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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366157 A ladder of length $l$ and mass $m$ is placed against a smooth vertical wall, but the ground is not smooth. Coefficient of friction between the ground and the ladder is $μ$ . The angle $θ$ at which the ladder will stay in equilibrium is

1

θ = \tan^{- 1} (\frac{1}{2 μ})

2

θ = \tan^{- 1} (μ)

3

θ = \tan^{- 1} (2 μ)

4 None of these

Explanation:

The forces on the ladder are shown in the figure. Net force along $y$ and $x$ directions are zero.
$m g = N_{1}$
$μ N_{1} = N_{2}$ or $N_{2} = μ m g$
Taking moment about $O$ , we get
supporting img
$μ m g l \sin θ = m g \frac{l}{2} \cos θ$
$\tan θ = \frac{1}{2 μ}$ or $θ = \tan^{- 1} (\frac{1}{2 μ})$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366158 A particle of mass $5 g$ is moving with a uniform speed of $3 \sqrt{2} c m / s$ in the $x - y$ plane along the line $y = x + 4$ . The magnitude of its angular momentum about the origin in $g c m^{2} / s$ is :

1

30 g c m^{2} / s

2

40 g c m^{2} / s

3

60 g c m^{2} / s

4

20 g c m^{2} / s

Explanation:

supporting img
$L_{0} = m v \cos 45^{\circ} \times 4$
$= 5 \times 3 \sqrt{2} \times \frac{1}{\sqrt{2}} \times 4 = 60 g c m^{2} / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366159 The position of a particle is given by: $\vec{r} = (\hat{i} + 2 \hat{j} - \hat{k})$ and momentum $\vec{P} = (3 \hat{i} + 4 \hat{j} - 2 \hat{k})$ . The angular momentum is perpendicular to

1

X

- axis

2

Y

- axis

3

Z

- axis

4 Line at equal angles to all the three axes

Explanation:

$\vec{L} = \vec{r} \times \vec{p} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 1 \\ 3 & 4 & - 2 \end{array} | = - \hat{j} - 2 \hat{k}$
$\vec{L}$ lies in $y z$ plane

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366160 Angular momentum $L$ of body with moment of inertia I and angular velocity $ω r a d / \sec$ is equal to

1

\frac{I}{ω}

2

I ω

3

I ω^{2}

4 None of these

Explanation:

$L = I ω$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366157 A ladder of length $l$ and mass $m$ is placed against a smooth vertical wall, but the ground is not smooth. Coefficient of friction between the ground and the ladder is $μ$ . The angle $θ$ at which the ladder will stay in equilibrium is

1

θ = \tan^{- 1} (\frac{1}{2 μ})

2

θ = \tan^{- 1} (μ)

3

θ = \tan^{- 1} (2 μ)

4 None of these

Explanation:

The forces on the ladder are shown in the figure. Net force along $y$ and $x$ directions are zero.
$m g = N_{1}$
$μ N_{1} = N_{2}$ or $N_{2} = μ m g$
Taking moment about $O$ , we get
supporting img
$μ m g l \sin θ = m g \frac{l}{2} \cos θ$
$\tan θ = \frac{1}{2 μ}$ or $θ = \tan^{- 1} (\frac{1}{2 μ})$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366158 A particle of mass $5 g$ is moving with a uniform speed of $3 \sqrt{2} c m / s$ in the $x - y$ plane along the line $y = x + 4$ . The magnitude of its angular momentum about the origin in $g c m^{2} / s$ is :

1

30 g c m^{2} / s

2

40 g c m^{2} / s

3

60 g c m^{2} / s

4

20 g c m^{2} / s

Explanation:

supporting img
$L_{0} = m v \cos 45^{\circ} \times 4$
$= 5 \times 3 \sqrt{2} \times \frac{1}{\sqrt{2}} \times 4 = 60 g c m^{2} / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366159 The position of a particle is given by: $\vec{r} = (\hat{i} + 2 \hat{j} - \hat{k})$ and momentum $\vec{P} = (3 \hat{i} + 4 \hat{j} - 2 \hat{k})$ . The angular momentum is perpendicular to

1

X

- axis

2

Y

- axis

3

Z

- axis

4 Line at equal angles to all the three axes

Explanation:

$\vec{L} = \vec{r} \times \vec{p} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 1 \\ 3 & 4 & - 2 \end{array} | = - \hat{j} - 2 \hat{k}$
$\vec{L}$ lies in $y z$ plane

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366160 Angular momentum $L$ of body with moment of inertia I and angular velocity $ω r a d / \sec$ is equal to

1

\frac{I}{ω}

2

I ω

3

I ω^{2}

4 None of these

Explanation:

$L = I ω$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366157 A ladder of length $l$ and mass $m$ is placed against a smooth vertical wall, but the ground is not smooth. Coefficient of friction between the ground and the ladder is $μ$ . The angle $θ$ at which the ladder will stay in equilibrium is

1

θ = \tan^{- 1} (\frac{1}{2 μ})

2

θ = \tan^{- 1} (μ)

3

θ = \tan^{- 1} (2 μ)

4 None of these

Explanation:

The forces on the ladder are shown in the figure. Net force along $y$ and $x$ directions are zero.
$m g = N_{1}$
$μ N_{1} = N_{2}$ or $N_{2} = μ m g$
Taking moment about $O$ , we get
supporting img
$μ m g l \sin θ = m g \frac{l}{2} \cos θ$
$\tan θ = \frac{1}{2 μ}$ or $θ = \tan^{- 1} (\frac{1}{2 μ})$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366158 A particle of mass $5 g$ is moving with a uniform speed of $3 \sqrt{2} c m / s$ in the $x - y$ plane along the line $y = x + 4$ . The magnitude of its angular momentum about the origin in $g c m^{2} / s$ is :

1

30 g c m^{2} / s

2

40 g c m^{2} / s

3

60 g c m^{2} / s

4

20 g c m^{2} / s

Explanation:

supporting img
$L_{0} = m v \cos 45^{\circ} \times 4$
$= 5 \times 3 \sqrt{2} \times \frac{1}{\sqrt{2}} \times 4 = 60 g c m^{2} / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366159 The position of a particle is given by: $\vec{r} = (\hat{i} + 2 \hat{j} - \hat{k})$ and momentum $\vec{P} = (3 \hat{i} + 4 \hat{j} - 2 \hat{k})$ . The angular momentum is perpendicular to

1

X

- axis

2

Y

- axis

3

Z

- axis

4 Line at equal angles to all the three axes

Explanation:

$\vec{L} = \vec{r} \times \vec{p} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 1 \\ 3 & 4 & - 2 \end{array} | = - \hat{j} - 2 \hat{k}$
$\vec{L}$ lies in $y z$ plane

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366160 Angular momentum $L$ of body with moment of inertia I and angular velocity $ω r a d / \sec$ is equal to

1

\frac{I}{ω}

2

I ω

3

I ω^{2}

4 None of these

Explanation:

$L = I ω$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366157 A ladder of length $l$ and mass $m$ is placed against a smooth vertical wall, but the ground is not smooth. Coefficient of friction between the ground and the ladder is $μ$ . The angle $θ$ at which the ladder will stay in equilibrium is

1

θ = \tan^{- 1} (\frac{1}{2 μ})

2

θ = \tan^{- 1} (μ)

3

θ = \tan^{- 1} (2 μ)

4 None of these

Explanation:

The forces on the ladder are shown in the figure. Net force along $y$ and $x$ directions are zero.
$m g = N_{1}$
$μ N_{1} = N_{2}$ or $N_{2} = μ m g$
Taking moment about $O$ , we get
supporting img
$μ m g l \sin θ = m g \frac{l}{2} \cos θ$
$\tan θ = \frac{1}{2 μ}$ or $θ = \tan^{- 1} (\frac{1}{2 μ})$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366158 A particle of mass $5 g$ is moving with a uniform speed of $3 \sqrt{2} c m / s$ in the $x - y$ plane along the line $y = x + 4$ . The magnitude of its angular momentum about the origin in $g c m^{2} / s$ is :

1

30 g c m^{2} / s

2

40 g c m^{2} / s

3

60 g c m^{2} / s

4

20 g c m^{2} / s

Explanation:

supporting img
$L_{0} = m v \cos 45^{\circ} \times 4$
$= 5 \times 3 \sqrt{2} \times \frac{1}{\sqrt{2}} \times 4 = 60 g c m^{2} / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366159 The position of a particle is given by: $\vec{r} = (\hat{i} + 2 \hat{j} - \hat{k})$ and momentum $\vec{P} = (3 \hat{i} + 4 \hat{j} - 2 \hat{k})$ . The angular momentum is perpendicular to

1

X

- axis

2

Y

- axis

3

Z

- axis

4 Line at equal angles to all the three axes

Explanation:

$\vec{L} = \vec{r} \times \vec{p} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 1 \\ 3 & 4 & - 2 \end{array} | = - \hat{j} - 2 \hat{k}$
$\vec{L}$ lies in $y z$ plane

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366160 Angular momentum $L$ of body with moment of inertia I and angular velocity $ω r a d / \sec$ is equal to

1

\frac{I}{ω}

2

I ω

3

I ω^{2}

4 None of these

Explanation:

$L = I ω$

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