Laws of Motion Test

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Laws of Motion Test - 40

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Question 1
1 / -0

Friction is a/an

A
self-adjusting force

B
necessary evil

C
important force in daily life

D
all the above
Solution
1. Static friction is considered a self adjusting force because it wants the object to remain at rest and not move. Therefore, if an external force is applied, the static force will be equal to the magnitude of external force, until it surpasses the threshold of the motion. Hence friction is a self-adjusting force.
2. If there is no friction between any two surfaces in contact , then we will not be able to walk properly on the earth. Hence friction is necessary, But too much friction will reduce the efficiency of machines. Hence friction is a necessary evil.
3. Friction allows us to hold anything, pick up anything , allows us to go anywhere . Same goes with riding a car or a bike. Thus friction is important force in our daily life.
Question 2
1 / -0

If the momentum of a body is doubled, the kinetic energy is

A
Halved

B
Unchanged

C
Doubled

D
Becomes 4 times

Solution

The correct answer is option (D).

Hint: Relate the formula of momentum and kinetic energy of any body.

Step 1: Find the relation between momentum and kinetic energy.
By formula,
$$p=mv$$ (where, $$p$$ is the momentum of the body)
$$v=\dfrac{p}{m}$$
Also,
$$K.E.=\dfrac{1}{2}mv^2$$
Substituting the value of $$v$$ in the above equation, we get,
$$K.E.=\dfrac{1}{2}m{(\dfrac{p}{m}})^2$$

$$K.E.=\dfrac{1}{2}m\dfrac{p^2}{m^2}$$

$$K.E.=\dfrac{p^2}{2m}$$
Hence, $$K.E.\propto p^2$$

Step 2: Calculate the final kinetic energy.
From the above relation, we can write,
$$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{p_2^2}$$
According to the question, $$p_2=2p_1$$
$$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{(2p_2)^2}$$

$$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{4p_1^2}$$

$$K.E._2=4K.E._1$$

Hence, the kinetic energy becomes four times.
The correct answer is option (D).
Question 3
1 / -0

A stone tied to a string is rotated in a vertical plane. If mass of the stone is $$m$$, the length of the string is $$r$$ and the linear speed of the stone is $$v$$,When the stone is at its lowest point, then the tension in the string will be :

A
$$\displaystyle \frac { m{ v }^{ 2 } }{ r } +mg$$

B
$$\displaystyle \frac { m{ v }^{ 2 } }{ r } -mg$$

C
$$\displaystyle \frac { mv }{ r } $$

D
$$\displaystyle mg$$

Solution

At the lowest point, as shown in the figure both mg and centrifugal force
$$ \dfrac { m{ v }^{ 2 } }{ r } \\ $$will act in the same direction so,$$T=mg+\dfrac { m{ v }^{ 2 } }{ r } \\ $$
Question 4
1 / -0

In a circus, a motor cyclist moves in a spherical cage of radius $$5$$m. The minimum velocity with which he must cross the highest point without loosing contact is __________. (Take $$g=9.8m/s^2$$)

A
$$3\ m/s$$

B
$$5\ m/s$$

C
$$6\ m/s$$

D
$$7\ m/s$$

Solution

He must not lost the contact at the highest point, and since he is not losing the contact he must be in the circular motion.
Equation of motion athe highest point is:
$$N+mg=\dfrac { m{ v }^{ 2 } }{ R } $$
For the minimum velocity the normal has to be zero.
$$\\ mg=\dfrac { m{ { v }^{ 2 } }_{ min } }{ R } \\ \Rightarrow { { v } }_{ min }=\sqrt { Rg } =\sqrt { 5\times 9.8 } =\sqrt { 49 } =7m/s$$
Question 5
1 / -0

Why is it advised to tie any luggage kept on the roof of a bus with a rope?

A
While the bus is moving, luggage tends to remain in inertia of motion state

B
When the bus stops, the luggage tends to resist the change and due to inertia of motion it move forward and may fall off

C
Both a and b are true

D
None of them is True

Solution

While the bus is moving, luggage tends to remain in inertia of motion state. When the bus stops, the luggage tends to resist the change and due to inertia of motion it moves forward and may fall off. That's why it is advised to tie any luggage kept on the roof of a bus with a rope.
Question 6
1 / -0

A molecule of mass $$m$$ of an ideal gas collides with the wall of the vessel with the velocity $$v$$ and returns back with the same velocity. The change in the linear momentum of the molecule will be:

A
$$4mv$$

B
$$8mv$$

C
$$2mv$$

D
$$-2mv$$

Solution

Initial momentum $$=mv$$

Final momentum $$=-mv$$

Change in momentum $$=-mv-(mv)=-2mv$$
Question 7
1 / -0

An aircraft executes a horizontal loop with a speed of $$150$$ m/s with its wings banked at an angle of $$\displaystyle { 12 }^{ o }$$. The radius of the loop is $$\displaystyle \left( g=10{ m }/{ { s }^{ 2 } } \right) $$

A
10.6 km

B
9.6 km

C
7.4 km

D
5.8 km

Solution

Using the relation for the radius (r) of loop.

$$\displaystyle \tan { \theta } =\frac { { v }^{ 2 } }{ rg } $$

or, $$\displaystyle \tan { { 12 }^{ o } } =\frac { { \left( 150 \right) }^{ 2 } }{ r\times 10 } $$

or, $$\displaystyle r=\frac { 2250 }{ 0.2125 } =10.6\times { 10 }^{ 3 }m=10.6\quad km$$
Question 8
1 / -0

A stone ties to a rope is rotated in a vertical circle with uniform speed. If the difference between the maximum and minimum tension in the rope is $$20\ N$$, mass of the stone in $$kg$$ is:
($$\displaystyle g=10{ ms }^{ -2 }$$)

A
$$1.0$$

B
$$1.5$$

C
$$0.5$$

D
$$0.75$$

Solution

Let the mass of the stone be $$m$$, and the angular speed with which it rotates be $$\omega$$.
Thus at the bottom most point,
$$T_b-mg=m\omega^2 l$$
And at the top most point,
$$T_t+mg=m\omega ^2 l$$
$$\implies T_b-T_t=2mg=20$$
$$\implies mg=10$$
$$\implies m=1\ kg$$
Question 9
1 / -0

The catcher prepares to receive a pitch from the pitcher. As the ball reaches and makes contact with his glove, the catcher pulls his hand backward. This action reduces the impact of the ball on the catcher's hand because:

A
The energy absorbed by his hand is reduced

B
The momentum of the pitch is reduced

C
The time of impact is increased

D
The time of impact is reduced

E
The force exerted on his hand remains the same

Solution

We know that,
impulse = change in momentum
$$F\times t=p_{2}-p_{1}$$
As $$p_{2}$$ is zero , because final velocity of the ball becomes zero , so RHS of this equation is constant ($$=p_{1}$$), when catcher pulls his hands backward, he increases the time of impact $$t$$ and due to this impact of ball decreases.
Question 10
1 / -0

In vertical circular motion, the ratio of kinetic energy to potential energy at the horizontal position is ____________.

A
$$5:2$$

B
$$2:1$$

C
$$3:2$$

D
$$2:3$$

Solution

We assume that the object in motion just manages to complete the vertical circle. In that case,
$$v_{top} = \sqrt{gR}$$
Total energy at the topmost point $$=\dfrac12mgR + 2mgR = \dfrac52mgR$$
At the horizontal position $$PE = mgR$$
Total energy $$= \dfrac52mgR$$
$$\therefore KE = (\dfrac52-1) = \dfrac32mgR$$

$$\therefore \dfrac{KE}{PE} = \dfrac32$$

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Re-Attempt Weekly Quiz Competition

Laws of Motion Test - 40

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Laws of Motion Test - 40

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SHARING IS CARING

Question 1

self-adjusting force

necessary evil

important force in daily life

all the above

Solution

Question 2

Halved

Unchanged

Doubled

Becomes 4 times

Solution

Question 3

$$\displaystyle \frac { m{ v }^{ 2 } }{ r } +mg$$

$$\displaystyle \frac { m{ v }^{ 2 } }{ r } -mg$$

$$\displaystyle \frac { mv }{ r } $$

$$\displaystyle mg$$

Solution

Question 4

$$3\ m/s$$

$$5\ m/s$$

$$6\ m/s$$

$$7\ m/s$$

Solution

Question 5

While the bus is moving, luggage tends to remain in inertia of motion state

When the bus stops, the luggage tends to resist the change and due to inertia of motion it move forward and may fall off

Both a and b are true

None of them is True

Solution

Question 6

$$4mv$$

$$8mv$$

$$2mv$$

$$-2mv$$

Solution

Question 7

10.6 km

9.6 km

7.4 km

5.8 km

Solution

Question 8

$$1.0$$

$$1.5$$

$$0.5$$

$$0.75$$

Solution

Question 9

The energy absorbed by his hand is reduced

The momentum of the pitch is reduced

The time of impact is increased

The time of impact is reduced

The force exerted on his hand remains the same

Solution

Question 10

$$5:2$$

$$2:1$$

$$3:2$$

$$2:3$$

Solution

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