Forces and Laws of Motion Test

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Forces and Laws of Motion Test - 41

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Question 1
1 / -0

A body of mass $$300\ g$$ is at rest. What force in Newton will you have to apply to move it through $$200\ cm$$ in $$10\ s$$?

A
Zero

B
$$12\ N$$

C
$$1.2\ N$$

D
$$0.012\ N$$

Solution

Given: $$S=200 cm=2 m, u=0\ m/s, t=10 s$$
Using the formula : $$S=ut+\dfrac{1}{2}at^2$$
$$\Rightarrow 2=0+0.5 \times a \times 10^2$$
$$\Rightarrow a=0.04\ m/s^2$$

$$\underline{\text{Required force:}}$$
Mass of the body $$m= 300\ g=0.3\ kg$$
From Newton's second law:
$$F=ma=0.3 \times 0.04=0.012\ N$$
Question 2
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In a rocket, fuel burns at the rate of $$1\ kg/s$$. This fuel is ejected from the rocket with a velocity of $$60\ km/s$$. Force exerted on the rocket in $$N$$ is:

A
$$60 N$$

B
$$600 N$$

C
$$6000 N$$

D
$$60000 N$$

Solution

Overall momentum of the system is conserved.
Therefore, change in momentum of the rocket will be equal to that of the fuel.
Change in momentum of the fuel will be
$$\Delta mv = 60000$$
Hence, change in momentum of the rocket will also be:
$$\bigtriangleup P = 60000$$
Thrust force $$F =\dfrac{\bigtriangleup p}{\bigtriangleup t}$$
$$F = 60000\ N$$
Question 3
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Which of the following units is used to measure thrust?

A
Dyne

B
Pascal

C
Newton per meter square

D
Dyne per centimeter

Solution

Force acting on an object perpendicular to the surface is called thrust. Hence, the unit of thrust will be same as that of Force. Its SI unit is "Newton" and in CGS system its unit is "Dyne".

Hence, option A is correct
Question 4
1 / -0

A man is standing on a boat in still water. If he walks towards the shore the boat will :

A
move away from the shore.

B
remain stationary.

C
move towards the shore.

D
sink.

Solution

When the man from the boat in still water jumps from the boat, the boat is pushed backward due to the force he exerts on the boat.
This is an example of Newton's third law. Here, the action is on the boat and the boat gives the reaction on the man.
Question 5
1 / -0

A force 100 N acts in a body mass 2 kg for 10 s. The change in the velocity of the body is :

A
100 m s$$^{-1}$$

B
250 m s$$^{-1}$$

C
500 m s$$^{-1}$$

D
1000 m s$$^{-1}$$

Solution

Given : $$F = 100 \ N$$ $$m =2 \ kg$$ $$t = 10 \ s$$
Change in momentum of the body $$m(\Delta v) = F.t$$
where $$\Delta v$$ is the change in velocity.
$$\implies \ 2\times \Delta v = 100\times 10$$
$$\implies \ \Delta v = 500 \ m/s$$
Question 6
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Two individual forces of magnitude $$F_1$$ and $$F_2$$ act on a body of mass 1 kg as shown in the figures (i) and (i). If product of net magnitudes of forces produced in fig (i) and (ii) is 27 N then find the magnitude of forces $$F_1$$ and $$F_2$$. (assume $$F_1$$ > $$F_2$$)

A
3 N, 6 N

B
6 N, 6 N

C
6 N, 3 N

D
3 N, 3 N

Solution

From figure 1, net force F' $$= F_1 - F_2$$
From figure 2 net force F" $$= F_1 + F_2$$
Given $$(F_1-F_2) (F_1+ F_2) = 27 N$$
$$F_1^2 - F_2^2 = 27$$
Factors of F $$=$$ 27 are 1, 2, 7, 3, 9
$$\Rightarrow $$ If $$F_1 + F_2 = 9 N$$
$$F_1-F_2 = 3N$$
$$\therefore F_1 = 6N $$ and $$F_2 = 3N$$
Question 7
1 / -0

S.I. unit of momentum is :

A
$$kg\ ms^{-1}$$

B
$$kg\ mg^{-2}$$

C
$$kg\ mg^2$$

D
$$kg\ m^{-1}s^{-1}$$

Solution

Hint: Momentum $$p=m\times v$$
Correct opton is Option (A) $$\bf{\mathrm{kg} \mathrm{m} / \mathrm{s}^{-1}}$$

Explanation:
$$\bullet$$ Momentum depends upon the mass and velocity.

$$\bullet$$ In terms of equation, the momentum of an object is equal to the mass of the object times velocity of object .
$$\bullet$$ Momentum = $$\bf\text { mass } \times \text { velocity }$$

$$\bullet$$The SI unit of momentum is $$\mathrm{kg} \mathrm{m} / \mathrm{s}^{-1}$$
Question 8
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A book is at rest on a table. What is the "reaction" force according to Newton's third law to the gravitational force by the earth on the book?

A
the normal force exerted by the table on the book

B
the normal force exerted by the table on the ground

C
the normal force exerted by the ground on the table

D
the gravitational force exerted on the earth by the book.
Question 9
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A loaded $$20,000 \ kg$$ coal wagon is moving on a level track at $$6$$ ms$$^{-1}$$. Suddenly $$5000 kg$$ of coal is dropped out of the wagon. The final speed of the wagon is:

A
$$6$$ ms$$^{-1}$$

B
$$8$$ ms$$^{-1}$$

C
$$4.8$$ ms$$^{-1}$$

D
$$4.5$$ ms$$^{-1}$$

Solution

Initial mass of the system $$m_i = 20000 \ kg$$
Final mass of the system $$m_f = 20000-5000= 15000 \ kg$$
Since no external force is acting on it, we can apply conservation of momentum,
$$m_iv_i=m_fv_f$$
$$20000\times 6=15000\times v_f$$
Final velocity, $$v_2=8\ m/s$$
Question 10
1 / -0

The time, in which a force of $$2\ N$$ produces a change as the momentum of $$0.4\ kg ms^{-1}$$ in the body whose mass is $$1\ kg$$ is :

A
$$0.2\ s$$

B
$$0.02\ s$$

C
$$0.5\ s$$

D
$$0.05\ s$$

Solution

According to newton's second equation of motion-
$$F = \dfrac{ \Delta P }{ t }$$
$$F$$: External force
$$\Delta P$$: Change in the momentum
$$t$$: Time taken
$$\Rightarrow 2 \displaystyle =\frac{0.4-0}{t}$$
$$\Rightarrow t = \dfrac{ 0.4 }{ 2 }$$
$$\Rightarrow t = 0.2\ s$$

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Forces and Laws of Motion Test - 41

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Forces and Laws of Motion Test - 41

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Question 1

Zero

$$12\ N$$

$$1.2\ N$$

$$0.012\ N$$

Solution

Question 2

$$60 N$$

$$600 N$$

$$6000 N$$

$$60000 N$$

Solution

Question 3

Dyne

Pascal

Newton per meter square

Dyne per centimeter

Solution

Question 4

move away from the shore.

remain stationary.

move towards the shore.

sink.

Solution

Question 5

100 m s$$^{-1}$$

250 m s$$^{-1}$$

500 m s$$^{-1}$$

1000 m s$$^{-1}$$

Solution

Question 6

3 N, 6 N

6 N, 6 N

6 N, 3 N

3 N, 3 N

Solution

Question 7

$$kg\ ms^{-1}$$

$$kg\ mg^{-2}$$

$$kg\ mg^2$$

$$kg\ m^{-1}s^{-1}$$

Solution

Question 8

the normal force exerted by the table on the book

the normal force exerted by the table on the ground

the normal force exerted by the ground on the table

the gravitational force exerted on the earth by the book.

Question 9

$$6$$ ms$$^{-1}$$

$$8$$ ms$$^{-1}$$

$$4.8$$ ms$$^{-1}$$

$$4.5$$ ms$$^{-1}$$

Solution

Question 10

$$0.2\ s$$

$$0.02\ s$$

$$0.5\ s$$

$$0.05\ s$$

Solution

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