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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365628 Assertion :
Torque is equal to rate of change of angular momentum.
Reason :
Angular momentum depends on moment of inertia.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Torque $τ = I α$
But $α = \frac{d ω}{d t}$ and $L = I ω$
$\Rightarrow τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I α$
So correct option is (2).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365629 Calculate the angular momentum of a body whose rotational energy is 10 joule. If the angular momentum vector coincides with the axis of rotation and its moment of inertia about this axis is $8 \times 10^{- 7} k g m^{2}$

1

4 \times 10^{- 3} k g m^{2} / s

2

6 \times 10^{- 3} k g m^{2} / s

3

2 \times 10^{- 3} k g m^{2} / s

4 None of these

Explanation:

Angular momentum is given by
$L = I ω (K = \frac{1}{2} I ω^{2})$
$\Rightarrow L = I \sqrt{\frac{2 K}{I}} = \sqrt{2 K I}$
$L = \sqrt{2 K I} = \sqrt{2 \times 10 \times 8 \times 10^{- 7}} = 4 \times 10^{- 3} k g m^{2} s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365630 A ring of mass $10 k g$ and diameter $0.4 m$ is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be

1

0.4 k g \times m^{2} / s

2

4.4 k g \times m^{2} / s

3

88 k g \times m^{2} / s

4

44 k g \times m^{2} / s

Explanation:

$I = m r^{2} \Rightarrow 10 \times (0.2)^{2} = 0.4 k g m^{2}$
$ω = 2 π n = 2 π \times \frac{2100}{60} r a d / s$
$∴ L = I ω = \frac{0.4 \times 2 π \times 210}{6} = 88 k g - m^{2} / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365631 A solid cylinder of mass $20 k g$ and radius $20 c m$ rotates about its axis with an angular speed of $100 r a d s^{- 1}$ . The angular momentum of the cylinder about its axis is

1

40 J s

2

400 J s

3

20 J s

4

200 J s

Explanation:

Here, $M = 20 k g$
$R = 20 c m = 20 \times 10^{- 2} m, ω = 100 r a d s^{- 1}$
Moment of inertia of the solid cylinder about its axis is
$I = \frac{M R^{2}}{2} = \frac{(20 k g) {(20 \times 10^{- 2} m)}^{2}}{2} = 0.4 k g m^{2}$
Angular momentum of the cylinder about its axis is
$L = I ω = (0.4 k g m^{2}) 100 r a d s^{- 1}) = 40 J s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365632 If the earth is treated as a sphere of radius $R$ and mass $M$ . Its angular momentum about the axis of rotation with period $T$ is

1

\frac{4 π M R^{2}}{5 T}

2

\frac{π M R^{2}}{T}

3

\frac{M R^{2} π}{T}

4

\frac{2 π M R^{2}}{5 T}

Explanation:

Angular momentum of earth about its axis of rotation,
$L = I ω = \frac{2}{5} M R^{2} \times \frac{2 π}{T} = \frac{4 π M R^{2}}{5 T}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365628 Assertion :
Torque is equal to rate of change of angular momentum.
Reason :
Angular momentum depends on moment of inertia.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Torque $τ = I α$
But $α = \frac{d ω}{d t}$ and $L = I ω$
$\Rightarrow τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I α$
So correct option is (2).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365629 Calculate the angular momentum of a body whose rotational energy is 10 joule. If the angular momentum vector coincides with the axis of rotation and its moment of inertia about this axis is $8 \times 10^{- 7} k g m^{2}$

1

4 \times 10^{- 3} k g m^{2} / s

2

6 \times 10^{- 3} k g m^{2} / s

3

2 \times 10^{- 3} k g m^{2} / s

4 None of these

Explanation:

Angular momentum is given by
$L = I ω (K = \frac{1}{2} I ω^{2})$
$\Rightarrow L = I \sqrt{\frac{2 K}{I}} = \sqrt{2 K I}$
$L = \sqrt{2 K I} = \sqrt{2 \times 10 \times 8 \times 10^{- 7}} = 4 \times 10^{- 3} k g m^{2} s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365630 A ring of mass $10 k g$ and diameter $0.4 m$ is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be

1

0.4 k g \times m^{2} / s

2

4.4 k g \times m^{2} / s

3

88 k g \times m^{2} / s

4

44 k g \times m^{2} / s

Explanation:

$I = m r^{2} \Rightarrow 10 \times (0.2)^{2} = 0.4 k g m^{2}$
$ω = 2 π n = 2 π \times \frac{2100}{60} r a d / s$
$∴ L = I ω = \frac{0.4 \times 2 π \times 210}{6} = 88 k g - m^{2} / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365631 A solid cylinder of mass $20 k g$ and radius $20 c m$ rotates about its axis with an angular speed of $100 r a d s^{- 1}$ . The angular momentum of the cylinder about its axis is

1

40 J s

2

400 J s

3

20 J s

4

200 J s

Explanation:

Here, $M = 20 k g$
$R = 20 c m = 20 \times 10^{- 2} m, ω = 100 r a d s^{- 1}$
Moment of inertia of the solid cylinder about its axis is
$I = \frac{M R^{2}}{2} = \frac{(20 k g) {(20 \times 10^{- 2} m)}^{2}}{2} = 0.4 k g m^{2}$
Angular momentum of the cylinder about its axis is
$L = I ω = (0.4 k g m^{2}) 100 r a d s^{- 1}) = 40 J s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365632 If the earth is treated as a sphere of radius $R$ and mass $M$ . Its angular momentum about the axis of rotation with period $T$ is

1

\frac{4 π M R^{2}}{5 T}

2

\frac{π M R^{2}}{T}

3

\frac{M R^{2} π}{T}

4

\frac{2 π M R^{2}}{5 T}

Explanation:

Angular momentum of earth about its axis of rotation,
$L = I ω = \frac{2}{5} M R^{2} \times \frac{2 π}{T} = \frac{4 π M R^{2}}{5 T}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365628 Assertion :
Torque is equal to rate of change of angular momentum.
Reason :
Angular momentum depends on moment of inertia.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Torque $τ = I α$
But $α = \frac{d ω}{d t}$ and $L = I ω$
$\Rightarrow τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I α$
So correct option is (2).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365629 Calculate the angular momentum of a body whose rotational energy is 10 joule. If the angular momentum vector coincides with the axis of rotation and its moment of inertia about this axis is $8 \times 10^{- 7} k g m^{2}$

1

4 \times 10^{- 3} k g m^{2} / s

2

6 \times 10^{- 3} k g m^{2} / s

3

2 \times 10^{- 3} k g m^{2} / s

4 None of these

Explanation:

Angular momentum is given by
$L = I ω (K = \frac{1}{2} I ω^{2})$
$\Rightarrow L = I \sqrt{\frac{2 K}{I}} = \sqrt{2 K I}$
$L = \sqrt{2 K I} = \sqrt{2 \times 10 \times 8 \times 10^{- 7}} = 4 \times 10^{- 3} k g m^{2} s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365630 A ring of mass $10 k g$ and diameter $0.4 m$ is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be

1

0.4 k g \times m^{2} / s

2

4.4 k g \times m^{2} / s

3

88 k g \times m^{2} / s

4

44 k g \times m^{2} / s

Explanation:

$I = m r^{2} \Rightarrow 10 \times (0.2)^{2} = 0.4 k g m^{2}$
$ω = 2 π n = 2 π \times \frac{2100}{60} r a d / s$
$∴ L = I ω = \frac{0.4 \times 2 π \times 210}{6} = 88 k g - m^{2} / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365631 A solid cylinder of mass $20 k g$ and radius $20 c m$ rotates about its axis with an angular speed of $100 r a d s^{- 1}$ . The angular momentum of the cylinder about its axis is

1

40 J s

2

400 J s

3

20 J s

4

200 J s

Explanation:

Here, $M = 20 k g$
$R = 20 c m = 20 \times 10^{- 2} m, ω = 100 r a d s^{- 1}$
Moment of inertia of the solid cylinder about its axis is
$I = \frac{M R^{2}}{2} = \frac{(20 k g) {(20 \times 10^{- 2} m)}^{2}}{2} = 0.4 k g m^{2}$
Angular momentum of the cylinder about its axis is
$L = I ω = (0.4 k g m^{2}) 100 r a d s^{- 1}) = 40 J s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365632 If the earth is treated as a sphere of radius $R$ and mass $M$ . Its angular momentum about the axis of rotation with period $T$ is

1

\frac{4 π M R^{2}}{5 T}

2

\frac{π M R^{2}}{T}

3

\frac{M R^{2} π}{T}

4

\frac{2 π M R^{2}}{5 T}

Explanation:

Angular momentum of earth about its axis of rotation,
$L = I ω = \frac{2}{5} M R^{2} \times \frac{2 π}{T} = \frac{4 π M R^{2}}{5 T}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365628 Assertion :
Torque is equal to rate of change of angular momentum.
Reason :
Angular momentum depends on moment of inertia.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Torque $τ = I α$
But $α = \frac{d ω}{d t}$ and $L = I ω$
$\Rightarrow τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I α$
So correct option is (2).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365629 Calculate the angular momentum of a body whose rotational energy is 10 joule. If the angular momentum vector coincides with the axis of rotation and its moment of inertia about this axis is $8 \times 10^{- 7} k g m^{2}$

1

4 \times 10^{- 3} k g m^{2} / s

2

6 \times 10^{- 3} k g m^{2} / s

3

2 \times 10^{- 3} k g m^{2} / s

4 None of these

Explanation:

Angular momentum is given by
$L = I ω (K = \frac{1}{2} I ω^{2})$
$\Rightarrow L = I \sqrt{\frac{2 K}{I}} = \sqrt{2 K I}$
$L = \sqrt{2 K I} = \sqrt{2 \times 10 \times 8 \times 10^{- 7}} = 4 \times 10^{- 3} k g m^{2} s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365630 A ring of mass $10 k g$ and diameter $0.4 m$ is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be

1

0.4 k g \times m^{2} / s

2

4.4 k g \times m^{2} / s

3

88 k g \times m^{2} / s

4

44 k g \times m^{2} / s

Explanation:

$I = m r^{2} \Rightarrow 10 \times (0.2)^{2} = 0.4 k g m^{2}$
$ω = 2 π n = 2 π \times \frac{2100}{60} r a d / s$
$∴ L = I ω = \frac{0.4 \times 2 π \times 210}{6} = 88 k g - m^{2} / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365631 A solid cylinder of mass $20 k g$ and radius $20 c m$ rotates about its axis with an angular speed of $100 r a d s^{- 1}$ . The angular momentum of the cylinder about its axis is

1

40 J s

2

400 J s

3

20 J s

4

200 J s

Explanation:

Here, $M = 20 k g$
$R = 20 c m = 20 \times 10^{- 2} m, ω = 100 r a d s^{- 1}$
Moment of inertia of the solid cylinder about its axis is
$I = \frac{M R^{2}}{2} = \frac{(20 k g) {(20 \times 10^{- 2} m)}^{2}}{2} = 0.4 k g m^{2}$
Angular momentum of the cylinder about its axis is
$L = I ω = (0.4 k g m^{2}) 100 r a d s^{- 1}) = 40 J s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365632 If the earth is treated as a sphere of radius $R$ and mass $M$ . Its angular momentum about the axis of rotation with period $T$ is

1

\frac{4 π M R^{2}}{5 T}

2

\frac{π M R^{2}}{T}

3

\frac{M R^{2} π}{T}

4

\frac{2 π M R^{2}}{5 T}

Explanation:

Angular momentum of earth about its axis of rotation,
$L = I ω = \frac{2}{5} M R^{2} \times \frac{2 π}{T} = \frac{4 π M R^{2}}{5 T}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365628 Assertion :
Torque is equal to rate of change of angular momentum.
Reason :
Angular momentum depends on moment of inertia.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Torque $τ = I α$
But $α = \frac{d ω}{d t}$ and $L = I ω$
$\Rightarrow τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I α$
So correct option is (2).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365629 Calculate the angular momentum of a body whose rotational energy is 10 joule. If the angular momentum vector coincides with the axis of rotation and its moment of inertia about this axis is $8 \times 10^{- 7} k g m^{2}$

1

4 \times 10^{- 3} k g m^{2} / s

2

6 \times 10^{- 3} k g m^{2} / s

3

2 \times 10^{- 3} k g m^{2} / s

4 None of these

Explanation:

Angular momentum is given by
$L = I ω (K = \frac{1}{2} I ω^{2})$
$\Rightarrow L = I \sqrt{\frac{2 K}{I}} = \sqrt{2 K I}$
$L = \sqrt{2 K I} = \sqrt{2 \times 10 \times 8 \times 10^{- 7}} = 4 \times 10^{- 3} k g m^{2} s^{- 1}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365630 A ring of mass $10 k g$ and diameter $0.4 m$ is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be

1

0.4 k g \times m^{2} / s

2

4.4 k g \times m^{2} / s

3

88 k g \times m^{2} / s

4

44 k g \times m^{2} / s

Explanation:

$I = m r^{2} \Rightarrow 10 \times (0.2)^{2} = 0.4 k g m^{2}$
$ω = 2 π n = 2 π \times \frac{2100}{60} r a d / s$
$∴ L = I ω = \frac{0.4 \times 2 π \times 210}{6} = 88 k g - m^{2} / s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365631 A solid cylinder of mass $20 k g$ and radius $20 c m$ rotates about its axis with an angular speed of $100 r a d s^{- 1}$ . The angular momentum of the cylinder about its axis is

1

40 J s

2

400 J s

3

20 J s

4

200 J s

Explanation:

Here, $M = 20 k g$
$R = 20 c m = 20 \times 10^{- 2} m, ω = 100 r a d s^{- 1}$
Moment of inertia of the solid cylinder about its axis is
$I = \frac{M R^{2}}{2} = \frac{(20 k g) {(20 \times 10^{- 2} m)}^{2}}{2} = 0.4 k g m^{2}$
Angular momentum of the cylinder about its axis is
$L = I ω = (0.4 k g m^{2}) 100 r a d s^{- 1}) = 40 J s$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365632 If the earth is treated as a sphere of radius $R$ and mass $M$ . Its angular momentum about the axis of rotation with period $T$ is

1

\frac{4 π M R^{2}}{5 T}

2

\frac{π M R^{2}}{T}

3

\frac{M R^{2} π}{T}

4

\frac{2 π M R^{2}}{5 T}

Explanation:

Angular momentum of earth about its axis of rotation,
$L = I ω = \frac{2}{5} M R^{2} \times \frac{2 π}{T} = \frac{4 π M R^{2}}{5 T}$

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