Forces and Laws of Motion Test

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Forces and Laws of Motion Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

A body is at rest on the surface of the earth. Which of the following statements is correct?

A
No force is acting on the body

B
Only weight of the body acts on it

C
Net downward force is equal to the net upward force

D
None of the above statement is correct

Solution
Question 2
1 / -0

What force is required to produce an acceleration of $$3 {m}/{{s}^{2}}$$ in an object of mass $$0.7 kg$$?

A
2.1

B
100

C
1.1

D
5

Solution

Use the formula,
$$F=ma$$
Given $$m=0.7kg$$
$$3m/s^2$$
The Answer is:
$$=0.7kg \times 3m/s^2$$
$$=2.1 dyne\ cm.$$
Hence,
option $$A$$ is correct answer.
Question 3
1 / -0

A Gardner waters the plants by a pipe of diameter $$1\ mm$$. The water comes out at the rate or $$10\ cm^{3}/sec$$. The reaction force exerted on the hand of the Gardner is:

A
$$0$$ N

B
$$1.27\times 10^{-2}N$$

C
$$1.27\times 10^{-4}N$$

D
$$0.127\ N$$

Solution

reaction force = $$Av^2p$$
$$F = v\dfrac{dm}{dt}$$
where $$v$$is the velocity of water
$$\dfrac{dm}{dt}$$ is the rate of mass flow
$$F = v\dfrac{dm}{dt} =vp\dfrac{dv}{dt}$$ where$$p$$ is the density
A= area of cross section then $$F = vpA\dfrac{dv}{dt} =pAv^2$$

in this question directly $$\dfrac{dv}{dt}$$ is given
$$\dfrac{dv}{dt} = Av$$
$$10=\dfrac{\pi d^2}{4}v$$
therefore $$v = \dfrac{40}{\pi d^2}$$

now $$F = \dfrac{40}{\pi d^2}p10$$ = $$\dfrac{400 \times 1 \times 7}{22 \times 10^{-2}}$$ = $$127.27 \times 10^{-2}$$dyne

1dyne = $$10^{-5}$$N
now $$F= \dfrac{127.27}{10^{-2}} \times 10^{-5} =0.127N $$

option is D
Question 4
1 / -0

A rocket of mass 1000 kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity $$100\, ms ^{-1}$$ with respect to the rocket. What is the minimum rate of burning of fuel, so as to just lift the rocket upwards against, the gravitational attraction? (Take $$g= 10\, ms^{-2} $$) :

A
$$50\, kgs^{-1}$$

B
$$100\, kgs^{-1}$$

C
$$200\, kgs^{-1}$$

D
$$400\, kgs^{-1}$$

Solution

Given, the velocity of exhaust gases with respect to rocket $$=100 ms^{-1}$$
The minimum force on the rocket to lift it
$$F_{min} =mg =1000 \times 10 =10000\,N$$

Hence, minimum rate of burning of fuel is given by
$$\frac{dm}{dt}=\frac{F_{min}}{v}=\frac{10000}{100}$$
$$=100 \, kgs^{-1}$$
Question 5
1 / -0

A ship of mass $$3 \times 10^7$$ kg, initially at rest, is pulled by a force of $$5 \times 10^4 N$$ through a distance of 3 m. Assuming that the resistance due to water is negligible, the speed of the ship is

A
$$1.5 m/s$$

B
$$60 m/s$$

C
$$0.1 m/s$$

D
$$5 m/s$$

Solution

Hint: Here we should know about Newton’s second law of motion & third equation of motion
Solution:
Step1: Find acceleration of ship ‘a’
According to Newton’s second law,
$$F = ma$$
Where, m=Mass & a=Acceleration
So, $$a = \dfrac{F}{m}$$
$$ \Rightarrow \dfrac{{5 \times {{10}^4}}}{{3 \times {{10}^7}}}$$…..(As force & mass are given)
$$a = \dfrac{5}{3} \times {10^{ - 3}}$$ m/s²
Step2: Find speed of the ship after moving distance of 3m
According Newton’s third law of motion,
$${v^2} = {u^2} + 2as$$
v=Final velocity, u=Initial velocity, a=Acceleration & s=Displacement
$$ \Rightarrow {0^2} + 2 \times \dfrac{5}{3} \times {10^{ - 3}} \times 3$$
$$ \Rightarrow \dfrac{{10}}{3} \times {10^{ - 3}} \times 3$$
$$ \Rightarrow {10^{ - 2}}$$m/s
$$ \Rightarrow 0.1$$m/s
Hence option (C) is correct.
Question 6
1 / -0

A rider on horse falls back when horse starts running all of a sudden because

A
Rider is taken back

B
Rider is suddenly afraid of falling

C
Inertia of rest keeps the upper part of body at rest while lower part of the body moves forward with the horse.

D
Inertia of motion keeps the upper part of body at rest while lower part of the body moves forward with the horse.

Solution

Hint: Use the concept of Newton's first law of motion.
Explanation:
$$\bullet$$When a raider sits on a horse, both the horse and the rider are resting. When a horse abruptly starts to run, it tends to stay at rest due to the inertia of the riders ( in the condition same as earlier). As a result, the rider stays in the same posture as the horse travels away and the rider falls back.
$$\bullet$$It should be observed that if the horse begins to go slowly, the rider will not fall back. Because the friction between the rider and the horse is sufficient to push the rider ahead.
Option C is correct.
Question 7
1 / -0

Two small spheres of masses $${M}_{1}$$ and $${M}_{2}$$ are suspended by weightless insulating threads of lengths $${L}_{1}$$ and $${L}_{2}$$, the spheres carry charges $${Q}_{1}$$ and $${Q}_{2}$$ respectively. The spheres are suspended such that they are in level with another and the threads are inclined to the vertical at angles of $${\theta}_{1}$$ and $${\theta}_{2}$$ as shown, which one of the following conditions is essential, if $${ \theta }_{ 1 }={ \theta }_{ 2 }:$$

A
$${ M }_{ 1 }\neq { M }_{ 2 }$$ but $${ Q }_{ 1 }={ Q }_{ 2 }$$

B
$${ M }_{ 1 }= { M }_{ 2 }$$

C
$${ Q }_{ 1 }={ Q }_{ 2 }$$

D
$${ L }_{ 1 }={ L }_{ 2 }$$

Solution

For a single change system:-
We can write from $$FBD$$,
$$T\,\cos\theta_1=Mg$$ $$T\sin\theta_1=Fe$$.
Now for both change, $$Fe$$ is equal in magnitude but, $$\theta_1=\theta_2$$ necessarily imply.
$$M_1=M_2$$
But, we can't say $$Q_1=Q_2$$ as both change face $$Fe=\dfrac{KQ_1Q_2}{d^2}$$ in equal magnitude but opposite direction [Newtons third law] so $$T\,\sin\theta$$ component would be equal for both.
Question 8
1 / -0

A free-floating astronaut A pushes another free-floating astronaut B in space. The mass of A is greater than that of B. The force exerted by the astronaut A on the astronaut B will be.

A
Greater than the force exerted by B on A

B
Equal to the force exerted by B on A

C
Equal to zero

D
Less than the force exerted by B on A

Solution

According to newton third law for every action there is an equal and opposite reaction.
The force on each other are actually action and reaction. So both forces are equal. So the magnitude of the force exerted by astronaut $$A$$ on $$B$$ is equal to the force exerted by $$B$$ on $$A$$.
Question 9
1 / -0

A small child tries to move a large rubber toy placed on the ground. The toy does not move but gets deformed under her pushing force $$\left( { F } \right) $$ which is obliquely upward as shown. Then

A
The resultant of the pushing force $$\left( { F } \right) $$, weight of the toy, normal force by the ground on the toy and the frictional force is zero

B
The normal force by the ground is equal and opposite to the weight of the toy

C
The pushing force $$\left( { F } \right) $$ of the child is balanced by the equal and opposite frictional force

D
The pushing force $$\left( { F } \right) $$ of the child is balanced by the total internal force in the toy generated due to deformation

Solution

According to newton second law
$$F=ma$$
So
Since there is no acceleration of body , $$a=0$$ hence $$F=0$$ therefore sum of all forces acting on body must be equal to zero. Therefore sum of pushing force, weight of toy and normal reaction by ground and friction by ground must be equal to zero.
Question 10
1 / -0

Which of the following is not the characteristics of matter?

A
The matter has a mass

B
Matter does not have inertia

C
Matter is affected by gravity

D
Matter cannot be destroyed

Solution

The matter is affected by inertia. Inertia is the tendency of the matter to be in a state of rest or motion.
Example: The book on the desk remains in rest position until it is picked up from that desk. Thus it is showing inertia.

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Forces and Laws of Motion Test - 49

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Forces and Laws of Motion Test - 49

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SHARING IS CARING

Question 1

No force is acting on the body

Only weight of the body acts on it

Net downward force is equal to the net upward force

None of the above statement is correct

Solution

Question 2

2.1

100

1.1

5

Solution

Question 3

$$0$$ N

$$1.27\times 10^{-2}N$$

$$1.27\times 10^{-4}N$$

$$0.127\ N$$

Solution

Question 4

$$50\, kgs^{-1}$$

$$100\, kgs^{-1}$$

$$200\, kgs^{-1}$$

$$400\, kgs^{-1}$$

Solution

Question 5

$$1.5 m/s$$

$$60 m/s$$

$$0.1 m/s$$

$$5 m/s$$

Solution

Question 6

Rider is taken back

Rider is suddenly afraid of falling

Inertia of rest keeps the upper part of body at rest while lower part of the body moves forward with the horse.

Inertia of motion keeps the upper part of body at rest while lower part of the body moves forward with the horse.

Solution

Question 7

$${ M }_{ 1 }\neq { M }_{ 2 }$$ but $${ Q }_{ 1 }={ Q }_{ 2 }$$

$${ M }_{ 1 }= { M }_{ 2 }$$

$${ Q }_{ 1 }={ Q }_{ 2 }$$

$${ L }_{ 1 }={ L }_{ 2 }$$

Solution

Question 8

Greater than the force exerted by B on A

Equal to the force exerted by B on A

Equal to zero

Less than the force exerted by B on A

Solution

Question 9

The resultant of the pushing force $$\left( { F } \right) $$, weight of the toy, normal force by the ground on the toy and the frictional force is zero

The normal force by the ground is equal and opposite to the weight of the toy

The pushing force $$\left( { F } \right) $$ of the child is balanced by the equal and opposite frictional force

The pushing force $$\left( { F } \right) $$ of the child is balanced by the total internal force in the toy generated due to deformation

Solution

Question 10

The matter has a mass

Matter does not have inertia

Matter is affected by gravity

Matter cannot be destroyed

Solution

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