System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 42

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Question 1
1 / -0

A body falling vertically downwards under gravity breaks in two parts of unequal masses. The centre of mass of the two parts taken together:-

A
Shifts horizontally towards heavier piece

B
Shifts horizontally towards lighter piece

C
Doesn't shift horizontally

D
Shift horizontally if initial speed is zero

Solution

Body is falling vertically downwards as there is no horizontal force in x-direction. So x cordinate of $$X_{cm}$$ not change
$$\dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2}=X_{cm}$$
$$X_{cm}=0$$
then,
$$m_1 x_1=-m_2 x_2$$
let, $$m_1<m_2$$
So, $$x_1>x_2$$
Question 2
1 / -0

Moment of inertia of a thin rod of mass $$m$$ and length $$l$$ about at axis passing though a point $$l/4$$ from one and perpendicular to the rod is:

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

B
$$\frac { { ml }^{ 2 } }{ 13 }$$

C
$$\frac { { 7ml }^{ 2 } }{ 48 }$$

D
$$\frac { { ml }^{ 2 } }{ 9 }$$

Solution
Question 3
1 / -0

Moment of momentum is called:

A
Torque

B
Impulse

C
Couple

D
Angular momentum

Solution

Angular Momentum $$L=\overset{\rightarrow}r\times\overset{\rightarrow}p$$ where $$p$$ is linear momentum. The Angular momentum is also called as Moment of momentum.
Question 4
1 / -0

The density of rod $$AB$$ increases linearly from $$A$$ to $$B$$. Its midpoint is $$O$$ and its centre of mass is at $$C$$. Four axes pass through $$A, B, O$$ and $$C$$, all perpendicular to the length of the rod. The moments of inertia of the rod about these axes are $${I}_{A}$$, $${I}_{B}$$,$${I}_{C}$$ respectively.

A
$${I}_{A}>{I}_{B}$$

B
$${I}_{A}<{I}_{B}$$

C
$${I}_{B}>{I}_{C}$$

D
$${I}_{O}<{I}_{C}$$

Solution

$$I_C$$ is least and $$C$$ lies between $$O$$ and $$B$$ as density is increasing from $$A$$ to $$B$$
$$I_A > I_B$$ more mass is concentrate towards $$B$$.
Question 5
1 / -0

A rubber ball of mass $$10 gm$$ and volume $$15 cm^3$$ dipped in water to a depth of $$10 m$$. Assuming density of water uniform throughout the depth, find
(a) the acceleration of the ball, and
(b) the time taken by it to the surface if it is released from rest. take $$g = 980 cm/s^2$$.

A
(a) $$5.9 m/s^2$$ (b) $$2.02$$sec.

B
(a) $$4.9 m/s^2$$ (b) $$2.02$$sec.

C
(a) $$4.9 m/s^2$$ (b) $$12.02$$sec.

D
(a) $$8.9 m/s^2$$ (b) $$2.02$$sec.

Solution

Given,

$$m=10gm$$

$$V=15cm^3$$

$$H=10m$$

$$g=980 cm/s^2$$

$$\rho=1kg/cm^2$$

The effective upward force on the ball,

$$F=F_B-mg =ma$$

$$a=\dfrac{1}{m}(F_B-mg)$$

$$a=\dfrac{1}{2}(\rho gh-mg)$$

$$a=\dfrac{1}{10}(15\times 1\times 980-10\times 980)$$

$$a=490cm/s^2=4.9 m/s^2$$

From the kinematics,

$$s=ut+\dfrac{1}{2}at^2$$ initial speed of ball is zero.

$$10=\dfrac{1}{2}\times 4.9\times t^2$$

$$t=2.02sec$$

Thus, the correct option is B.
Question 6
1 / -0

Using the parallel axes theorem, find the $$M.l.$$ of a sphere of mass $$m$$ about an axis that touches it tangentially. Give that $$I_{cm}=\dfrac {2}{5}mr^{2}$$.

A
$$\dfrac {7}{5}\ mr^{2}$$

B
$$\dfrac {7}{15}\ mr^{2}$$

C
$$\dfrac {8}{15}\ mr^{2}$$

D
$$\dfrac {9}{15}\ mr^{2}$$

Solution
Question 7
1 / -0

Find the angular momentum of a particle of mass $$m$$ describing a circle of radius $$r$$ with angular speed $$\omega$$.

A
$$2m{ \omega }r^{ 2 }$$.

B
$$m{ \omega }r^{ 2 }$$.

C
$$3m{ \omega }r^{ 2 }$$.

D
$$4m{ \omega }r^{ 2 }$$.

Solution

We know,
Angular momentum$$(L)=I\omega$$

For a particle of mass $$m$$ revolving around Radius $$R$$

$$I=mR^2$$ (about centre of revolution)

$$L=mR^2\omega$$

Hence option $$\textbf B$$ s correct answer.
Question 8
1 / -0

Two particles $$A$$ and $$B$$ initially at rest move towards each other under a mutual force of attraction. At the instant when velocity of $$A$$ is $$v$$ and that of $$B$$ is $$2v$$, the velocity of centre of mass of the system

A
$$v$$

B
$$2v$$

C
$$3v$$

D
$$zero$$

Solution

The velocity of the center of mass will be zero. The initial velocity of the particles were zero because both $$A$$ and $$B$$ were at rest so momentum will be zero. and it is given that there is no external force hence no change in momentum of center of mass and it will remain zero.
By the law of conservation of momentum, the momentum is zero, and mass is not zero, velocity must be zero

Hence, The option $$D$$ is the correct answer
Question 9
1 / -0

A uniform rod of mass $$6M$$ and length $$6I$$ is bent to make an equilateral hexagon. Its $$M.I.$$ about an axis passing through the centre of mass and perpendicular to the place of hexagon is:

A
$$22mI^{2}$$

B
$$6mI^{2}$$

C
$$4mI^{2}$$

D
$$mI^{2}/12$$

Solution

We have,

$$ mass=6m, length =6L$$

Since each side of the hexagon has side 2L and mass m. The mole of each side of the hexagon about an axis perpendicular to the plane of the hexagon through the center of the side is:

$$I_1=\dfrac{M(2L)^2}{6}=\dfrac{2ML^2}{3}$$

The distance between the center of the hexagon and the center of the rod is $$|sqrt{3L}. By using parallel
axis theorem the mole about the axis is:

$$I'=\dfrac{2ML^2}{3}+m(\sqrt{3L})^2=\dfrac{11ML^2}{3}$$

So, the Total mole of the system about the axis is:

$$I=6I'=6\times\dfrac{11ML^2}{3}$$

$$=22ML^2$$
Question 10
1 / -0

The $$M.I.$$ of a thin rod of length $$\ell$$ about the perpendicular axis through its centre is $$I$$. The $$M.I.$$ of the square structure made by four such rods about a perpendicular axis to the plane and through the centre will be :

A
$$4\ I$$

B
$$8\ I$$

C
$$12\ I$$

D
$$16\ I$$

Solution

Given Moment of inertia (MOI) , $$I=\dfrac{1}{12}ml^2$$ $$\Rightarrow ml^2= 12I$$
Using parallel axis theorem of moment, $$I=I_{cm}+ mx^2$$, Where $$I_{cm} $$ is MOI about centre of mass and $$x$$ is the distance between centre of mass and axis of rotation.
$$\therefore $$ Net MOI $$= 4\left ( \dfrac{1}{12}ml^2+m\left (\dfrac{l}{2}\right )^2\right) =\dfrac{4}{3}ml^2=16I$$

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System of Particles and Rotational Motion Test - 42

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System of Particles and Rotational Motion Test - 42

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SHARING IS CARING

Question 1

Shifts horizontally towards heavier piece

Shifts horizontally towards lighter piece

Doesn't shift horizontally

Shift horizontally if initial speed is zero

Solution

Question 2

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

$$\frac { { ml }^{ 2 } }{ 13 }$$

$$\frac { { 7ml }^{ 2 } }{ 48 }$$

$$\frac { { ml }^{ 2 } }{ 9 }$$

Solution

Question 3

Torque

Impulse

Couple

Angular momentum

Solution

Question 4

$${I}_{A}>{I}_{B}$$

$${I}_{A}<{I}_{B}$$

$${I}_{B}>{I}_{C}$$

$${I}_{O}<{I}_{C}$$

Solution

Question 5

(a) $$5.9 m/s^2$$ (b) $$2.02$$sec.

(a) $$4.9 m/s^2$$ (b) $$2.02$$sec.

(a) $$4.9 m/s^2$$ (b) $$12.02$$sec.

(a) $$8.9 m/s^2$$ (b) $$2.02$$sec.

Solution

Question 6

$$\dfrac {7}{5}\ mr^{2}$$

$$\dfrac {7}{15}\ mr^{2}$$

$$\dfrac {8}{15}\ mr^{2}$$

$$\dfrac {9}{15}\ mr^{2}$$

Solution

Question 7

$$2m{ \omega }r^{ 2 }$$.

$$m{ \omega }r^{ 2 }$$.

$$3m{ \omega }r^{ 2 }$$.

$$4m{ \omega }r^{ 2 }$$.

Solution

Question 8

$$v$$

$$2v$$

$$3v$$

$$zero$$

Solution

Question 9

$$22mI^{2}$$

$$6mI^{2}$$

$$4mI^{2}$$

$$mI^{2}/12$$

Solution

Question 10

$$4\ I$$

$$8\ I$$

$$12\ I$$

$$16\ I$$

Solution

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