Laws of Motion Test

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Laws of Motion Test - 36

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Question 1
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Directions For Questions

A small particle of mass $$m$$ attached with a light inextensible thread of length $$L$$ is moving in a vertical circle. In the given case particle is moving in complete vertical circle and ratio of its maximum to minimum velocity is $$2:1$$.

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Minimum velocity of the particle is

A
$$\displaystyle 4\sqrt{\frac{gL}{3}}$$

B
$$\displaystyle 2\sqrt{\frac{gL}{3}}$$

C
$$\displaystyle \sqrt{\frac{gL}{3}}$$

D
$$\displaystyle 3\sqrt{\frac{gL}{3}}$$

Solution

The maximum and minimum velocities will be at A and B respectively.
Work-energy theorem, $$-mg(2L)= \dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$$
Given: $$u:v= 2:1$$
$$-mg(2L)= \dfrac{1}{2}mv^2-\dfrac{1}{2}m(2v)^2$$
$$\implies v= 2\sqrt{\dfrac{gL}{3}}$$
Question 2
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A turn has a radius of $$10 m$$. If a vehicle goes round it at an average speed of $$18$$ km/h, what should be the proper angle of banking?

A
$$\displaystyle \tan ^{-1}\left ( 1/4 \right )$$

B
$$\displaystyle \tan ^{-1}\left ( 1/2 \right )$$

C
$$\displaystyle \tan ^{-1}\left ( 3/4 \right )$$

D
$$\displaystyle \tan ^{-1}\left ( 5/7 \right )$$

Solution

Given : $$r =10$$ m
Average speed of the vehicle $$v = 18 km/hr = 18 \times \dfrac{5}{18} = 5$$ $$m/s$$
Let the proper banking angle be $$\theta$$
Using $$tan\theta = \dfrac{v^2}{gr}$$
$$\therefore$$ $$tan\theta =\dfrac{5^2}{10 \times 10}$$ $$\implies \theta = tan^{-1} (1/4)$$
Question 3
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A mass is performing vertical circular motion (see figure). If the average velocity of the particle is increased, then at which point is the maximum breaking possibility of the string:

A
A

B
B

C
C

D
D

Solution

Tension at any point in vertical motion is given by :
$$T=\dfrac{mv^{2}}{l}+mg\cos\theta$$
where $$\theta=$$ angular displacement from lowest point ,
$$l=$$ length of string
$$m=$$ mass of string
It is clear that tension at the lowest point ($$B$$) is greatest than at other points ($$A,C,D$$). If we increase average velocity, tension will increase at lowest point, therefore at point B, string has maximum possibility of break.
Question 4
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A stone of mass $$1\;kg$$ is tied to the end of a string of $$1\;m$$ long. It is whirled in a vertical circle. If the velocity of the stone at the top be $$4\;m/s$$. What is the tension in the string?

A
$$\;6\;N$$

B
$$\;16\;N$$

C
$$\;5\;N$$

D
$$\;10\;N$$

Solution

Given : $$v =4$$ $$m/s$$ $$m =1 kg$$ $$r = 1$$ m
Using circular motion equation : $$\dfrac{mv^2}{r} =T + mg$$
$$\implies$$ $$ T =\dfrac{mv^2}{r} -mg $$

$$\therefore$$ $$ T =\dfrac{1 \times 4^2}{1} -1 \times 10 $$ $$\implies T = 6 $$ $$N$$
Question 5
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An instrument box placed on a table is given a push. It is observed that it stops its motion after some time. What is the force that stopped its motion?

A
Gravitational force

B
Tension force

C
Friction force

D
Mechanical force

Solution

As a body is set in motion on surface (not frictionless) a frictional force starts acting on the body in the direction opposite to the motion of the body.
Question 6
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A car moves on a circular road, describing equal angles about the centre in equal intervals of times which of the statements about the velocity of car

A
velocity is constant

B
magnitude of velocity is constant but the direction of velocity change

C
both magnitude and velocity change

D
velocity is directed towards the centre of circle

Solution

Hint: Apply the formula of angular velocity.
Explanation:
$$\bullet$$The car covers equal angles in equal intervals of time, so its angular velocity is constant.
By formula,
$$v=\omega r$$
$$\bullet$$The circular road has a constant $$r$$ and according to the question, $$\omega$$ is also constant.
Therefore, $$v=\text{constant}$$
$$\bullet$$Magnitude of the linear velocity is thus constant but since it is moving in a circular road, its direction is constantly changing.

Hence, the magnitude of velocity is constant but the direction of velocity changes.
The correct answer is option (B).
Question 7
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Directions For Questions

Reena is riding a motorised scooter along a straight path. The combined mass of Reena and her scooter is $$80\ kg$$. The frictional forces that are acting total to $$45\ N$$. What is the magnitude of the driving force being provided by the motor if:

...view full instructions

She is moving with a constant speed of $$10 ms^{-1}$$?

A
$$45\ N$$

B
$$50\ N$$

C
$$75\ N$$

D
$$20\ N$$

Solution

Given,
$$m=80kg$$
Friction force, $$f=45N$$
Constant velocity, $$v=10m/s$$ So, the acceleration is zero.
Driving force, $$f_D=ma-f$$
$$f_D=80\times 0-45=-45N$$
The magnitude of the driving force,
$$|f_D|=45N$$
The correct option is A.
Question 8
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In which of the following cases the net force is not equal to zero?

A
A kite skillfully held stationary in the sky

B
A ball falling freely from a height

C
A helicopter hovering above the ground

D
A cork floating on the surface of water

Solution

According to newton's 1st law of motion, A body maintains its states ie Rest or motion with uniform velocity until and unless there is no external force apply on that.
So, A kite skillfully held stationary in the sky, A helicopter hovering above the ground, A cork floating on the surface of water, All these examples are maintaining state of rest so resultant or net force on it will be zero.
A freely falling ball have acceleration that will equal to $$g$$ So, Net force on it will not be equal to zero.
Question 9
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A boy is whirling a stone tied at one end such that the stone is in uniform circular motion.Which of the following statement is correct?

A
The velocity of stone at A is equal to the velocity of stone B

B
The speed of stone at A is equal to the speed of stone at B.

C
The centripetal force is radially outward in the string.

D
All of the above statements are correct.

Solution

In uniform circular motion speed is constant while velocity being a vector quantity is constantly changing as its direction keeps changing. Centripetal force acts inwards towards the center to counterbalance the centrifugal force acting outwards from the center.
Question 10
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A force of $$50 \ dynes$$ is acted on a body of mass $$5 g$$ which is at rest for an interval of $$3 s$$, then impulse is:

A
$$0.15\times 10^{-13}Ns$$

B
$$0.98\times 10^{-3}Ns$$

C
$$1.5\times 10^{-3}Ns$$

D
$$2.5\times 10^{-3}Ns$$

Solution

$$Impulse\: J = F\Delta t = 50 \times 10^{-5}\times 3 = 1.5 \times 10^{-3} Ns$$