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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366174 A particle of mass $m$ is projected with a velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about an axis of projection when the particle is of maximum height $h$ is

1 zero

2

\frac{m v^{3}}{4 \sqrt{2} g}

3

\frac{m v^{2}}{\sqrt{2} g}

4

m (2 g h^{3})

Explanation:

Maximum height attained
$= \frac{v^{2} \sin^{2} θ}{2 g} = \frac{v^{2} \sin^{2} 45^{\circ}}{2 g} = \frac{v^{2}}{4 g}$
$(∵ θ = 45^{\circ})$
At highest point, momentum
$= m v \cos 45^{\circ} = \frac{m v}{\sqrt{2}}$
Angular momentum
$= \frac{m v}{\sqrt{2}} \times \frac{v^{2}}{4 g} = \frac{m v^{3}}{4 \sqrt{2} g}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366175 A ball of mass $160 g$ is thrown up at an angle of $60^{\circ}$ to the horizontal at a speed of $10 {ms}^{- 1}$ . The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly $(g = 10 {ms}^{- 2})$

1

3.0 kg m^{2} / s

2

3.46 kg m^{2} / s

3

1.73 kg m^{2} / s

4

6.0 kg m^{2} / s

Explanation:

The angular momentum of the particle at the heighest point is
supporting img
$L = (m u \cos 60^{\circ}) H$
$\begin{aligned} = (\frac{m u}{2}) \frac{u^{2} \sin^{2} 60^{\circ}}{2 g} \\ = \frac{m u^{3} \sin^{2} 60^{\circ}}{4 g} = 3.0 kg m^{2} / s \end{aligned}$

JEE - 2014

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366176 The angular momentum of a rotating body changes from $4 A_{o}$ to $A_{o}$ in $4 \sec$ . The torque acting on the body is :

1

4 A_{0}

2

\frac{3}{4} A_{0}

3

\frac{3}{2} A_{0}

4

3 A_{0}

Explanation:

$τ = \frac{Δ L}{Δ t}$
Given, $Δ L = 4 A_{0} - A_{0} = 3 A_{0}, Δ t = 4 \sec$
$\Rightarrow τ = \frac{3}{4} A_{0}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366177 A particle of mass $m$ is projected horizontally from a top of a tower of height $H$ with a velocity $u$ . The angular momentum of the particle with respect to base of the tower $P$ is
supporting img

1

\frac{m u H}{2}

2

m u H

3

2 m u H

4

\frac{m u H}{4}

Explanation:

The angular momentum of the particle with respect to $P$ is
supporting img
$\begin{aligned} L_{p} = m v_{y} R \\ v_{y} = \sqrt{2 g H} \\ R = u T = u \sqrt{\frac{2 H}{g}} \\ \Rightarrow L_{p} = m \sqrt{2 g H} \sqrt{\frac{2 H}{g}} u = 2 m u H \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366178 Angular momentum of a body changes when

1 Direction of velocity changes

2 Torque acts on a body

3 Force acts on a body

4 None of these

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366174 A particle of mass $m$ is projected with a velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about an axis of projection when the particle is of maximum height $h$ is

1 zero

2

\frac{m v^{3}}{4 \sqrt{2} g}

3

\frac{m v^{2}}{\sqrt{2} g}

4

m (2 g h^{3})

Explanation:

Maximum height attained
$= \frac{v^{2} \sin^{2} θ}{2 g} = \frac{v^{2} \sin^{2} 45^{\circ}}{2 g} = \frac{v^{2}}{4 g}$
$(∵ θ = 45^{\circ})$
At highest point, momentum
$= m v \cos 45^{\circ} = \frac{m v}{\sqrt{2}}$
Angular momentum
$= \frac{m v}{\sqrt{2}} \times \frac{v^{2}}{4 g} = \frac{m v^{3}}{4 \sqrt{2} g}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366175 A ball of mass $160 g$ is thrown up at an angle of $60^{\circ}$ to the horizontal at a speed of $10 {ms}^{- 1}$ . The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly $(g = 10 {ms}^{- 2})$

1

3.0 kg m^{2} / s

2

3.46 kg m^{2} / s

3

1.73 kg m^{2} / s

4

6.0 kg m^{2} / s

Explanation:

The angular momentum of the particle at the heighest point is
supporting img
$L = (m u \cos 60^{\circ}) H$
$\begin{aligned} = (\frac{m u}{2}) \frac{u^{2} \sin^{2} 60^{\circ}}{2 g} \\ = \frac{m u^{3} \sin^{2} 60^{\circ}}{4 g} = 3.0 kg m^{2} / s \end{aligned}$

JEE - 2014

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366176 The angular momentum of a rotating body changes from $4 A_{o}$ to $A_{o}$ in $4 \sec$ . The torque acting on the body is :

1

4 A_{0}

2

\frac{3}{4} A_{0}

3

\frac{3}{2} A_{0}

4

3 A_{0}

Explanation:

$τ = \frac{Δ L}{Δ t}$
Given, $Δ L = 4 A_{0} - A_{0} = 3 A_{0}, Δ t = 4 \sec$
$\Rightarrow τ = \frac{3}{4} A_{0}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366177 A particle of mass $m$ is projected horizontally from a top of a tower of height $H$ with a velocity $u$ . The angular momentum of the particle with respect to base of the tower $P$ is
supporting img

1

\frac{m u H}{2}

2

m u H

3

2 m u H

4

\frac{m u H}{4}

Explanation:

The angular momentum of the particle with respect to $P$ is
supporting img
$\begin{aligned} L_{p} = m v_{y} R \\ v_{y} = \sqrt{2 g H} \\ R = u T = u \sqrt{\frac{2 H}{g}} \\ \Rightarrow L_{p} = m \sqrt{2 g H} \sqrt{\frac{2 H}{g}} u = 2 m u H \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366178 Angular momentum of a body changes when

1 Direction of velocity changes

2 Torque acts on a body

3 Force acts on a body

4 None of these

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366174 A particle of mass $m$ is projected with a velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about an axis of projection when the particle is of maximum height $h$ is

1 zero

2

\frac{m v^{3}}{4 \sqrt{2} g}

3

\frac{m v^{2}}{\sqrt{2} g}

4

m (2 g h^{3})

Explanation:

Maximum height attained
$= \frac{v^{2} \sin^{2} θ}{2 g} = \frac{v^{2} \sin^{2} 45^{\circ}}{2 g} = \frac{v^{2}}{4 g}$
$(∵ θ = 45^{\circ})$
At highest point, momentum
$= m v \cos 45^{\circ} = \frac{m v}{\sqrt{2}}$
Angular momentum
$= \frac{m v}{\sqrt{2}} \times \frac{v^{2}}{4 g} = \frac{m v^{3}}{4 \sqrt{2} g}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366175 A ball of mass $160 g$ is thrown up at an angle of $60^{\circ}$ to the horizontal at a speed of $10 {ms}^{- 1}$ . The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly $(g = 10 {ms}^{- 2})$

1

3.0 kg m^{2} / s

2

3.46 kg m^{2} / s

3

1.73 kg m^{2} / s

4

6.0 kg m^{2} / s

Explanation:

The angular momentum of the particle at the heighest point is
supporting img
$L = (m u \cos 60^{\circ}) H$
$\begin{aligned} = (\frac{m u}{2}) \frac{u^{2} \sin^{2} 60^{\circ}}{2 g} \\ = \frac{m u^{3} \sin^{2} 60^{\circ}}{4 g} = 3.0 kg m^{2} / s \end{aligned}$

JEE - 2014

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366176 The angular momentum of a rotating body changes from $4 A_{o}$ to $A_{o}$ in $4 \sec$ . The torque acting on the body is :

1

4 A_{0}

2

\frac{3}{4} A_{0}

3

\frac{3}{2} A_{0}

4

3 A_{0}

Explanation:

$τ = \frac{Δ L}{Δ t}$
Given, $Δ L = 4 A_{0} - A_{0} = 3 A_{0}, Δ t = 4 \sec$
$\Rightarrow τ = \frac{3}{4} A_{0}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366177 A particle of mass $m$ is projected horizontally from a top of a tower of height $H$ with a velocity $u$ . The angular momentum of the particle with respect to base of the tower $P$ is
supporting img

1

\frac{m u H}{2}

2

m u H

3

2 m u H

4

\frac{m u H}{4}

Explanation:

The angular momentum of the particle with respect to $P$ is
supporting img
$\begin{aligned} L_{p} = m v_{y} R \\ v_{y} = \sqrt{2 g H} \\ R = u T = u \sqrt{\frac{2 H}{g}} \\ \Rightarrow L_{p} = m \sqrt{2 g H} \sqrt{\frac{2 H}{g}} u = 2 m u H \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366178 Angular momentum of a body changes when

1 Direction of velocity changes

2 Torque acts on a body

3 Force acts on a body

4 None of these

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366174 A particle of mass $m$ is projected with a velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about an axis of projection when the particle is of maximum height $h$ is

1 zero

2

\frac{m v^{3}}{4 \sqrt{2} g}

3

\frac{m v^{2}}{\sqrt{2} g}

4

m (2 g h^{3})

Explanation:

Maximum height attained
$= \frac{v^{2} \sin^{2} θ}{2 g} = \frac{v^{2} \sin^{2} 45^{\circ}}{2 g} = \frac{v^{2}}{4 g}$
$(∵ θ = 45^{\circ})$
At highest point, momentum
$= m v \cos 45^{\circ} = \frac{m v}{\sqrt{2}}$
Angular momentum
$= \frac{m v}{\sqrt{2}} \times \frac{v^{2}}{4 g} = \frac{m v^{3}}{4 \sqrt{2} g}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366175 A ball of mass $160 g$ is thrown up at an angle of $60^{\circ}$ to the horizontal at a speed of $10 {ms}^{- 1}$ . The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly $(g = 10 {ms}^{- 2})$

1

3.0 kg m^{2} / s

2

3.46 kg m^{2} / s

3

1.73 kg m^{2} / s

4

6.0 kg m^{2} / s

Explanation:

The angular momentum of the particle at the heighest point is
supporting img
$L = (m u \cos 60^{\circ}) H$
$\begin{aligned} = (\frac{m u}{2}) \frac{u^{2} \sin^{2} 60^{\circ}}{2 g} \\ = \frac{m u^{3} \sin^{2} 60^{\circ}}{4 g} = 3.0 kg m^{2} / s \end{aligned}$

JEE - 2014

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366176 The angular momentum of a rotating body changes from $4 A_{o}$ to $A_{o}$ in $4 \sec$ . The torque acting on the body is :

1

4 A_{0}

2

\frac{3}{4} A_{0}

3

\frac{3}{2} A_{0}

4

3 A_{0}

Explanation:

$τ = \frac{Δ L}{Δ t}$
Given, $Δ L = 4 A_{0} - A_{0} = 3 A_{0}, Δ t = 4 \sec$
$\Rightarrow τ = \frac{3}{4} A_{0}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366177 A particle of mass $m$ is projected horizontally from a top of a tower of height $H$ with a velocity $u$ . The angular momentum of the particle with respect to base of the tower $P$ is
supporting img

1

\frac{m u H}{2}

2

m u H

3

2 m u H

4

\frac{m u H}{4}

Explanation:

The angular momentum of the particle with respect to $P$ is
supporting img
$\begin{aligned} L_{p} = m v_{y} R \\ v_{y} = \sqrt{2 g H} \\ R = u T = u \sqrt{\frac{2 H}{g}} \\ \Rightarrow L_{p} = m \sqrt{2 g H} \sqrt{\frac{2 H}{g}} u = 2 m u H \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366178 Angular momentum of a body changes when

1 Direction of velocity changes

2 Torque acts on a body

3 Force acts on a body

4 None of these

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366174 A particle of mass $m$ is projected with a velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about an axis of projection when the particle is of maximum height $h$ is

1 zero

2

\frac{m v^{3}}{4 \sqrt{2} g}

3

\frac{m v^{2}}{\sqrt{2} g}

4

m (2 g h^{3})

Explanation:

Maximum height attained
$= \frac{v^{2} \sin^{2} θ}{2 g} = \frac{v^{2} \sin^{2} 45^{\circ}}{2 g} = \frac{v^{2}}{4 g}$
$(∵ θ = 45^{\circ})$
At highest point, momentum
$= m v \cos 45^{\circ} = \frac{m v}{\sqrt{2}}$
Angular momentum
$= \frac{m v}{\sqrt{2}} \times \frac{v^{2}}{4 g} = \frac{m v^{3}}{4 \sqrt{2} g}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366175 A ball of mass $160 g$ is thrown up at an angle of $60^{\circ}$ to the horizontal at a speed of $10 {ms}^{- 1}$ . The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly $(g = 10 {ms}^{- 2})$

1

3.0 kg m^{2} / s

2

3.46 kg m^{2} / s

3

1.73 kg m^{2} / s

4

6.0 kg m^{2} / s

Explanation:

The angular momentum of the particle at the heighest point is
supporting img
$L = (m u \cos 60^{\circ}) H$
$\begin{aligned} = (\frac{m u}{2}) \frac{u^{2} \sin^{2} 60^{\circ}}{2 g} \\ = \frac{m u^{3} \sin^{2} 60^{\circ}}{4 g} = 3.0 kg m^{2} / s \end{aligned}$

JEE - 2014

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366176 The angular momentum of a rotating body changes from $4 A_{o}$ to $A_{o}$ in $4 \sec$ . The torque acting on the body is :

1

4 A_{0}

2

\frac{3}{4} A_{0}

3

\frac{3}{2} A_{0}

4

3 A_{0}

Explanation:

$τ = \frac{Δ L}{Δ t}$
Given, $Δ L = 4 A_{0} - A_{0} = 3 A_{0}, Δ t = 4 \sec$
$\Rightarrow τ = \frac{3}{4} A_{0}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366177 A particle of mass $m$ is projected horizontally from a top of a tower of height $H$ with a velocity $u$ . The angular momentum of the particle with respect to base of the tower $P$ is
supporting img

1

\frac{m u H}{2}

2

m u H

3

2 m u H

4

\frac{m u H}{4}

Explanation:

The angular momentum of the particle with respect to $P$ is
supporting img
$\begin{aligned} L_{p} = m v_{y} R \\ v_{y} = \sqrt{2 g H} \\ R = u T = u \sqrt{\frac{2 H}{g}} \\ \Rightarrow L_{p} = m \sqrt{2 g H} \sqrt{\frac{2 H}{g}} u = 2 m u H \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366178 Angular momentum of a body changes when

1 Direction of velocity changes

2 Torque acts on a body

3 Force acts on a body

4 None of these