Motion Test

Result

Motion Test - 7

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Question 1
1 / -0

In case of uniform circular motion, the acceleration is

A
zero

B
constant in magnitude and directed radially inwards

C
variable in magnitude

D
variable in magnitude but tangential to the circle

Solution

In case of a circular motion,

In uniform circular motion, direction of velocity (tangential) as well direction of acceleration(radially inward) keeps on changing.
Magnitude of velocity and magnitude of acceleration is constant.
Question 2
1 / -0

A particle has an initial velocity of 9 m/s due east and a constant acceleration of 2 m/s² due west. The distance covered by the particle in the 5^thsecond of its motion is

A
0

B
0.5 m

C
2 m

D
None of these

Solution

To find the distance, it is to be determined first if the particle has changed its direction during the course of motion. This can be determine by knowing the time of zero velocity. If this time lies within the interval of interest, the time interval should be divided into two parts, the distance should be calculated seperately and then the total distance is the summation of these distances.

u = 9, a = -2, v = 0

Displacement between, t = 4 to t = 4.5

Displacement between, t = 4.5 to t = 5

s = |0.25| + |- 0.25| = 0.5 m
Question 3
1 / -0

A particle moves along the x- axis as follows:

It starts from rest at t = 0 from a point x = 0 and comes to rest at t = 1 at a point x = 1. No other information is available about its motion for the intermediate time (0 < t < 1). If α denotes the instantaneous acceleration of the particle, then

A
α cannot remain positive for all t in the interval 0 ≤ t ≤ 1

B
|α| cannot exceed 2 at any point in its path

C
|α| can never be ≥ 4

D
α must change sign during the motion, but no other assertion can be made with the information given

Solution

The motion of the particle is simple harmonic. The particle is at one extreme position x = 0 at t = 0 and at the other extreme position at x = 1 unit at t = 1s.
Therefore, the time period of the motion is T = 2s and its amplitude A = unit.
Therefore, angular frequency is ω = 2π/T = 2π/2 = π rad s^-1.

∴ Maximum acceleration is α_max= ω²A = (π)²× 4.93.
Thus, the value of α can exceed 2 and can be greater than 4 at some points in the path of the particle.
Further, in simple harmonic motion, α changes sign and cannot remain positive in the time interval 0 ≤ t ≤ 1s.
Question 4
1 / -0

A person travels along a straight road for the first half-length with a constant speed v₁ and the second half-length with a constant speed v₂. The average speed v is

A
(v₁ + v₂)/2

B
2v₁v₂/(v₁ + v₂)

C
(v₁v₂)^1/2

D
(v₂/v₁)^1/2

Solution

A person travels along a straight road.
Average speed =
Let distance travelled = 2d
Time to cover first half-length = and time to cover second half-length =
Average speed =
Question 5
1 / -0

The displacement s of a point moving in a straight line is given by s = 8t² + 5t - 5, where s being in cm and t in s. The initial velocity of the particle is

A
5 cm/s

B
16 cm/s

C
19 cm/s

D
zero

Solution

Given, s = 8t²+ 5t - 5
Velocity \'v\' = ds/dt = 16t + 5
At t = 0 sec, initial velocity = 16(0) +5 = 5 cm/s
Question 6
1 / -0

A ball is dropped from a height of 90 m. Its velocity just before touching the floor is (g = 10 ms^-2)

A
900 ms^-1

B
30ms^-1

C
30/ms^-1

D
1800 ms^-1

Solution

Height = 90 m
Initial potential energy = mgh = m (g) (90) J
Just before touching the floor it has no potential energy and all P.E. is converted to K.E.
Let velocity at this moment is V.
mg(90) =
v =
v =
= m/s
Question 7
1 / -0

A ball is thrown upwards with a velocity v. It attains a height of 50 m and comes back to the thrower. Choose the correct statement. (g = 10 m/s²)

A
The total distance covered is zero

B
The total distance covered is 50 m

C
The displacement is 104 m

D
The time taken is seconds

Solution

s = ut + 1/2 gt²
s = 1/2 x 10 x t²
t = √10 seconds
For return, t = √10 seconds

Hence, total time = 2√10 seconds
Question 8
1 / -0

The acceleration due to gravity on the Moon is one-sixth that on the Earth. A high-jumper can jump 2 m on Earth. What distance can he jump on the Moon?

A
2 m

B
m

C
12 m

D
3 m

Solution

The initial potential energies on both the Earth and the Moon will be same. So, by equating the energies, we get,
mgh = mg'h'
g × 2 = g/6 × h'
h' = 12 m
Question 9
1 / -0

A particle starts moving from the position of rest under a constant acceleration. If it travels a distance x in the first 10 seconds and distance y in the next 10 seconds, then

A
y = x

B
y = 2x

C
y = 3x

D
y = 4x

Solution

Let constant acceleration = a
Distance travelled in time t seconds =
Then, distance travelled in first 10 seconds is x, where x = a(100) = 50a
Distance travelled in the first 20 seconds = = 200a
Hence, distance travelled in the next 10 seconds (y)= 200a – 50a

y = 150a = 3(50a) = 3x
Question 10
1 / -0

Tripling the speed of a motor car multiplies the distance needed for stopping it by

A
3

B
6

C
9

D
12

Solution

v² - u² = 2as
As v =0; -u² = 2as
s = u²/2a (since a is negative so negative sign cancels each other)
s is proportional to u²
s = ku²
Hence, tripling the speed makes the distance 9 times for stopping.
Question 11
1 / -0

A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. Where is the ball at the time T/2 seconds?

A
At h/4 m from the ground

B
At h/2 m from the ground

C
At 3h/4 m from the ground

D
Depends upon the mass and volume of the ball

Solution

Height = h metres, Time = T seconds
Since in the beginning, u = 0 m/s

h = a =
When time is , then
H =
H =
H = from the top of the tower
Hence, from the bottom, it would be h - =
Question 12
1 / -0

A pebble is thrown vertically upwards from a bridge with an initial velocity of 4.9 m/s. It strikes the water after 2 s. The height of the bridge is

A
19.6 m

B
14.7 m

C
9.8 m

D
4.9 m

Solution
Question 13
1 / -0

A bomb is released by a horizontally flying aeroplane. The trajectory of the bomb is a

A
straight line

B
parabola

C
hyperbola

D
circle

Solution

It would be a parabola.

The bomb will have some inital velocity when it will be droped from the plane. also, it will be acted by gravity. As a result the final trajectory will be that of a parabola.
Question 14
1 / -0

A river is flowing at the rate of 50 km/h. A person, who can row the boat at 10 km/h, wants to reach a point just opposite on the other bank of the river. He should row

A
upstream

B
straight

C
inclined to the bank upstream

D
inclined to the bank downstream

Solution

To reach B from A, a person will have to swim inclined upstream because he has to reach the exact opposite point. The current will cause him to reach the point.
Question 15
1 / -0

An observer moves with a constant speed along a line joining two stationary objects. He will observe that the two objects

a. have the same speed
b. have the same velocity
c. move in the same direction
d. move in opposite directions

Which of the above statements are correct?

A
Only a and b

B
Only a, b and c

C
Only c and d

D
Only b, c and d

Solution

Since velocity of both objects is zero, thus the relative velocity and speed will be same for both objects also the direction will be the same.
Question 16
1 / -0

The velocity of light emitted by a source S, observed by an observer O, who is at rest with respect to S is c. If the observer moves towards S with velocity v, the velocity of light as observed will be

A
c + v

B
c - v

C
[1 - (c²/v²)]^1/2

D
c

Solution

Speed of light will remain same i.e., `c`, because The speed of light is independent of the motion of the observer.
Question 17
1 / -0

A wooden block is dropped from the top of a cliff 100 m high and simultaneously a bullet of mass 10 g is fired from the foot of the cliff upwards with a velocity of 100 m/s. The bullet and wooden block will meet each other after a time

A
10 s

B
0.5 s

C
1 s

D
7 s

Solution

Let them meet after time t
Then, h₁ + h₂ = 100 m -- 1
h₁ = g t² -- 2
h₂ = 100 t – --- 3
From equations 1 , 2 and 3 we have
100 t = 100
t = 1 second
Question 18
1 / -0

A parachutist steps from an aircraft, falls freely for two seconds, and then opens his parachute. Which of the following acceleration-time (a - t) graphs best represents his downward acceleration during the first 5 seconds?

A

B

C

D

Solution

Initially the object is falling down with constant acceleration. As the parachute opens its surface area is increases as a result upthrust on the parachute increases.
For a moment, upthrust will increase the weight of the person and direction of the acceleration will change. Slowly, it will be balanced by the weight and acceleration becomes zero.
Question 19
1 / -0

The area of the shaded portion of the graph represents

A
average acceleration

B
maximum KE

C
momentum

D
displacement

Solution

The area of the shaded portion of the graph represents displacement.
Question 20
1 / -0

Rotatory motion of a body is produced by applying a/an

A
couple

B
impulse of a force

C
force

D
impact force

Solution

Rotatory motion of a body is produced by applying a pair of parallel forces acting in opposite directions. These forces are called a couple.