Laws of Motion Online Objective Test | Laws of Motion Online Objective Test

# Laws of Motion # CBSE 11TH PHYSICS Prepare / Learn

Q1. In which of the following cases is the net force zero

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

B. a coin falling under gravity

C. a rocket just before lift off

D. a feather falling under gravity

Answer : Option A
Explaination / Solution:

As no external fields like, electric,magnetic and gravitational field acting on electron the net force is zero

# Laws of Motion # CBSE 11TH PHYSICS Prepare / Learn

Q2. 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

B. 0.05 N vertically down

C. 0.5 N vertically down

D. 0.5 N vertically up

Answer : Option C
Explaination / Solution:

F = mg = 0.05 x 10 = 0.5 N

# Laws of Motion # CBSE 11TH PHYSICS Prepare / Learn

Q3. 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 up

B. 0

C. 0.5 N vertically down

D. 0.05 N vertically down

Answer : Option C
Explaination / Solution:

F = mg = 0.05 x 10 = 0.5 N downwards

# Laws of Motion # CBSE 11TH PHYSICS Prepare / Learn

Q4. 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

B. 0.5 N vertically down

C. 0.5 N vertically up

D. 0.05 N vertically down

Answer : Option B
Explaination / Solution:

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

# Laws of Motion # CBSE 11TH PHYSICS Prepare / Learn

Q5. Neglecting air resistance, in which of the following cases the net force acting on a stone of mass 0.1 kg is not acting downward?

just after it is dropped from the window of a train running at a constant velocity of 36 km/h

just after it is dropped from the window of a train accelerating with 1 m

s^{- 2}

C. lying on the floor of a train which is accelerating with 1 m S^-2 , the stone being at rest relative to the train

D. just after it is dropped from the window of a stationary train

Answer : Option C
Explaination / Solution:

Weight of the stone is balanced by the reaction of the floor. The only acceleration is provided by the horizontal motion of the train.

a = 1ms-2

force in horizontal direction

F = ma = 0.1 x 1 = 0.1N

# Laws of Motion # CBSE 11TH PHYSICS Prepare / Learn

Q6.

Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg lying on the floor of a train which is accelerating with 1 m $s^{- 2}$ , the stone being at rest relative to the train. Neglect air resistance.

A. 1.0 N in the direction of motion

B. 0.1 N along the direction of motion

C. 0.2 opposite to the direction of motion

D. 0.1 N opposite to the direction of motion

Answer : Option B
Explaination / Solution:

Weight of the stone is balanced by the reaction of the floor. The only acceleration is provided by the horizontal motion of the train.

a = 1ms-2

force in horizontal direction

F = ma = 0.1 x 1 = 0.1N

# Laws of Motion # CBSE 11TH PHYSICS Prepare / Learn

Q7. One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v, T is the tension in the string, the net force on the particle directed towards the centre is

A. T

B. T+mv2l

C. T−mv2l

D. 0

Answer : Option A
Explaination / Solution:

According to Newton's third law of motion, the net force on a rotating particle is equal to Tension in the String. As the action (ie force towards the centre) and reaction (tension in the string) are equal in magnitude and opposite in direction.

# Laws of Motion # CBSE 11TH PHYSICS Prepare / Learn

Q8.

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m $s^{- 1}$ . How long does the body take to stop?

A. 4.0 s

B. 6.0 s

C. 4.0 s

D. 5.0 s

Answer : Option B
Explaination / Solution:

F=mdvdtdt=mdvF=20×1550=6s

# Laws of Motion # CBSE 11TH PHYSICS Prepare / Learn

Q9. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m

s^{- 1}

to 3.5 m

s^{- 1}

in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

A. 0.07 m

s^{- 2}

, 0.16 N in the direction of motion

B. 0.08 m

s^{- 2}

, 0.14 N in

C. 0.18 N along the direction of motion

D. 0.09 m

s^{- 2}

, 0.12 N in the direction of motion

Answer : Option C
Explaination / Solution:

$F = m \frac{d v}{d t} = 3 (\frac{3.5 - 2}{25}) = 0.18 N$

# Laws of Motion # CBSE 11TH PHYSICS Prepare / Learn

Q10. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the net force acting on the body.

A. 14 N at an angle of 37

^{\circ}

to the 8 N force

B. 16 N at an angle of 46

^{\circ}

to the 8 N force

C. 12 N at an angle of 36

^{\circ}

^{\circ}

towards the 8 N force

D. 12 N at an angle of 38

^{\circ}

to the 8 N force

Answer : Option C
Explaination / Solution:

Let F1 = 8N & F2 = 6 N.

then the resultant force is $F = \sqrt{F_{1}^{2} + F_{2}^{2}}$

= $\sqrt{8^{2} + 6^{2}}$

= 10N

The direction is determined as $t a n θ = \frac{F_{2}}{F_{1}}$

= 6/8

therefore $θ = 36^{o} 52^{o}$

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