System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

A body of mass 5 kg undergoes a change in speed from 30 to 40 m/s. Its momentum would increase by :

A
50 kg m/s

B
75 kg m/s

C
150 kg m/s

D
350 kg m/s

Solution

speed is increased by $$10$$m/s so the momentum will change by $$ 50 $$kg m/s
so the answer is A.
Question 2
1 / -0

The moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through one end is $$I$$. The same rod is bent into a ring and its moment of inertia about the diameter is $${I}_{1}$$. The ratio $$\dfrac{I}{{I}_{1}}$$ is

A
$$\dfrac{4\pi}{3}$$

B
$$\dfrac{8{\pi}^{2}}{3}$$

C
$$\dfrac{5\pi}{3}$$

D
$$\dfrac{8{\pi}^{2}}{5}$$

Solution

Moment of inertial of a thin uniform rod about perpendicular axis through it one end
$$I = \dfrac{1}{3} M{l}^{2}$$ ....(i)
Same rod is bent into a ring
$$\therefore l = 2\pi r \Rightarrow \dfrac{l}{r} = 2\pi$$ ....(ii)
Moment of inertia of ring about diameter
$${I}_{1} = \dfrac{M{R}^{2}}{2}$$ .....(iii)
Dividing equation (i) by (ii), we get
$$\dfrac{I}{{I}_{1}} = \dfrac{2}{3} \dfrac{M{l}^{2}}{M{R}^{2}} = \dfrac{2}{3} \dfrac{{l}^{2}}{{R}^{2}}$$
$$\dfrac{I}{{I}_{1}} = \dfrac{2}{3} {\left(2\pi\right)}^{2} = \dfrac{8{\pi}^{2}}{3}$$ [using equation (ii)]
Question 3
1 / -0

Around the centre of gravity ______ vanishes. Fill in the blank.

A

Resultant acceleration due to gravity force

B
Resultant velocity due to gravity force

C
Resultant torque due to gravity force

D
None

Solution

Resultant torque due to gravity force vanishes around the centre of gravity because perpendicular distance between the gravitational force and the point about that toque is calculated becomes very very small or almost zero.
Question 4
1 / -0

A large number of particles are placed around the origin, each at a distance R from the origin. The distance of the centre of mass of the system from the origin is:

A
equal to R

B
less than equal to R

C
greater than R

D
greater than equal to R

Solution

All the particles may be considered to lie on a circle of radius $$R$$ centered at the origin, so the center of mass will not lie outside the circle. Therefore, the distance of the center of mass from the origin is less than or equal to R.
Question 5
1 / -0

A body of mass $${ m }_{ 1 }=4 \ kg$$ moves at $$5i \ m/s$$ and another body of mass $${ m }_{ 2 }=2 \ kg$$ moves at $$10 i\ m/ s$$. The kinetic energy of centre of mass is

A
$$\dfrac{200 }{ 3} J$$

B
$$\dfrac{500 }{ 3} J$$

C
$$\dfrac{400 }{ 3} J$$

D
$$\dfrac{800 }{ 3} J$$

Solution

$$v_c=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$$

$$v_c=\dfrac{40}{6}ms^{-1}$$

Kinetic energy $$=\dfrac{1}{2}mv^2=\dfrac{1}{2}\times6\times(\dfrac{40}{6})^2$$

$$K_c=\dfrac{400}{3}J$$
Question 6
1 / -0

The ratio of angular speeds of minute hand and hour hand of a watch is.

A
$$1:12$$

B
$$6:1$$

C
$$12:1$$

D
$$1:6$$

Solution

Hint: The rate of change of angular displacement is known as angular speed.
Explanation:
Step 1: Formula used:
We know that angular speed is given by:
$$ \omega = \dfrac{angular \ displacement}{time \ taken} $$

Step 2: Find angular velocities in both cases:

Time taken by minute hand to rotate by an angle of $$2\pi \ radian$$, $$t_m = 60\ minute $$
$$\therefore$$ Angular speed of minute hand $$\omega_{min}=\displaystyle\frac{2\pi}{60}\frac{rad}{min}$$

Time taken by hour hand to rotate by an angle of $$2\pi \ radian$$, $$t_{hr} = 12 \ hr$$ $$ = 12 \times 60 \ min$$
$$\therefore$$ Angular speed of hour hand $$w_{hr} = \dfrac{2\pi}{12 \times 60}$$ $$rad/min$$

Step 3: Conclusion:

$$\omega_{min}=\displaystyle\frac{2\pi}{60}\frac{rad}{min}$$
and $$\omega_{hr}=\displaystyle\frac{2\pi}{12\times 60}\frac{rad}{min}$$
$$\therefore \displaystyle\frac{\omega_{min}}{\omega_{hr}}=\frac{2\pi /60}{2\pi /12\times 60}$$
$$=\displaystyle\frac{12}{1}$$

Option C is correct.
Question 7
1 / -0

The moment of inertia of a solid sphere about an axis passing through centre of gravity is $$\dfrac {2}{5}MR^{2}$$, then its radius of gyration about a parallel axis at a distance $$2R$$ from first axis is

A
$$5 R$$

B
$$225\sqrt {R}$$

C
$$\dfrac {5}{2}R$$

D
$$\sqrt {\dfrac {12}{5}}R$$

Solution

The radius of gyration K of a body about a given axis of rotation is that radial distance from the axis, the square of which on being multiplied by the total mass of the body gives the moment of inertia of the body about that axis.
Thus,$$I=MK2=\Sigma mR2$$
$$where M is the total mass of the body.
This means that$$K=\sqrt {\dfrac{I}{M} }$$

According to theorem of parallel axis
$$I=I_CG+M(2R)^2$$
where, $$ICG$$ is moment of inertia about an axis through centre of gravity.
$$\therefore I =25MR^2+4MR^2=225MR^2$$ or$$MK^2=225MR^2$$
$$\therefore K=225\sqrt{R}$$
The total mass of a body may be supposed to be concentrated at a radial distance K from the axis of rotation, so far as the moment of inertia of the body about that axis is concerned.
Question 8
1 / -0

The radius of gyration of a body about an axis at a distance 6 cm from its centre of mass is 10 cm. Then, its radius of gyration about a parallel axis through its centre of mass will be :

A
80 cm

B
8 cm

C
0.8 cm

D
80 m

Solution

Let radius of gyration for the axis not passing through center of mass be $$r$$ and that for the axis passing through the center of mass be $$k$$ and the distance between the two parallel axes be $$a$$.
Parallel axes theorem gives:
$$m{ r }^{ 2 }=m({ k }^{ 2 }+{ a }^{ 2 })\Rightarrow { r }^{ 2 }={ k }^{ 2 }+{ a }^{ 2 }$$.
$$\Rightarrow{k}=\sqrt{10^{2}-6^{2}}=8cm$$.
Thus, option B is the correct answer.
Question 9
1 / -0

A body of mass $$1\ kg$$ is thrown with a velocity of $$\displaystyle 10{ ms }^{ -1 }$$ at an angle of $$\displaystyle { 60 }^{ \circ }$$ with the horizontal. Its momentum at the highest point is:

A
$$\displaystyle 2\ kg\ { ms }^{ -1 }$$

B
$$\displaystyle 3\ kg\ { ms }^{- 1 }$$

C
$$\displaystyle 4\ kg\ { ms }^{ -1 }$$

D
$$\displaystyle 5\ kg\ { ms }^{ -1 }$$

Solution

At the highest point, its velocity in the upward direction is zero but its velocity in the horizontal direction remains constant at every instant. Hence , only the horizontal component of velocity is responsible for the momentum of the body at the highest point.
Velocity of the body $$v = 10$$ $$ms^{-1}$$
Velocity component of the body in the horizontal direction $$v_H = v cos 60^o = 10 \times 0.5 = 5$$ $$ms^{-1}$$
$$\therefore$$ Momentum $$P = mv_H = 1 \times 5 =5$$ $$kg$$ $$ms^{-1}$$
Question 10
1 / -0

What torque will increase angular velocity of a solid disc of mass $$16kg$$ and diameter $$1m$$ from zero to $$2$$rpm in $$8s$$?

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

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

C
$$\cfrac { \pi }{ 3 } N-m$$

D
$$\pi N-m$$

Solution

Torque $$\left( \tau \right) =I\alpha $$
$$\tau =\cfrac { 1 }{ 2 } \times M{ R }^{ 2 }\times \cfrac { 2\pi \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ t } $$
$$\therefore$$ $$\tau =16\times { \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }\times \pi \cfrac { \left( 2-0 \right) }{ 8 } =\pi N-m$$

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System of Particles and Rotational Motion Test - 33

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System of Particles and Rotational Motion Test - 33

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Question 1

50 kg m/s

75 kg m/s

150 kg m/s

350 kg m/s

Solution

Question 2

$$\dfrac{4\pi}{3}$$

$$\dfrac{8{\pi}^{2}}{3}$$

$$\dfrac{5\pi}{3}$$

$$\dfrac{8{\pi}^{2}}{5}$$

Solution

Question 3

Resultant acceleration due to gravity force

Resultant velocity due to gravity force

Resultant torque due to gravity force

None

Solution

Question 4

equal to R

less than equal to R

greater than R

greater than equal to R

Solution

Question 5

$$\dfrac{200 }{ 3} J$$

$$\dfrac{500 }{ 3} J$$

$$\dfrac{400 }{ 3} J$$

$$\dfrac{800 }{ 3} J$$

Solution

Question 6

$$1:12$$

$$6:1$$

$$12:1$$

$$1:6$$

Solution

Question 7

$$5 R$$

$$225\sqrt {R}$$

$$\dfrac {5}{2}R$$

$$\sqrt {\dfrac {12}{5}}R$$

Solution

Question 8

80 cm

8 cm

0.8 cm

80 m

Solution

Question 9

$$\displaystyle 2\ kg\ { ms }^{ -1 }$$

$$\displaystyle 3\ kg\ { ms }^{- 1 }$$

$$\displaystyle 4\ kg\ { ms }^{ -1 }$$

$$\displaystyle 5\ kg\ { ms }^{ -1 }$$

Solution

Question 10

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

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

$$\cfrac { \pi }{ 3 } N-m$$

$$\pi N-m$$

Solution

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