System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 39

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Question 1
1 / -0

The radius of gyration of a solid sphere of radius $$R$$ about a certain axis is also equal to $$R$$. If $$r$$ is the distance between the axis and the centre of the sphere, then $$r$$ is equal to:

A
$$R$$

B
$$0.5R$$

C
$$\sqrt { 0.6 } R$$

D
$$\sqrt { 0.3 } R$$

Solution

The radius of gyration of a solid sphere of radius $$R$$ about a certain axis is equal to $$R$$.
$$k=R$$
$$Mk^2=MR^2$$
$$I=I_{cm}+mr^2$$
$$MR^2=\dfrac{2}{5}MR^2+Mr^2$$
$$r^2=\dfrac{3}{5}R^2$$
$$r=\sqrt{0.6}R$$
The correct option is C.
Question 2
1 / -0

A rigid body rotates about a fixed axis with variable angular velocity equal to (a-bt) at time t where a and b are constants. The angle through which it rotates before it comes to rest is___?

A
$$\dfrac{a^2}{b}$$

B
$$\dfrac{a^2}{2b}$$

C
$$\dfrac{a^2}{4b}$$

D
$$\dfrac{a^2}{2b^2}$$

Solution

Let the angle rotated be $$\theta$$

Given,

$$\omega=a-bt$$

$$\dfrac{d\theta}{dt}=a-bt$$

Also,

It will come to rest when, $$\omega=a-bt=0$$

$$t_f=\dfrac{a}{b}$$

$$\displaystyle \int_0^\theta d\theta=\int_0^{\dfrac{a}{b}} a-bt$$

$$\theta=\dfrac{a^2}{b}-\dfrac{a^2}{2b}=\dfrac{a^2}{2b}$$

Option $$\textbf B$$ is the correct answer
Question 3
1 / -0

The cylinders P and Q are of equal of mas and length but made of metals with densities $${\rho _P}$$ and $${\rho _Q}\,\,\left( {{\rho _P} > {\rho _Q}} \right).$$ If their moment of inertia about an axis passing through centre and normal to the circular face be $${I_P}$$ and $${I_Q}$$, then:

A
$${I_P} = {I_Q}$$

B
$${I_P} > {I_Q}$$

C
$${I_P} < {I_Q}$$

D
$${I_P} \leqslant {I_Q}$$

Solution

Given: The cylinders P and Q are of equal of mas and length but made of metals with densities $$\rho_P$$ and $$\rho_Q(\rho_P>\rho_Q)$$. If their moment of inertia about an axis passing through centre and normal to the circular face be $$I_P$$ and $$I_Q$$,
To find the relation between $$I_P$$ and $$I_Q$$
Solution:
Let mass be m,
We know, $$m=\rho V$$ (as mass=density$$\times$$volume)
$$\implies m=\rho\times \pi r^2l\\\implies r^2=\dfrac {m}{\pi \rho l}.......(i)$$
Moment of inertia about given axis is,
$$I=\dfrac 12mr^2=\dfrac 12m\times \dfrac {m}{\pi \rho l}\\\implies I=\dfrac {m^2}{2\pi \rho l}\\\implies I\propto \dfrac 1\rho\\\implies \dfrac {I_P}{I_Q}=\dfrac {\rho_Q}{\rho_P}$$
Given,
$$\rho_P>\rho_Q\\\implies I_P<I_Q$$
Question 4
1 / -0

The reduce mass of two particles having masses m and 2 m is

A
2 m

B
3 m

C
$$\dfrac {2 m}{3}$$

D
$$\dfrac { m}{2}$$

Solution

Given,
$$m_1=m$$
$$m_2=2m$$

The reduced mass of two particle system is given by
$$\mu=\dfrac{m_1m_2}{m_1+m_2}$$

$$\mu=\dfrac{m\times 2m}{m+2m}$$

$$\mu=\dfrac{2m}{3}$$
The correct option is C.
Question 5
1 / -0

Two bodies of masses 10 kg and 2 kg are moving with velocities $$2\hat { i } -7\hat { k } +3\hat { j }\ m{ s }^{ -1 }$$ and $$-10\hat { i } +35\hat { k } -3\hat { j }\ m{ s }^{ -1 }$$ respectively. The velocity of their centre of mass is

A
$$2\hat { i }\ m{ s }^{ -1 }$$

B
$$2\hat { j }\ m{ s }^{ -1 }$$

C
$$\left( 2\hat { j } +2\hat { k } \right) m{ s }^{ -1 }$$

D
$$\left( 2\hat { i } +2\hat { j } +2\hat { k } \right) m{ s }^{ -1 }$$

Solution

$$m_1 = 10\ kg\quad \vec {v_1}=2\hat i -7\hat k+3\hat j\ m/s$$
$$m_2=2\ kg \quad \vec {v_2}=-10\hat i +35\hat k-3\hat j\ m/s$$
velocity of centre of mass should be
$$\vec {v}=\dfrac {m_1 \vec {v_1}+m_2 \vec {v_2}}{m_1 +m_2}$$
or $$\vec {v}=\dfrac {20\hat i-70\hat k+30\hat j+(-20\hat i+70\hat k-6\hat j)}{10+2}$$
$$\Rightarrow \ \boxed {\vec {v}=\dfrac {24\ \hat j}{12}=2\hat j\ m/s}$$
Question 6
1 / -0

A ball of mass m moving with a constant velocity u strikes against a ball of same mass at rest. If e is the coefficient of restitution, then what will be the ration of velocity of two balls after collision?

A
$$\dfrac { 1-e }{ 1+e } $$

B
$$\dfrac { e-1 }{ e+1 } $$

C
$$\dfrac { 1+e }{ 1-e } $$

D
$$\dfrac { e+1 }{ e-1 } $$

Solution

momentum should be conserved
so $$mu+0=mv_1 +mv_2$$
$$\Rightarrow \ v_1 +v_2 =u.....(1)$$
restitution coefficient should be given
as $$e=\dfrac {v_2 -v_1}{u_1 -u_2}=\dfrac {v_2 -v_1}{u-0}$$
$$\Rightarrow \ v_2 -v_1 =e\ u.....(2)$$
adding $$(1)$$ & $$(2)$$ we get
$$2v_2 =u(e+1).....(3)$$
subtracting $$(1)$$ & $$(2)$$ we get
$$2v_1=u(1-e)....(4)$$
dividing $$(4)$$ by $$(3)$$
we get
$$\boxed {\dfrac {v_1}{v_2}=\dfrac {1-e}{1+e}}$$
Question 7
1 / -0

Four identical rods are joined end to end to form a square. The mass of each rod is $$M$$. The moment of inertia of the system about one of the diagonals is:

A
$$\cfrac { 2M{ l }^{ 2 } }{ 3 } $$

B
$$\cfrac { 13M{ l }^{ 2 } }{ 3 } $$

C
$$\cfrac { M{ l }^{ 2 } }{ 6 } $$

D
$$\cfrac { 13M{ l }^{ 2 } }{ 6 } $$

Solution
Question 8
1 / -0

The moment of inertia of a solid sphere about an axis passing through the centre of gravity is $$1/2M{R}^{2}$$, then its radius of gyration about a parallel axis at a distance $$2R$$ from first axis is:

A
$$5R$$

B
$$\sqrt { \cfrac { 22 }{ 5 } } R$$

C
$$\cfrac { 5 }{ 2 } R\quad $$

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

Solution
Question 9
1 / -0

Four identical rods are joined end to end to form a square. The mass of each rod is $$M$$. The moment of inertia of the system about an axis passing through the point of intersecion of diagonals and perpendicular to the plane of the square is:

A
$$\cfrac { 4M{ l }^{ 2 } }{ 3 } $$

B
$$\cfrac { 13M{ l }^{ 2 } }{ 3 } $$

C
$$\cfrac { M{ l }^{ 2 } }{ 6 } $$

D
$$\cfrac { 13M{ l }^{ 2 } }{ 6 } $$

Solution

To the moment of inertia about axis passing through c and perpendicular to the plane of rods we use perpendicular axis theorum 4 times
$$I_c= 4(\dfrac{ml^2}{12}+\dfrac{ml^2}{4})=\dfrac{4ml^2}{3}$$

hence Option A is correct
Question 10
1 / -0

A coin is placed at the edge of a horizontal disc rotating about a vertical axis through its axis through speed $$ 2 rad s^{-1}$$. The radius of the disc is 50 cm. Find the minimum coefficient of friction between disc and coin so that the coin does not slip (g=10$$ms^{-2}$$).

A
0.1

B
0.2

C
0.3

D
0.4

Solution

Friction $$f=mr\omega^2=\mu mg\\ \Rightarrow\cfrac{50}{100}\times4=\mu.10\\ \Rightarrow \mu=0.2$$

The coin does not slip if centripetal force necessary for rotation is provided by friction

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System of Particles and Rotational Motion Test - 39

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System of Particles and Rotational Motion Test - 39

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SHARING IS CARING

Question 1

$$R$$

$$0.5R$$

$$\sqrt { 0.6 } R$$

$$\sqrt { 0.3 } R$$

Solution

Question 2

$$\dfrac{a^2}{b}$$

$$\dfrac{a^2}{2b}$$

$$\dfrac{a^2}{4b}$$

$$\dfrac{a^2}{2b^2}$$

Solution

Question 3

$${I_P} = {I_Q}$$

$${I_P} > {I_Q}$$

$${I_P} < {I_Q}$$

$${I_P} \leqslant {I_Q}$$

Solution

Question 4

2 m

3 m

$$\dfrac {2 m}{3}$$

$$\dfrac { m}{2}$$

Solution

Question 5

$$2\hat { i }\ m{ s }^{ -1 }$$

$$2\hat { j }\ m{ s }^{ -1 }$$

$$\left( 2\hat { j } +2\hat { k } \right) m{ s }^{ -1 }$$

$$\left( 2\hat { i } +2\hat { j } +2\hat { k } \right) m{ s }^{ -1 }$$

Solution

Question 6

$$\dfrac { 1-e }{ 1+e } $$

$$\dfrac { e-1 }{ e+1 } $$

$$\dfrac { 1+e }{ 1-e } $$

$$\dfrac { e+1 }{ e-1 } $$

Solution

Question 7

$$\cfrac { 2M{ l }^{ 2 } }{ 3 } $$

$$\cfrac { 13M{ l }^{ 2 } }{ 3 } $$

$$\cfrac { M{ l }^{ 2 } }{ 6 } $$

$$\cfrac { 13M{ l }^{ 2 } }{ 6 } $$

Solution

Question 8

$$5R$$

$$\sqrt { \cfrac { 22 }{ 5 } } R$$

$$\cfrac { 5 }{ 2 } R\quad $$

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

Solution

Question 9

$$\cfrac { 4M{ l }^{ 2 } }{ 3 } $$

$$\cfrac { 13M{ l }^{ 2 } }{ 3 } $$

$$\cfrac { M{ l }^{ 2 } }{ 6 } $$

$$\cfrac { 13M{ l }^{ 2 } }{ 6 } $$

Solution

Question 10

0.1

0.2

0.3

0.4

Solution

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