Motion - Olympiad Level MCQ, Class 9 Science


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


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This mock test of Motion - Olympiad Level MCQ, Class 9 Science for Class 9 helps you for every Class 9 entrance exam. This contains 25 Multiple Choice Questions for Class 9 Motion - Olympiad Level MCQ, Class 9 Science (mcq) to study with solutions a complete question bank. The solved questions answers in this Motion - Olympiad Level MCQ, Class 9 Science quiz give you a good mix of easy questions and tough questions. Class 9 students definitely take this Motion - Olympiad Level MCQ, Class 9 Science exercise for a better result in the exam. You can find other Motion - Olympiad Level MCQ, Class 9 Science extra questions, long questions & short questions for Class 9 on EduRev as well by searching above.
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

Solution:

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.

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.

Solution:
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.

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

Solution:
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
QUESTION: 4

Distance of the moon from the earth is 4 × 108 m. The time taken by a radar signal transmittedfrom the earth to reach the moon is

Solution:
speed = distance / time
or
time = distance / speed
here,
distance between earth and moon = 4 x 10^8 m
speed of radar signal = 3 x 108 m/s [speed of light ]
so,
the time taken by radar signal to reach the moon will be
t =  4 x 10^8 / 3 x 10^8
thus,
t = 1.33 s
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

Solution:
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

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 :-

Solution:
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

Solution:
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?

Solution:

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

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 :-

Solution:
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.
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 :-

Solution:

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

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 :-

Solution:
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
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 :-

Solution:

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

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 :-

Solution: 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.
QUESTION: 14

Acceleration of a body projected upwards with a certain velocity is

Solution:
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

Solution:
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
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

Solution:
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
QUESTION: 17

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

Solution:

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

Solution: Linear velocity V = angular velocity × radius. Angular velocity = [30×2π]/60 R= 14s. So V = 44 cm/s.
QUESTION: 19

The relationship between average speed, time and distance is

Solution:
QUESTION: 20

A body moving along a circular path has

Solution:
QUESTION: 21

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

Solution:
QUESTION: 22

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

Solution:
QUESTION: 23

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

Solution:
QUESTION: 24

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

Solution:
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 :-

Solution: