Laws Of Motion Test

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Laws Of Motion Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

A stone of mass 5 kg is attached to a string of 10 m length and is whirled in a horizontal circle. The string can withstand a maximum tension of 160 N. The maximum velocity of revolution that can be given to the stone without breaking the string is

A
17.88 m/s

B
16 m/s

C
20 m/s

D
19.4 m/s

Solution

Max tension = Max. Centripetal force
160= mv^2 / r => v^2 = 160 *10 / 5 => V= 17.88.
Question 2
1 / -0

The acceleration of the particle performing uniform circular motion is

A
ω²/r

B
0

C
vr

D
v²/r

Solution

We know that centripetal force is required for an object to travel in an circular motion thus the acceleration of the object in circular motion will be (mv²/r) / m
Question 3
1 / -0

It is advised to drive along the slippery road slowly, because:

A
Normal reaction gets reduced

B
advice is not correct

C
water on the road opposes the motion

D
Friction is reduced

Solution

It is advised to drive along the slippery road slowly, because friction is reduced and if the driver is driving at a larger speed there is a risk of skidding.
Question 4
1 / -0

Centripetal force always acts at 90 degrees to the velocity, and away from the centre of the circle.

A
true

B
cannot predict

C
false

D
none of these

Solution

Centripetal always act towards centre.as centrifugal act away from centre.
Question 5
1 / -0

Railway tracks are banked at the curves so that the necessary centripetal force may be obtained from the horizontal component of the reaction on the train.

A
true

B
false

C
cannot predict

D
none of these

Solution

True. Railway tracks are banked at the curves so that the necessary centripetal force may be obtained from the horizontal component of the reaction on the train.
Question 6
1 / -0

Which of the following is called a fictitious force?

A
Gravitational force

B
Frictional force

C
Centrifugal force

D
Centripetal force

Solution

Centrifugal force, a fictitious force, peculiar to a particle moving on a circular path, that has the same magnitude and dimensions as the force that keeps the particle on its circular path (the centripetal force) but points in the opposite direction.
Question 7
1 / -0

At which place of the earth, the centripetal force is maximum

A
At the earth surface

B
At the equator

C
At the north pole

D
At the south pole

Solution

The shape of earth is geoid due to which radius of earth is 14km shorter at equator than on poles. And as centripetal force, f = mv² /r
We get f is maximum at the equator.
Question 8
1 / -0

The maximum speed at which a car can turn round a curve of 400 m radius on a level road is: (Given: coefficient of friction between the tyres and road is 0.1, and g = 10 m/s²)

A
20 m/s

B
14 m/s

C
26 m/s

D
21 m/s

Solution

When the car moves in a curvature of radius R, the force acting towards the centre of curvature must be equal to mv²/R
where m is mass of the car, v is the speed of car
But when we draw the free body diagram of the car the only force towards the centre of curvature is the friction force which is f = μmg
Hence by comparing we get v = √μgR
Substituting values gives, v= 20ms^-1
Question 9
1 / -0

The angle through which the outer edge is raised above the inner edge is called

A
angle of inclination

B
angle of repose

C
angle of banking

D
angle of declination

Solution

A banked turn (or banking turn) is a turn or change of direction in which the vehicle banks or inclines, usually towards the inside of the turn. The bank angle is the angle at which the vehicle is inclined about its longitudinal axis with respect to the horizontal.
Question 10
1 / -0

A model aeroplane is tethered to a post and held by a fine line. It flies in a horizontal circle. Then the line breaks. What direction will it fly in?

A
In a circular path, as before

B
Directly to the centre of the circle

C
In a straight line at a tangent

D
Directly to the centre of the circle

Solution

It is because at every point of circle an object has two acceleration (Tangential & Angular acceleration).At every point a body experience a tangential force which is perpendicular to the radius of the circle, when the string breaks the centripetal force disappear ( the radially inward force which holds a body in a circular motion) Hence the only Tangential force act on the body & it goes in that way.

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Re-Attempt Weekly Quiz Competition

Laws Of Motion Test - 23

Result

Laws Of Motion Test - 23

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SHARING IS CARING

Question 1

17.88 m/s

16 m/s

20 m/s

19.4 m/s

Solution

Question 2

ω2/r

0

vr

v2/r

Solution

Question 3

Normal reaction gets reduced

advice is not correct

water on the road opposes the motion

Friction is reduced

Solution

Question 4

true

cannot predict

false

none of these

Solution

Question 5

true

false

cannot predict

none of these

Solution

Question 6

Gravitational force

Frictional force

Centrifugal force

Centripetal force

Solution

Question 7

At the earth surface

At the equator

At the north pole

At the south pole

Solution

Question 8

20 m/s

14 m/s

26 m/s

21 m/s

Solution

Question 9

angle of inclination

angle of repose

angle of banking

angle of declination

Solution

Question 10

In a circular path, as before

Directly to the centre of the circle

In a straight line at a tangent

Directly to the centre of the circle

Solution

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ω²/r

v²/r