s

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