System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 21

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Question 1
1 / -0

Rolling of ball on the ground is the instance of __________ as well as ____________

A
Periodic motion, rotational motion

B
Oscillatory motion, rotational motion

C
Curvilinear motion, rotational motion

D
Rectilinear motion, rotational motion

Solution

During the rolling of ball on the ground, the center of mass of the ball exhibits translational (rectilinear) motion whereas the ball exhibits rotational motion in center of mass frame. Thus rolling of ball on the ground is the instance of rectilinear motion as well as rotational motion.
Question 2
1 / -0

Two particles $$A$$ and $$B$$ initially at rest move towards each other under a mutual force of attraction. The speed of center of mass at the instant when the speed of $$A$$ is $$v$$ and the speed of $$B$$ is $$2v$$ is

A
$$v$$

B
Zero

C
$$2v$$

D
$$\dfrac{3v}{2}$$

Solution

Under mutual force of attraction, the velocity of COM is zero.
Question 3
1 / -0

The centre of mass of a body:

A
Lies always at the geometrical center

B
Lies always inside the body

C
Lies always outside the body

D
Lies within or outside the body

Solution

The centre of mass of a body can lie within or outside the body.
For example, centre of mass of a uniform rod lies at its geometrical centre which lies within the rod whereas centre of mass of a uniform ring lies at its geometrical centre which lies outside the ring.
Question 4
1 / -0

Reference to ability of an object to return to its original position after it has been tilted slightly is termed as:

A
Stability

B
Equilibrium

C
Centre of gravity

D
Torque

Solution

stability of an object depends on the ability of an object to return back to its original equlibirum, when disturbed
Question 5
1 / -0

Radius of gyration of a thin uniform rod of mass M length L about an axis perpendicular to it at one of its ends is:

A
$$L$$

B
$$L/2$$

C
$$L/ 2 \sqrt(3)$$

D
$$L/\sqrt(3)$$

Solution

The moment of inertia of a rod of mass m about an axis perpendicular to it at one of its ends is given by $$I = ML^2/3$$. Comparing this with the general equation for the moment of inertia, we get, $$I=MK^2$$, K is the radius of gyration, we get $$K=L/ \sqrt{3}$$

The correct options is option d
Question 6
1 / -0

Find the new coordinates of the point (3, 4), if the origin is shifted to the point (1, 3).

A
(2, 1)

B
(1, 2)

C
(1, -2)

D
(2, -1)

Solution

Initially the origin was (0,0) and the point was (3,4). If the origin is now shifted to a new point (1,3), then the coordinates of the new point will be (X,Y) with respect to the new origin(h=1, k=3). Thus, (3,4) = (X+1, Y+3). Solving, we get, X=2 and Y=1
Question 7
1 / -0

Angular momentum and angular velocity for a rolling sphere of radius R are related by the formula

A
$$L = I\omega$$; I is the moment of inertia of the sphere

B
$$L = I\omega^2$$ I is the moment of inertia of the sphere

C
$$L = \dfrac{I}{\omega}$$, I is the moment of inertia of the sphere

D
$$L = I\omega-f \times R$$; f is the frictional force and R is the radius of the sphere and I is the moment of inertia of the sphere

Solution

Angular momentum is related to angular velocity using $$L = I\omega$$;
Question 8
1 / -0

If you place pivot at center of a meter rule, weight has no

A
property

B
Concern

C
Turning effect

D
Magnitude

Solution

The weight of an object is concentrated at the centre of gravity and hence a pivot at the centre of the metre scale does not record any changes in mass. Hence the scale does not turn around
Question 9
1 / -0

If the K.E. of a body is increased by $$300\%$$, its momentum will increase by _______.

A
$$100\%$$

B
$$150\%$$

C
$$\sqrt {300}\%$$

D
$$175\%$$

Solution

Kinetic energy is given by
K=V²/(2M)……………(1)
When KE increases by 300%, the value of KE becomes 4 times it's original value.
from equation (1) that v becomes 2V

So, V increases by 100%.
Question 10
1 / -0

The momentum P and kinetic energy E of a body of mass m are related as :

A
$$P = \sqrt {2mE} $$

B
$$P = \dfrac{1}{2}mE$$

C
$$P = \dfrac{{2m}}{E}$$

D
$$P=2mE$$

Solution

A. $$P=\sqrt{2mE}$$

lets,

$$m= $$ mass of a body

$$v=$$ velocity of body.

The kinetic energy, $$E=\dfrac{1}{2}mv^2$$. . . . .. . (1)

The momentum, $$P=mv$$. . . . . . . (2)

In equation (1), multiply and divide by $$m$$

$$E=\dfrac{P^2}{2m}$$

$$P^2=2mE$$

$$P=\sqrt{2mE}$$

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System of Particles and Rotational Motion Test - 21

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System of Particles and Rotational Motion Test - 21

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Question 1

Periodic motion, rotational motion

Oscillatory motion, rotational motion

Curvilinear motion, rotational motion

Rectilinear motion, rotational motion

Solution

Question 2

$$v$$

Zero

$$2v$$

$$\dfrac{3v}{2}$$

Solution

Question 3

Lies always at the geometrical center

Lies always inside the body

Lies always outside the body

Lies within or outside the body

Solution

Question 4

Stability

Equilibrium

Centre of gravity

Torque

Solution

Question 5

$$L$$

$$L/2$$

$$L/ 2 \sqrt(3)$$

$$L/\sqrt(3)$$

Solution

Question 6

(2, 1)

(1, 2)

(1, -2)

(2, -1)

Solution

Question 7

$$L = I\omega$$; I is the moment of inertia of the sphere

$$L = I\omega^2$$ I is the moment of inertia of the sphere

$$L = \dfrac{I}{\omega}$$, I is the moment of inertia of the sphere

$$L = I\omega-f \times R$$; f is the frictional force and R is the radius of the sphere and I is the moment of inertia of the sphere

Solution

Question 8

property

Concern

Turning effect

Magnitude

Solution

Question 9

$$100\%$$

$$150\%$$

$$\sqrt {300}\%$$

$$175\%$$

Solution

Question 10

$$P = \sqrt {2mE} $$

$$P = \dfrac{1}{2}mE$$

$$P = \dfrac{{2m}}{E}$$

$$P=2mE$$

Solution

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