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Motion - Olympiad Level MCQ, Class 9 Science - Class 9 MCQ


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25 Questions MCQ Test - Motion - Olympiad Level MCQ, Class 9 Science

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Motion - Olympiad Level MCQ, Class 9 Science - Question 1

A car travels first one-third of a certain distance with a speed of 10 km/hr, the next one-third of the distance with the speed of 20 km/hr and the last one-third distance with a speed of 60 km/hr. The average speed of the car for the whole journey is

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 1

Let the total distance be 3x km. Now, average speed is given by 

average speed = (total distance travelled) / (total time taken)

So, let times taken for the first, second and third halves of the journeys be t1, t2, and t3 respectively. Then, 

t1 = x/10

t2 = x/20

t3 = x/60

So, average speed = x / (t1 + t2 + t3)

= x / (x/10 + x/20 + x/60)

= 3/ ((6 + 3 + 1)/60)

= (3*60)/10

= 18 km/hr

The correct option is A.

Motion - Olympiad Level MCQ, Class 9 Science - Question 2

A motor ship covers the distance of 300 km between two localities on a river in 10 hrs downstreamand in 12 hrs upstream. Find the flow velocity of the river assuming that these velocities are constant.

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 2
Let the velocity of flow in the river be x kmph and the speed of the motor ship be s kmph.
While going downstream the ship takes 300/(s+x) = 10 hours, or
s + x = 30 …(1)
While going upstream the ship takes 300/(s-x) = 12 hours, or
s - x = 25 …(2)
Add (1) and (2)
2s = 55, or
s = 27.5 kmph and x = 30–27.5 = 2.5 kmph.
The velocity of flow of the river is 2.5 kmph and the speed of the motor ship is 27.5 kmph, in still water.

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Motion - Olympiad Level MCQ, Class 9 Science - Question 3

Driver of a train travelling at 115 km/hr sees on a same track, 100m infront of him, a slow traintravelling in the same direction at 25 km/hr. The least retardation that must be applied to fastertrain to avoid a collision is

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 3
Velocity of train 'B' with respect to train 'A' = (115−25)km/h
VBA=−90km/h =−25m/s
Distance between the trains is 100m. Using the equations of motion
v^2=u^2−2as ,v=0
u^2=2as
625=2xax100
Then the least retardation that must be appiled to faster train to avoid a collision, a=3.125m/s^2
Motion - Olympiad Level MCQ, Class 9 Science - Question 4

The moon is 4×108m away from the earth. A radar signal with a velocity of 3.08×108m/s transmitted from the earth will reach the moon in about :

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 4

speed = distance / time

or

time = distance / speed

here,

distance between earth and moon = 4 x 108 m

speed of radar signal = 3 x 108m/s [speed of light ]

so,

the time taken by radar signal to reach the moon will be

t =  4 x 108 / 3 x 108

thus,

t = 1.33 s

Motion - Olympiad Level MCQ, Class 9 Science - Question 5

A stone is dropped into a well in which the level of water is h, below the top of the well. If v is velocity of sound, then time T after which the splash is heard is equal to

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 5
Here T is the time taken by the stone to reach to the water level plus time taken by the sound to come out from the water level to top of the well. 

Now, if T1 is the time taken by the stone to reach the water level then,
h = 1/2 x g x T1^2
=> T1 = (2h/g)^1/2

Now since v is the velocity of the sound, time taken by velocity to reach the top is
T2 = h/v

So, total time is T = T1 + T2 = (2h/g)^1/2 + h/v

Motion - Olympiad Level MCQ, Class 9 Science - Question 6

A stone weighing 3 kg falls from the top of a tower 100 m high and buries itself 2 m deepin the sand. The time of penetration is :-

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 6

Given:
- Weight of the stone (m) = 3 kg
- Height of the tower (h) = 100 m
- Depth of the penetration in the sand (d) = 2 m
We need to calculate the time of penetration.
Step 1: Calculating the Initial Velocity (u)
To calculate the initial velocity, we can use the equation:
v^2 = u^2 + 2gh
where,
v = final velocity (which is 0 as the stone comes to rest)
u = initial velocity
g = acceleration due to gravity (which is approximately 9.8 m/s^2)
h = height of the tower
Substituting the values, we have:
0 = u^2 + 2 * 9.8 * 100
0 = u^2 + 1960
u^2 = -1960
Since velocity cannot be negative, we discard the negative value. Therefore,
u = √1960 ≈ 44.27 m/s
Step 2: Calculating the Time of Penetration (t)
We can use the equation of motion:
s = ut + (1/2)at^2
where,
s = distance traveled (which is the sum of the height of the tower and the depth of penetration)
u = initial velocity
t = time of penetration
a = acceleration due to gravity (which is approximately 9.8 m/s^2)
Substituting the values, we have:
100 + 2 = 44.27 * t + (1/2) * 9.8 * t^2
102 = 44.27t + 4.9t^2
4.9t^2 + 44.27t - 102 = 0
Solving this quadratic equation, we find that:
t ≈ 0.0903 sec or t ≈ -2.703 sec
Since time cannot be negative, we discard the negative value. Therefore,
t ≈ 0.0903 sec
Hence, the time of penetration is approximately 0.09 seconds.
Therefore, the correct answer is option A: 0.09 sec.
Motion - Olympiad Level MCQ, Class 9 Science - Question 7

The velocity of a body at any instant is 10 m/s. After 5 sec, velocity of the particle is 20 m/s. The velocity at 3 seconds before is

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 7
Given information:
- The velocity of a body at any instant is 10 m/s.
- After 5 seconds, the velocity of the particle is 20 m/s.
To find:
- The velocity at 3 seconds before.

Let's assume the initial velocity of the body as u and the final velocity as v.
From the given information, we have:
- Initial velocity, u = 10 m/s
- Final velocity, v = 20 m/s
- Time, t = 5 seconds
We can use the formula for velocity:
v = u + at
where a is the acceleration. Since the problem does not provide any information about acceleration, we can assume it to be constant.
Substituting the given values, we have:
20 = 10 + a * 5
Simplifying the equation, we get:
10 = 5a
a = 2 m/s²
Now, we can find the velocity at 3 seconds before using the formula:
v = u + at
where t is the time before the given time.
Substituting the values:
- Initial velocity, u = 10 m/s
- Time, t = 5 - 3 = 2 seconds
- Acceleration, a = 2 m/s²
v = 10 + 2 * 2
v = 10 + 4
Therefore, the velocity at 3 seconds before is 14 m/s. Hence, option B is correct.
Motion - Olympiad Level MCQ, Class 9 Science - Question 8

A body covers 200 cm in the first 2 sec.and 220 cm in next 4 sec. What is the velocity of thebody at the' end of 7th second?

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 8

Let 'u' be the initial velocity and 'a' the acceleration.

So we have the distance formula
s = ut + 1/2 at^2

In first 2 seconds,
s = 200
Put the values in the formula.
200 = u x 2 + 1/2 x a x (2)^2
200 = 2u + 1/2 x 4 x a
200 = 2u + 2a
100 = u + a -----(I)

In next 4 seconds
Let the distance traveled be 'y'
Time = 2+4 = 6 sec
So according to the formula ,
y = u x 6 + 1/2a x (6)^2
y = 6u + 1/2 x 36 x a
y = 6u + 18a ------(ii)

Now we know that 220 cm was traveled in between 2 sec - 6 sec
y - 200 = 220
y = 420

We know y = 6u + 18a (from [ii])

So, 6u + 18a = 420
u + 3a = 70 ------(iii)

Equating (I) and (iii)
-2a = 30
a = -15 cm/s^2

u = 100 - (-15)
u = 100 + 15 = 115 cm/sec

Now we know v = u + at

We have to find the "v" after 7th second

So v = 115 + (-15) x 7
v = 115 - 105
v = 10 cm/sec

Motion - Olympiad Level MCQ, Class 9 Science - Question 9

A boat takes 2 hrs. to travel 8 km and back in still water lake with water velocity of 4 km/hr, then the time taken for going upstream of 8 km and coming back is :-

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 9
Here the speed of water is still. So, they will have equal velocity and equal time taken. 
☆ Total time = 2 hrs 
Time for upstream = 1 hr
Downstream = 1 hr 
☆ Their velocity is 8 kmph each.
So, new downstream velocity will be (8+4) = 12 kmph 
new upstream velocity will be (8-4) = 4 kmph 
TIME TAKEN = DISTANCE / VELOCITY
(distance is 8 km in both cases ) =( 8/12 + 8/4 ) hours =( 2/3 + 2 ) hrs = 160 mins = 2 hr 40 mins.
Motion - Olympiad Level MCQ, Class 9 Science - Question 10

If two bodies of different masses m1 and m2 are dropped from differnet heights h1 and h2, thenratio of the times taken by the two to drop through these distances is :-

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 10

For a freely falling object :
a=g;

Initial velocity=u=0 m/s

g=9.8m/s²

from second equation of motion:S=ut+1/2 at²


for First object we get :


h1= gT1²/2

⇒T1=√2h1/g


For second object we get h2=gT2²/2

⇒T2=√2h2/g

T1/T2 =√h1/h2

But here h1= a and h2 =b
so, T1/T2=√a/b

Motion - Olympiad Level MCQ, Class 9 Science - Question 11

A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If theball is in contact with the floor for 0.01 sec, then average acceleration during contact is :-

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 11
When it is dropped from 10m,
Initial height = 10m
initial velocity = 0
velocity just before hotting ground = √2gh = √2*9.8*10 = 14.07 m/s (downward)

after rebound,
maximum height reached = 2.5m
final velocity at top = 0
initial velocity(just after rebound) = √2gh = √2*9.8*2.5 = √49 = 7 m/s (upward)

assuming downward as positive direction
So velocity just before hitting ground = +14.07 m/s 
velocity just after hitting ground = -7 m/s 
change in velocity = +14.07 - (-7) = 21.07 m/s
time = 0.01s
acceleration = change in velocity/time = 21.07/0.01 = 2107 m/s^2 ~ 2100m/s^2
Motion - Olympiad Level MCQ, Class 9 Science - Question 12

A stone is thrown vertically upward with an initial velocity u from the top of a tower, reaches the ground with a velocity 3u. The height of the tower is :-

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 12

H = (v² - u²) / (2g)
H = (9u² - u²) / (2g)
H = 8u² / (2g)
H = 4u²/g
Height of the tower is 4u²/g

Motion - Olympiad Level MCQ, Class 9 Science - Question 13

If a ball is thrown up with a certain velocity. It attains a height of 40 m and comes back to thethrower, then :-

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 13
Since the ball was thrown from a height and returns to the same height therefore initial position = final position. since displacement = Straight line distance between initial and final position therefore displacement = 0. since the ball attains the height of 40 m therefore while returningit covers the same distance.therefore distance = 2× 40 m = 80 m.
Motion - Olympiad Level MCQ, Class 9 Science - Question 14

Acceleration of a body projected upwards with a certain velocity is

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 14
Acceleration of a body projected upwards with a certain velocity is
To determine the acceleration of a body projected upwards with a certain velocity, we need to consider the forces acting on the body and apply Newton's second law of motion.
Newton's Second Law of Motion:
The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. It can be stated as:
F = ma
Where:
F is the net force acting on the object,
m is the mass of the object,
a is the acceleration of the object.
For a body projected upwards, the following forces are acting:
1. Gravitational force (weight) acting downwards.
2. Air resistance (if present) acting upwards.
Analysis:
1. When a body is projected upwards, the initial velocity is in the upward direction.
2. The only force acting on the body is the gravitational force (weight) acting downwards.
3. The gravitational force is always directed towards the center of the Earth and is equal to the mass of the body multiplied by the acceleration due to gravity (9.8 m/s^2).
4. As the body moves upwards, the gravitational force opposes its motion, causing it to decelerate.
5. Therefore, the acceleration of the body projected upwards is negative (-9.8 m/s^2).
Conclusion:
The acceleration of a body projected upwards with a certain velocity is -9.8 m/s^2 (Option B).
Motion - Olympiad Level MCQ, Class 9 Science - Question 15

If a body of mass 0.10 kg is moving on circular path of diameter 1.0 m at the rate of 10 revolutionsper 31.4 sec, then centripetal force acting on the body (n = 3.14) is

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 15
first get velocity tangent to circle v


it goes pi D meters in 3.14 seconds


v = pi D/3.14 = pi * 1/ 3.14 = 1 m/s


centripetal acceleration = Ac =v^2/R =1*1/.5

= 2 m/s^2


F = m Ac = .1 * 2 = .2 Newtons
Motion - Olympiad Level MCQ, Class 9 Science - Question 16

The earth's radius is 6400 km. It makes one revolution about its own axis in 24 hrs. The centripetalacceleration of a point on its equator is nearly

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 16
Time period of earth’s rotation is, T =  24 h = 86400 s

Angular velocity, ω = 2π/T

Radius of earth, r = 6400 km = 640000 m

Now, centripetal acceleration = ω^2r = (2π/T)^2r = 0.034 m/s^2 = 3.4 cm/s^2
Motion - Olympiad Level MCQ, Class 9 Science - Question 17

The acceleration of a point on the rim of flywheel 1 m in diameter, if it makes 1200 revolutions per minute is

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 17

Motion - Olympiad Level MCQ, Class 9 Science - Question 18

A phonograph record on turn table rotates at 30 rpm. The linear speed of a point on the record atthe needle at the beginning of the recording when it is at a distance of 14 cm from the centre is

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 18
Linear velocity V = angular velocity × radius. Angular velocity = [30×2π]/60 R= 14s. So V = 44 cm/s.
Motion - Olympiad Level MCQ, Class 9 Science - Question 19

The relationship between average speed, time and distance is

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 19
Relationship between average speed, time, and distance:

  • Average speed = Distance / Time: This is the correct relationship between average speed, distance, and time. It states that the average speed is calculated by dividing the total distance traveled by the time taken to cover that distance.

  • Time = Distance / Average speed: This is not the correct relationship. Time cannot be determined by dividing distance by average speed. Instead, it is obtained by dividing distance by average speed.

  • Distance = Average speed × Time: This is not the correct relationship. Distance cannot be determined by multiplying average speed by time. Instead, it is obtained by multiplying average speed by time.


Therefore, the correct relationship between average speed, time, and distance is:
Average speed = Distance / Time (Option C).
Motion - Olympiad Level MCQ, Class 9 Science - Question 20

A body moving along a circular path has

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 20
Explanation:

When a body moves along a circular path, it experiences certain characteristics:



  • Constant speed: The body moves at a constant rate around the circle, covering equal distances in equal intervals of time.

  • Constant velocity: The body's velocity, which includes both speed and direction, remains constant as long as it moves at a uniform rate along the circular path.

  • No radial acceleration: The body does not experience any radial acceleration because the distance from the center of the circle remains constant.

  • No tangential velocity: The body's tangential velocity, which is the velocity along the tangent of the circular path, changes as it moves along the circle, but it does not become zero.


Based on these characteristics, we can conclude that a body moving along a circular path has constant speed. Therefore, option A is correct.

Motion - Olympiad Level MCQ, Class 9 Science - Question 21

A rubber ball dropped from a certain height is an example of

Motion - Olympiad Level MCQ, Class 9 Science - Question 22

If the velocity of a body does not change, its acceleration is

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 22
Answer:
Given: The velocity of a body does not change.
To find: The acceleration of the body.

When the velocity of a body does not change, it means that the body is moving with a constant velocity. In this case, the acceleration of the body is zero.
Explanation:
Acceleration is defined as the rate of change of velocity. If the velocity of a body does not change, it means there is no change in the rate at which the body is moving. Therefore, the acceleration of the body is zero.
Conclusion:
If the velocity of a body does not change, its acceleration is zero.
Motion - Olympiad Level MCQ, Class 9 Science - Question 23

When the distance an object travels is directly proportional to the length of time, it is said to travel with

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 23
Explanation:
When the distance an object travels is directly proportional to the length of time, it means that for every unit of time, the object travels a constant distance. This indicates that the object is moving with a constant speed. Here's a detailed explanation:
Direct Proportion:
In a direct proportion, two quantities are related in such a way that an increase or decrease in one quantity results in a corresponding increase or decrease in the other quantity. In this case, the distance traveled by the object is directly proportional to the time taken.
Constant Speed:
When an object travels with a constant speed, it means that it covers equal distances in equal intervals of time. In other words, the object maintains a steady rate of motion, covering the same amount of distance for each unit of time.
Example:
To illustrate this concept, let's consider a car traveling on a straight road. If the car travels at a constant speed of 60 kilometers per hour, it will cover a distance of 60 kilometers in one hour, 120 kilometers in two hours, and so on. The distance traveled by the car is directly proportional to the time elapsed.
Conclusion:
Therefore, when the distance an object travels is directly proportional to the length of time, it indicates that the object is traveling with a constant speed. So, the correct answer is option B: constant speed.
Motion - Olympiad Level MCQ, Class 9 Science - Question 24

A body moves on three quarters of a circle of radius r. The displacement and distance travelled by it are:-

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 24
Analysis:
To solve this problem, we need to understand the concepts of displacement and distance.
Displacement:
Displacement is a vector quantity that represents the change in position of an object. It is the straight-line distance between the initial and final positions of the object, along with its direction.
Distance:
Distance is a scalar quantity that represents the total path length traveled by an object. It does not take into account the direction of motion, only the magnitude.

Given that the body moves on three quarters of a circle of radius r, we can analyze the displacement and distance traveled as follows:
Displacement:
The body moves on three quarters of a circle, which means it starts and ends at the same point. Therefore, the displacement is zero.
Distance:
To calculate the distance traveled, we need to find the length of the three-quarters of the circumference of the circle.
- The circumference of a circle is given by 2πr.
- Three-quarters of the circumference is equal to (3/4) * 2πr = (3π/2)r.
Therefore, the distance traveled by the body is (3π/2)r.
Conclusion:
Based on the analysis, we can conclude that the correct answer is option B: displacement = 0 and distance = (3π/2)r.
Motion - Olympiad Level MCQ, Class 9 Science - Question 25

For the motion on a straight line path with constant acceleration the ratio of the magnitude of the displacement to the distance covered is :-

Detailed Solution for Motion - Olympiad Level MCQ, Class 9 Science - Question 25

To solve this problem, we need to understand the definitions of displacement and distance covered in the context of motion with constant acceleration.
- Displacement: Displacement refers to the change in position of an object. It is a vector quantity and is measured in units of length (meters, kilometers, etc.). Displacement takes into account both the magnitude (distance) and direction of the change in position.
- Distance Covered: Distance covered refers to the total length of the path traveled by an object. It is a scalar quantity and is measured in units of length. Distance covered only considers the magnitude of the path traveled and does not take into account the direction.
Now, let's analyze the given options and determine the correct ratio:
A: = 1
This option states that the ratio of displacement to distance covered is equal to 1. This means that the magnitude of displacement and the distance covered are the same. Since displacement is a vector quantity and takes into account both magnitude and direction, this option is correct.
B:
This option suggests that the ratio of displacement to distance covered is greater than 1. However, this is not possible as displacement cannot be greater than the total distance covered.
C:
This option suggests that the ratio of displacement to distance covered is less than 1. However, this is also not possible as displacement cannot be less than the total distance covered.
D: < 1
This option suggests that the ratio of displacement to distance covered is less than 1, which is not possible.
Therefore, the correct answer is option A: = 1.
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