System of Particles and Rotational Motion Test

Result

System of Particles and Rotational Motion Test - 27

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Four identical particles each of mass $$"m"$$ are arranged at the corners of a square of side length $$"L"$$. If one of the masses is doubled the shift in the centre of mass of the system w.r.t. diagonally opposite mass.

A
$$\dfrac{L}{\sqrt{2}}$$

B
$$\dfrac {3\sqrt {2}L}{5}$$

C
$$\dfrac {L}{4\sqrt{2}}$$

D
$$\dfrac {L}{5\sqrt{2}}$$

Solution
Question 2
1 / -0

The momentum of a system is defined

A
As the product of mass of the system and the velocity of centre of mass

B
As the vector sum of the momentum of individual particles

C
For bodies undergoing translational, rotational and oscillatory motion

D
All of the above

Solution

$$v_{cm}=\dfrac{m_1v_1+m_2v_2+.....+m_nv_n }{m_1+m_2+m_3+.....+m_n}$$;

$$\therefore {m_1+m_2+m_3+.....+m_n})v_{cm} =({m_1v_1+m_2v_2+.....+m_nv_n })$$;

Option(A) is LHS of the above equation.
Option(B) is RHS of the above equation.
Hence, (A) and(B) are same.
Option(C) is correct since momentum can be defined for a body having translational motion, it may have additionally rotational and oscillatory motion.
Question 3
1 / -0

The point through which the total weight appears to act for any orientation of the object is ______.

A
centre of gravity.

B
centre of momentum

C
centre of force

D
none of the above

Solution

the point where the total weight appears to act for any orientation of the object is centre of gravity .
Question 4
1 / -0

A uniform meter scale balances horizontally on a knife-edge placed at 55 cm mark, when a mass of 25 g is supported from one end, then the mass of the scale is :

A
200 g

B
225 g

C
350 g

D
275 g

Solution

Since the metre rule is uniform, the CM will be at the middle i.e. 50 cm. Then, mass of ruler $$(w)\times (55-50)=25(100-55)$$
$$w=\dfrac {25\times 45}{5}=225 g$$
Question 5
1 / -0

A couple always tends to produce :

A
Linear motion

B
Rotatory motion

C
Both linear and rotatory motion

D
Vibratory motion

Solution

A couple has the force balanced and thus does not produce the acceleration.
Throgh they produce a torque causing rotational motion.
Question 6
1 / -0

The momentum of a system with respect to centre of mass-

A
Is zero only if the system is moving uniformly.

B
Is zero only if no external force acts on the system.

C
Is always zero.

D
Can be zero in certain conditions.

Solution

For a system, total momentum is always zero in the cm frame by definition since the momentum of the cm is zero in this frame.
Question 7
1 / -0

The center of mass of a system of particle is at the origin. it follows that :

A
the number of particles to the right of the origin is equal to the number of particles to left.

B
the total mass of the particles to the right of the origin is same as the total to the left of the origin.

C
the number of particles on X-axis should be equal to the number of particles on Y-axis.

D
if there is a particle on the X-axis, there must be at least one particle on the negative X-axis.

E
none

Solution

The number of particle on both sides of the origin need not be same. For example, 1 particle of mass m at a distance of say 4 units towards the right of origin and 2 particles of equal mass m at a distance 8 units also have the center of mass at the origin.
Thus A, B and C are incorrect.
Also if a particle lies on positive X axis and two other particles are placed towards the left of the origin such that the three particles make an equilateral triangle, the center is again at origin.
Thus option D is incorrect
Question 8
1 / -0

The diameter of a solid disc is $$0.5m$$ and its mass is $$16kg$$.
What torque is required to increase its angular velocity about an axis perpendicular to its plane from zero to $$120$$ rotations/minute in $$8$$ seconds?

A
$$\cfrac { \pi }{ 4 } N/m$$

B
$$\cfrac { \pi }{ 2 } N/m$$

C
$$\pi N/m$$

D
$$2\pi N/m$$

Solution

We have torque as $$\tau=I\alpha$$.
Also the final angular velocity $$\omega$$ is given as 120 rotations/min or 2 rotations/second or $$4\pi$$ radians/sec.
Thus we have $$\alpha$$ as $$\displaystyle\frac{4\pi-0}{8}=\displaystyle\frac{\pi}{2}rad/s^2$$
And moment of inertia I for solid disc is $$\displaystyle\frac{mr^2}{2}$$
Thus we get torque as
$$\tau=(\displaystyle\frac{mr^2}{2}\times \displaystyle\frac{\pi}{2})$$
Substituting the values we get-
$$\tau=\displaystyle \frac{\pi}{4} N/m$$
Question 9
1 / -0

A ball kept in a closed container moves in it making collision with the walls. The container is kept on a smooth surface. The velocity of the centre of mass of :

A
the ball remains fixed

B
the ball relative to container remains fixed.

C
the container remains fixed

D
both container and ball remain fixed.

Solution

Since the container is closed, the collisions made by ball with the walls of container will not affect the mass of container and in turn there is no change in velocity of center of mass hence, both container and ball remain fixed.
Question 10
1 / -0

A car sometimes overturns while taking a turn. When it overturns, :

A
the inner wheel leaves the ground first.

B
the outer wheel leaves ground first.

C
both the wheels leave the ground simultaneously.

D
either of the wheels leaves the ground first.

Solution

The inner wheels leave the ground first as the center of gravity is some height above ground it will experience outside pseudo force which result in slight raising of inner wheels using outer wheels as axis.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

System of Particles and Rotational Motion Test - 27

Result

System of Particles and Rotational Motion Test - 27

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\dfrac{L}{\sqrt{2}}$$

$$\dfrac {3\sqrt {2}L}{5}$$

$$\dfrac {L}{4\sqrt{2}}$$

$$\dfrac {L}{5\sqrt{2}}$$

Solution

Question 2

As the product of mass of the system and the velocity of centre of mass

As the vector sum of the momentum of individual particles

For bodies undergoing translational, rotational and oscillatory motion

All of the above

Solution

Question 3

centre of gravity.

centre of momentum

centre of force

none of the above

Solution

Question 4

200 g

225 g

350 g

275 g

Solution

Question 5

Linear motion

Rotatory motion

Both linear and rotatory motion

Vibratory motion

Solution

Question 6

Is zero only if the system is moving uniformly.

Is zero only if no external force acts on the system.

Is always zero.

Can be zero in certain conditions.

Solution

Question 7

the number of particles to the right of the origin is equal to the number of particles to left.

the total mass of the particles to the right of the origin is same as the total to the left of the origin.

the number of particles on X-axis should be equal to the number of particles on Y-axis.

if there is a particle on the X-axis, there must be at least one particle on the negative X-axis.

none

Solution

Question 8

$$\cfrac { \pi }{ 4 } N/m$$

$$\cfrac { \pi }{ 2 } N/m$$

$$\pi N/m$$

$$2\pi N/m$$

Solution

Question 9

the ball remains fixed

the ball relative to container remains fixed.

the container remains fixed

both container and ball remain fixed.

Solution

Question 10

the inner wheel leaves the ground first.

the outer wheel leaves ground first.

both the wheels leave the ground simultaneously.

either of the wheels leaves the ground first.

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.