Motion in a Plane Test

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Motion in a Plane Test - 9

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Question 1
4 / -1

The term centripetal acceleration was proposed by

A
Huygens

B
Kepler

C
Newton

D
Galileo

Solution

The term centripetal acceleration was proposed by Newton.
Question 2
4 / -1

Centripetal acceleration is

A
A constant vector

B
A constant scalar

C
A magnitude changing vector

D
Not a constant vector

Solution

Centripetal acceleration, a_c = v²/R
where v is the speed of an object and R is the radius of the circle. It is always directed towards the center of the circle. Since v and R are constants for a given uniform circular motion, therefore the magnitude of centripetal acceleration is also constant. However, the direction of centripetal acceleration changes continuously. Therefore, centripetal acceleration is not a constant vector.
Question 3
4 / -1

What is approximately the centripetal acceleration (in units of acceleration due to gravity on earth, g = 10 m s^-2) of an air-craft flying at a speed of 400 m s^-1 through a circular arc of radius 0.6 km?

A
26.7

B
16.9

C
13.5

D
30.2

Solution

Here, v = 400ms⁻¹,
r = 0.6 km = 0.6 × 10³m,
g = 10ms⁻²
Centripetal acceleration,

= 26.7 × 10ms⁻²
In the units of (g = 10ms⁻²), the centripetal acceleration is 26.7.
Question 4
4 / -1

Velocity vector and acceleration vector in a uniform circular motion are related as

A
Both in the same direction

B
Perpendicular to each other

C
Both in opposite direction

D
Not related to each other

Solution

In a uniform circular motion, the acceleration is directed towards the centre while velocity is acting tangentially. Therefore, velocity vector is perpendicular to acceleration vector.
Question 5
4 / -1

A body executing uniform circular motion has its position vector and acceleration vector

A
Along the same direction

B
In opposite direction

C
Normal to each other

D
Not related to each other

Solution

In Circular motion, the Position vector is always outward and passes through the center. whereas the and centripetal acceleration is always toward the center.
∴ Both is the opposite direction.
Question 6
4 / -1

For a particle performing uniform circular motion, choose the incorrect statement form the following.

A
Magnitude of particle velocity (speed) remains constant.

B
Particle velocity remains directed perpendicular to radius vector.

C
Direction of acceleration keeps changing as particle moves.

D
Magnitude of acceleration does not remain constant.

Solution

For a particle performing uniform circular motion, magnitude of the acceleration remains constant.
Question 7
4 / -1

Which of the following statements is incorrect?

A
In one dimension motion, the velocity and the acceleration of an object are always along the same line

B
In two or three dimensions, the angle between velocity and acceleration vectors may have any value between 0°and 180°

C
The kinematic equations for uniform acceleration can be applied in case of a uniform circular motion

D
The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant

Solution

The kinematic equations for uniform acceleration do not apply in case of uniform circular motion because in this case the magnitude of acceleration is constant but its direction is changing.
Question 8
4 / -1

A particle is moving on a circular path of radius r with uniform speed v. What is the displacement of the particle after it has described an angle of 60°?

A
r√2

B
r√3

C
r

D
2r

Solution

Let the particle is moving from point A to B
Displacement of the particle = AB
According to cosine formula

Displacement AB = x = r
Question 9
4 / -1

A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference and returns to the centre along QO as shown in the figure.

If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour) is

A
0, 1

B

C

D
0, 21.4

Solution

Since the initial position coincides with the final position. So, the net displacement of the cyclist = zero
Question 10
4 / -1

A cyclist is riding with a speed of 27 km h^-1. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m s^-1 every second. The net acceleration of the cyclist on the circular turn is

A
0.68 m s^-2

B
0.86 m s^-2

C
0.56 m s^-2

D
0.76 m s^-2

Solution

Given here,
Speed of the cyclist, v = 27 km/h = 7.5m/s
Radius of the circular tun on the road, r = 80 m
Therefore the centripetal acceleration of the cyclist at the moment when velocity is 7.5 m/s is,

When brakes are applied, the speed is decreased at the rate of 0.5m/s every second.
Therefore the tangential acceleration is,
a_r = - 0.5 m/s²
Therefore the total acceleration of the cyclist is,