System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

In case of torque of a couple, if the axis is changed by displacing it parallel to itself, torque will

A
Increase

B
Decrease

C
Remain constant

D
None of these

Solution

Torque of a couple $$\tau = d\times F$$
where $$d$$ is the distance between the two equal force $$F$$
Thus, the distance between the forces remains the same even on changing the axis. So, the torque will remain the same.
Question 2
1 / -0

Two particles of mass 1 kg and 0.5 kg are moving in the same direction with speed of 2 m/s and 6 m/s respectively on a smooth horizontal surface. The speed of centre of mass of the system is

A
$$\displaystyle\frac{10}{3}$$ m/s

B
$$\displaystyle\frac{10}{7}$$ m/s

C
$$\displaystyle\frac{11}{2}$$ m/s

D
$$\displaystyle\frac{12}{3}$$ m/s

Solution

As we know that center of mass = $$\dfrac{{m}_{1}\times{v}_{1}+{m}_{2}\times{v}_{2}}{{m}_{1}+{m}_{2}}$$

so C.O.M=$$\dfrac{{1}\times{2}+{0.5}\times{6}}{{1}+{0.5}}$$=$$\dfrac{10}{3}$$m/sec
Question 3
1 / -0

The passengers in a boat are not allowed to stand because:

A
This will raise the centre of gravity and the boat will be rocked

B
This will lower centre of gravity and the boat will rocked

C
The effective weight of system increases

D
Of surface tension effects

Solution

when passengers stands up on the boat from the sitting situation , the centre of gravity moves upwards and it increases the chances of toppling of boat .
so the answer is A.
Question 4
1 / -0

Three identical uniform rods, each of length l, are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is.

A
$$\displaystyle \frac{l}{\sqrt{3}}$$

B
$$\displaystyle \frac{l}{\sqrt{2}}$$

C
$$\displaystyle \frac{l}{\sqrt{5}}$$

D
$$\displaystyle \frac{l}{\sqrt{7}}$$

Solution

Let axis of rotation is passing through point 'A'.
Moments of Inertia (MOI) of rod AB about point 'A'= $$ \dfrac{ml^2}{3}$$,
MOI of rod AC about point 'A'= $$ \dfrac{ml^2}{3}$$,
From the figure AP= $$lsin(60^o)= \dfrac{\sqrt{3}}{2} l$$,
Using Parallel axis of theorem of MOI,
MOI of rod BC about point 'A'= $$ \dfrac{ml^2}{12}$$ + $$m ( \dfrac{\sqrt{3}}{2} l)^2$$ $$=\dfrac{5}{6} ml^2$$
By adding all the MOI ,
MOI = $$\dfrac{3}{2} ml^2$$
$$\therefore $$ radius of gyration 'r'
$$ \Rightarrow 3mr^2= \dfrac{3}{2} ml^2$$,
$$\Rightarrow r= \dfrac{l}{\sqrt{2}}$$
Question 5
1 / -0

Moment of inertia of thin uniform rod of length $$\displaystyle l$$ is equal to:

A
$$\displaystyle \frac { { ml }^{ 2 } }{ 6 } $$

B
$$\displaystyle \frac { { ml }^{ 2 } }{ 12 } $$

C
$$\displaystyle \frac { { ml }^{ 2 } }{ 4 } $$

D
$$\displaystyle \frac { { ml }^{ 2 } }{ 24 } $$

Solution

Let a small element on the rod at a distance $$x$$ from $$CM$$.
Mass of the small element of rod $$dm = \dfrac{m}{l}dx$$
Moment of this element about $$CM$$, $$dI = dm x^2 $$
$$\therefore$$ Total moment of inertia about $$CM$$ $$I = \int^{l/2}_{-l/2} x^2dm = \int^{l/2}_{-l/2} \dfrac{m}{l}x^2 dx$$
$$\therefore$$ $$I = \dfrac{m}{l} \times \dfrac{x^3}{3}\bigg|_{-l/2}^{l/2} = \dfrac{m}{l} \times \dfrac{2(l/2)^3}{3}$$
$$\implies$$ Moment of inertia about $$CM$$, $$I = \dfrac{ml^2}{12}$$
Question 6
1 / -0

Find the radius of gyration of a hollow uniform sphere of radius R about its tangent.

A
$$\sqrt{\dfrac{5}{3}} \ R$$

B
$$\sqrt{\dfrac{2}{3}} \ R$$

C
$$\sqrt{\dfrac{3}{5}} \ R$$

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

Solution

Moment of inertia of a hollow sphere about a tangent, $$I\,=\, \displaystyle\frac{5}{3}MR^2$$
$$MK^2\,=\, \displaystyle\frac{5}{3}MR^2\, \quad\, \Rightarrow\, K\, =\, \sqrt{\displaystyle\frac{5}{3}}R$$
Question 7
1 / -0

A uniform disc of mass M and radius R is mounted on an axle supported in friction less bearings. A light cord is wrapped around the rim of the disc and a steady downward pull T is exerted on the cord. The angular acceleration of the disc is:

A
$$\displaystyle \frac{MR}{2 T}$$

B
$$\displaystyle \frac{2T}{MR}$$

C
$$\displaystyle \frac{T}{MR}$$

D
$$\displaystyle \frac{MR}{T}$$

Solution

$$\because \vec{\tau} = \vec R \times \vec F$$
$$\Rightarrow \tau = RF sin \theta$$
Here, $$F= T$$ (downward pull) and $$\theta = 90^o$$
$$\therefore \tau = RT$$
Also $$\tau = 1 \alpha$$
$$\therefore I \alpha = TR$$
For a disc. $$I = \displaystyle \frac{MR^2}{2}$$
$$\therefore \displaystyle TR = \frac{MR^2}{2}. \alpha$$
$$\Rightarrow \displaystyle \alpha = \frac{2T}{MR}$$
Question 8
1 / -0

The momentum of a body of mass $$0.5$$ kg dropped from a certain height (h), when reaches the ground is $$10$$ N s. The value h is ___ m. ( g$$\displaystyle {= 10\ ms }^{ -2 }$$)

A
$$20$$

B
$$40$$

C
$$10$$

D
$$80$$

Solution

Initial velocity of body $$u = 0$$
Velocity with which it reaches the ground   $$V = \sqrt{2gh}$$
Momentum of the body $$P = 10 \ N \ s$$
Mass   $$m = 0.5 \ kg$$
Using    $$P = mV$$
Or $$10 = 0.5\times \sqrt{2gh}$$
$$\implies \ h = 20 \ m$$
Question 9
1 / -0

A body of mass 100 g is moving with a velocity of 15 m/s. The momentum associated with the ball will be :

A
0.5 kg m/s

B
1.5 kg m/s

C
2.5 kg m/s

D
3.2 Ns

Solution

momentum is multiple of mass and velocity so
mass of object in kg is $$0.1 kg$$
momentum = $$ 1.5$$ kg m/s
so the answer is B.
Question 10
1 / -0

If a body of mass M collides against a wall with velocity V and rebounds with the same speed, then its change of momentum will be

A
Zero

B
MV

C
2MV

D
-2MV

Solution

Initial velocity of body before collision   $$u = -V$$
where minus sign indicates the body moves towards the wall.
Initial momentum of the body $$P_i = Mu = -MV$$
Final velocity of body after collision   $$v =V$$
where positive sign indicates the body moves away from the wall.
Final momentum of the body $$P_f = Mv = MV$$
Change in momentum   $$\Delta P = P_f - P_i = MV - (-MV) = 2MV$$

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System of Particles and Rotational Motion Test - 32

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System of Particles and Rotational Motion Test - 32

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Question 1

Increase

Decrease

Remain constant

None of these

Solution

Question 2

$$\displaystyle\frac{10}{3}$$ m/s

$$\displaystyle\frac{10}{7}$$ m/s

$$\displaystyle\frac{11}{2}$$ m/s

$$\displaystyle\frac{12}{3}$$ m/s

Solution

Question 3

This will raise the centre of gravity and the boat will be rocked

This will lower centre of gravity and the boat will rocked

The effective weight of system increases

Of surface tension effects

Solution

Question 4

$$\displaystyle \frac{l}{\sqrt{3}}$$

$$\displaystyle \frac{l}{\sqrt{2}}$$

$$\displaystyle \frac{l}{\sqrt{5}}$$

$$\displaystyle \frac{l}{\sqrt{7}}$$

Solution

Question 5

$$\displaystyle \frac { { ml }^{ 2 } }{ 6 } $$

$$\displaystyle \frac { { ml }^{ 2 } }{ 12 } $$

$$\displaystyle \frac { { ml }^{ 2 } }{ 4 } $$

$$\displaystyle \frac { { ml }^{ 2 } }{ 24 } $$

Solution

Question 6

$$\sqrt{\dfrac{5}{3}} \ R$$

$$\sqrt{\dfrac{2}{3}} \ R$$

$$\sqrt{\dfrac{3}{5}} \ R$$

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

Solution

Question 7

$$\displaystyle \frac{MR}{2 T}$$

$$\displaystyle \frac{2T}{MR}$$

$$\displaystyle \frac{T}{MR}$$

$$\displaystyle \frac{MR}{T}$$

Solution

Question 8

$$20$$

$$40$$

$$10$$

$$80$$

Solution

Question 9

0.5 kg m/s

1.5 kg m/s

2.5 kg m/s

3.2 Ns

Solution

Question 10

Zero

MV

2MV

-2MV

Solution

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