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

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

Q1. The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.Find out the deceleration produced and the retarding force on a vehicle.

– 2.3 m

s^{- 2}

, 1000 N

– 2.3 m

s^{- 2}

, 1162.5 N

C. – 2.2 m

s^{- 2}

, 900 N

D. – 2.4 m

s^{- 2}

, 1100 N

Answer : Option B
Explaination / Solution:

V= final velocity = 0

V0 = initial velocity = 36 km/h = 10 m/s

$a = \frac{v - v_{0}}{t} = \frac{0 - 10}{4} = - 2.5 \frac{m}{s^{2}}$

F= ma = -465 x 2.5 = -1162.5N (negative sign indicates the retarding force)

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

Q2. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s-2. Calculate the initial thrust (force) of the blast.

A. 300000 N

B. 365000 N

C. 378000 N

D. 350000 N

Answer : Option A
Explaination / Solution:

Initial thrust = upthrust required to impart acceleration + uthrust to overcome gravity

= $m a + m g = m (a + g) = 20000 (5 + 10) = 300000 N$

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

Q3. A body of mass 0.40 kg moving initially with a constant speed of 10 m /s to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s.

A. -60 m

B. -50 m

C. -55 m

D. 50 m

Answer : Option B
Explaination / Solution:

Let $u = 10 m s^{- 1}$ , $F = - 8 N, T = 30 s a n d m = 0.4 k g$

$a = \frac{F}{m} = \frac{- 8}{0.4} = - 20 m / s^{2}$

At t= -5 s, there is no force acting on the particle,

so,

$x = u t = 10 \times (- 5) = - 50 m$

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

Q4.

A truck starts from rest and accelerates uniformly at 2.0 m $s^{- 2}$ . At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the magnitude of velocity (in m $s^{- 1}$ ) of the stone at t = 11s? (Neglect air resistance.)

A. 18.5

B. 19.6

C. 22.4 m/s

D. 21.0

Answer : Option C
Explaination / Solution:

During first 10 , the horizontal component of the velocity is vx = u + at = 0+2x10 = 20 m/s

From 10 to 11 s, the vertical component of the velocity is vy=u + gt = 0 + 10x1 = 10m/s

Relative velocity is,

$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{20^{2} + 10^{2}} = \sqrt{500} = 22.4 m / s$

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

Q5.

A truck starts from rest and accelerates uniformly at 2.0 m $s^{- 2}$ . At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the magnitude of acceleration (in m $s^{- 2}$ ) of the stone at t = 11s? (Neglect air resistance.)

A. 14

B. 12

C. 8

D. 10

Answer : Option D
Explaination / Solution:

When the stone is dropped from the truck , the horizontal force acting on it become zero after stone is released. Stone continue to move under the influence of gravity only. So that acceleration of the stone is equal to the gravitational acceleration g.

a = g = 10ms-2 acting vertically downward.

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

Q6.

A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m $s^{- 1}$ .What is the trajectory of the bob if the string is cut when the bob is at one of its extreme positions

A. bob will go down in a parabolic path

B. bob will fall vertically downwards

C. bob will go upwards

D. bob will go downwards at an angle

Answer : Option B
Explaination / Solution:

At the extreme position of the oscillation, the speed of the bob is zero. So the bob is momentarily at rest. if the string is cut, the bob will fall vertically downwards.

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

Q7. A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m

s^{- 1}

.What is the trajectory of the bob if the string is cut when the bob is at its mean position?

A. bob will go upwards

B. bob will go down in a parabolic path

C. bob will fall vertically downwards

D. bob will fall vertically downwards

Answer : Option B
Explaination / Solution:

At the mean position, the bob has a horizontal velocity. so when the string is cut, it will fall along a parabolic path under the effect of gravity.

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

Q8. A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform speed of 10 m

s^{- 1}

, what would be the reading on the scale?

A. 35 kg

B. 75 kg

C. 105 kg

D. 70 kg

Answer : Option D
Explaination / Solution:

Mass of the man, m = 70 kg

Acceleration, a = 0

UsingNewton’s second law of motion,

wecan write the equation of motion as:

R – mg = ma

Where, ma is the net force acting on the man.

As the lift is moving at a uniform speed, acceleration a = 0

∴ R = mg = 70 × 10 = 700 N

∴ Reading on the weighing scale = 700 / g = 700 / 10 = 70 kg

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

Q9. A man of mass 70 kg stands on a weighing scale in a lift which is moving downwards with a uniform acceleration of 5 m

s^{- 2}

what would be the reading on the scale?

A. 35 kg

B. 53 kg

C. 38 kg

D. 45 kg

Answer : Option A
Explaination / Solution:

When the lift moves downward with acceleration = 5 ms-2 the net force acting downward

$\begin{aligned} m g - N = m a \\ N = m g - m a \\ N = m (g - a) \\ N = 70 (10 - 5) \\ N = 350 N \end{aligned}$

R = 350 N

therefore reading on the scale = $\frac{350}{g} = \frac{350}{10} = 35 k g$

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

Q10.

A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m $s^{- 2}$ what would be the reading on the scale?

A. 150 kg

B. 130 kg

C. 120 kg

D. 105 kg

Answer : Option D
Explaination / Solution:

When the lift moves upward with acceleration = 5 ms-2 the net force acting upward

$\begin{aligned} R - m g = m a \\ R = m g + m a \\ R = m (g + a) \\ R = 70 (10 + 5) \\ R = 1050 N \end{aligned}$

R = 1050 N

(We experience weight due to reaction)

therefore Apparent weight = $= \frac{1050}{g} = \frac{1050}{10} = 105 k g$

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