Laws of Motion Test

Result

Laws of Motion Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

Rolling friction is

A
equal to static friction

B
less than static friction

C
greater than static friction

D
greater than or equal to static

Solution

As the area of contact is less in the case of rolling than in the case of sliding. In rolling surface of contact do not rub each other. The velocity of the point of contact with respect to the surface remains zero all the time although the Centre of mass of the wheel moves forward. Hence rolling friction is negligible in comparison to static friction.
Question 2
1 / -0

The centripetal force required for circular motion of a car such as taking a sharp turn is provided by

A
friction

B
gravitational force

C
rolling friction

D
contact force

Solution

it is the force of friction that acts between tyres and road.
Question 3
1 / -0

In which of the following cases is the net force non zero?

A
a kite skillfully held stationary in the sky

B
a drop of rain falling down with a constant speed

C
a cork of mass 10 g floating on water

D
a coin falling under gravity

Solution

The force acting on a coin is due to gravitational force which causes an acceleration in downward direction.
Question 4
1 / -0

In which of the following cases is the net force zero

A
a coin falling under gravity

B
a rocket just before lift off

C
a feather falling under gravity

D
a high-speed electron in space far from all material objects, and free of electric and magnetic fields

Solution

As no external fields like, electric,magnetic and gravitational field acting on electron the net force is zero
Question 5
1 / -0

A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, during its upward motion, Ignore air resistance

A
0.05 N vertically down

B
0.5 N vertically up

C
0.5 N vertically down

D
0

Solution

F = mg

= 0.05 x 10 = 0.5 N
Question 6
1 / -0

A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, during its downward motion, Ignore air resistance

A
0.5 N vertically down

B
0.05 N vertically down

C
0.5 N vertically up

D
0

Solution

F = mg = 0.05 x 10 = 0.5 N downwards
Question 7
1 / -0

A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, if the pebble was thrown at an angle of 45° with the horizontal direction, Ignore air resistance

A
0.05 N vertically down

B
0.5 N vertically down

C
0.5 N vertically up

D
0

Solution

F = mg = 0.05 x 10 = 0.5 N. The horizontal component of the velocity remain constant.

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Laws of Motion Test - 8

Result

Laws of Motion Test - 8

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SHARING IS CARING

Question 1

equal to static friction

less than static friction

greater than static friction

greater than or equal to static

Solution

Question 2

friction

gravitational force

rolling friction

contact force

Solution

Question 3

a kite skillfully held stationary in the sky

a drop of rain falling down with a constant speed

a cork of mass 10 g floating on water

a coin falling under gravity

Solution

Question 4

a coin falling under gravity

a rocket just before lift off

a feather falling under gravity

a high-speed electron in space far from all material objects, and free of electric and magnetic fields

Solution

Question 5

0.05 N vertically down

0.5 N vertically up

0.5 N vertically down

0

Solution

Question 6

0.5 N vertically down

0.05 N vertically down

0.5 N vertically up

0

Solution

Question 7

0.05 N vertically down

0.5 N vertically down

0.5 N vertically up

0

Solution

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