QBManager

Rotational Motion

149965 A thin circular ring of mass $m$ and radius $R$ is rotating about its axis perpendicular to the plane of the ring with a constant angular velocity $ω$ . Two point particles each of mass $M$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $ω / 2$ . Then, the ratio $m / M$ is

1 1

2 2

3

\frac{1}{2}

4

\sqrt{2}

5

\frac{1}{\sqrt{2}}

Explanation:

B
$τ_{external} = 0$
$∴$ Angular momentum is conserved.
$L_{i} = L_{f}$
$I_{ring} ω = I_{ring} ω^{'} + 2 {MR}^{2} ω^{'}$
${mR}^{2} ω = {mR}^{2} \frac{ω}{2} + 2 {MR}^{2} \frac{ω}{2}$
$m - \frac{m}{2} = M$
$\frac{m}{2} = M$
$\frac{m}{M} = 2$

Kerala CEE -2018

Rotational Motion

149966 A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?

1 Rotational kinetic energy

2 Moment of inertia

3 Angular velocity

4 Angular momentum

Explanation:

D In free space,
$τ_{external} = 0$
$\frac{dL}{dt} = 0$
$L =$ constant
Taking equal masses sphere, if radius is increased. Then moment of inertia, rotational kinetic energy and angular velocity will change but due to the law of conservation of momentum, angular momentum will not change.
So,
$L_{1} = L_{2}$

NEET- 2018

Rotational Motion

149968 Angular momentum is

1 A scalar

2 A polar vector

3 A scalar as well as vector

4 An axial vector
SRM JEE-2018]

Explanation:

D Axial vectors are those vectors which represent rotational effect and act along the axis of rotation.
Eg.: Angular velocity, torque, angular momentum etc are Axial vectors.
Angular momentum- Property characterizing the rotary inertia of an object or system of objects in motion about an axis that may or not pass through the object or system.
Formula-
$L = mvr in vector form \vec{L} = \vec{r} \times \vec{p}$
$where -$
$L = angular momentum.$
$m = mass$
$v = velocity$
$r = radius.$
$SI Unit - kg . m^{2} s^{- 1}$
$Dimension - [{ML}^{2} T^{- 1}]$

Rotational Motion

149969 Two particles of masses $m_{1}$ and $m_{2}$ have equal kinetic energies. The ratio of their momentum is:

1

m_{1} : m_{2}

2

m_{2} : m_{1}

3

\sqrt{m_{1}} : \sqrt{m_{2}}

4

m_{1}^{2} : m_{2}^{2}

Explanation:

C We know,
$Kinetic energies (K.E.) = \frac{1}{2} {mv}^{2}$
$K.E. = \frac{p^{2}}{2 m}$
Where $m =$ mass of the particle
For particle (i),
$K.E.)_{1} = \frac{p_{1}^{2}}{2 m_{1}}$
and for particle (ii),
Given,
$K.E.)_{2} = \frac{p_{2}^{2}}{2 m_{2}}$
$K.E.)_{1} = (K.E.)_{2}$
$\frac{p_{1}^{2}}{2 m_{1}} = \frac{p_{2}^{2}}{2 m_{2}}$
Now,
$p_{1}, p_{2}$ are momentum of particles and $m_{1}$ and $m_{2}$ are mass of particles respectively.
$∴ \vec{p} = 2 q l$
$l e f t (\frac{p_{1}}{p_{2}} r i g h t)^{2} = l e f t (\frac{m_{1}}{m_{2}} r i g h t$
$\frac{p_{1}}{p_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$
$p_{1} : p_{2} = \sqrt{m_{1}} : \sqrt{m_{2}}$

Karnataka CET-2018

Rotational Motion

149971 Assertion: Torque on a body can be zero even if there is a net force on it.
Reason: Torque and force on a body are always perpendicular.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

B Torque on a body $τ = \vec{r} \times \vec{F}$
If $\vec{F} = 0$ then $\vec{τ} = 0$
Hence, assertion is correct
And torque is always perpendicular to radius from vector but it is not reason behind torque to be zero.

AIIMS-26.05.2018(M)

Rotational Motion

149965 A thin circular ring of mass $m$ and radius $R$ is rotating about its axis perpendicular to the plane of the ring with a constant angular velocity $ω$ . Two point particles each of mass $M$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $ω / 2$ . Then, the ratio $m / M$ is

1 1

2 2

3

\frac{1}{2}

4

\sqrt{2}

5

\frac{1}{\sqrt{2}}

Explanation:

B
$τ_{external} = 0$
$∴$ Angular momentum is conserved.
$L_{i} = L_{f}$
$I_{ring} ω = I_{ring} ω^{'} + 2 {MR}^{2} ω^{'}$
${mR}^{2} ω = {mR}^{2} \frac{ω}{2} + 2 {MR}^{2} \frac{ω}{2}$
$m - \frac{m}{2} = M$
$\frac{m}{2} = M$
$\frac{m}{M} = 2$

Kerala CEE -2018

Rotational Motion

149966 A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?

1 Rotational kinetic energy

2 Moment of inertia

3 Angular velocity

4 Angular momentum

Explanation:

D In free space,
$τ_{external} = 0$
$\frac{dL}{dt} = 0$
$L =$ constant
Taking equal masses sphere, if radius is increased. Then moment of inertia, rotational kinetic energy and angular velocity will change but due to the law of conservation of momentum, angular momentum will not change.
So,
$L_{1} = L_{2}$

NEET- 2018

Rotational Motion

149968 Angular momentum is

1 A scalar

2 A polar vector

3 A scalar as well as vector

4 An axial vector
SRM JEE-2018]

Explanation:

D Axial vectors are those vectors which represent rotational effect and act along the axis of rotation.
Eg.: Angular velocity, torque, angular momentum etc are Axial vectors.
Angular momentum- Property characterizing the rotary inertia of an object or system of objects in motion about an axis that may or not pass through the object or system.
Formula-
$L = mvr in vector form \vec{L} = \vec{r} \times \vec{p}$
$where -$
$L = angular momentum.$
$m = mass$
$v = velocity$
$r = radius.$
$SI Unit - kg . m^{2} s^{- 1}$
$Dimension - [{ML}^{2} T^{- 1}]$

Rotational Motion

149969 Two particles of masses $m_{1}$ and $m_{2}$ have equal kinetic energies. The ratio of their momentum is:

1

m_{1} : m_{2}

2

m_{2} : m_{1}

3

\sqrt{m_{1}} : \sqrt{m_{2}}

4

m_{1}^{2} : m_{2}^{2}

Explanation:

C We know,
$Kinetic energies (K.E.) = \frac{1}{2} {mv}^{2}$
$K.E. = \frac{p^{2}}{2 m}$
Where $m =$ mass of the particle
For particle (i),
$K.E.)_{1} = \frac{p_{1}^{2}}{2 m_{1}}$
and for particle (ii),
Given,
$K.E.)_{2} = \frac{p_{2}^{2}}{2 m_{2}}$
$K.E.)_{1} = (K.E.)_{2}$
$\frac{p_{1}^{2}}{2 m_{1}} = \frac{p_{2}^{2}}{2 m_{2}}$
Now,
$p_{1}, p_{2}$ are momentum of particles and $m_{1}$ and $m_{2}$ are mass of particles respectively.
$∴ \vec{p} = 2 q l$
$l e f t (\frac{p_{1}}{p_{2}} r i g h t)^{2} = l e f t (\frac{m_{1}}{m_{2}} r i g h t$
$\frac{p_{1}}{p_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$
$p_{1} : p_{2} = \sqrt{m_{1}} : \sqrt{m_{2}}$

Karnataka CET-2018

Rotational Motion

149971 Assertion: Torque on a body can be zero even if there is a net force on it.
Reason: Torque and force on a body are always perpendicular.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

B Torque on a body $τ = \vec{r} \times \vec{F}$
If $\vec{F} = 0$ then $\vec{τ} = 0$
Hence, assertion is correct
And torque is always perpendicular to radius from vector but it is not reason behind torque to be zero.

AIIMS-26.05.2018(M)

Rotational Motion

149965 A thin circular ring of mass $m$ and radius $R$ is rotating about its axis perpendicular to the plane of the ring with a constant angular velocity $ω$ . Two point particles each of mass $M$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $ω / 2$ . Then, the ratio $m / M$ is

1 1

2 2

3

\frac{1}{2}

4

\sqrt{2}

5

\frac{1}{\sqrt{2}}

Explanation:

B
$τ_{external} = 0$
$∴$ Angular momentum is conserved.
$L_{i} = L_{f}$
$I_{ring} ω = I_{ring} ω^{'} + 2 {MR}^{2} ω^{'}$
${mR}^{2} ω = {mR}^{2} \frac{ω}{2} + 2 {MR}^{2} \frac{ω}{2}$
$m - \frac{m}{2} = M$
$\frac{m}{2} = M$
$\frac{m}{M} = 2$

Kerala CEE -2018

Rotational Motion

149966 A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?

1 Rotational kinetic energy

2 Moment of inertia

3 Angular velocity

4 Angular momentum

Explanation:

D In free space,
$τ_{external} = 0$
$\frac{dL}{dt} = 0$
$L =$ constant
Taking equal masses sphere, if radius is increased. Then moment of inertia, rotational kinetic energy and angular velocity will change but due to the law of conservation of momentum, angular momentum will not change.
So,
$L_{1} = L_{2}$

NEET- 2018

Rotational Motion

149968 Angular momentum is

1 A scalar

2 A polar vector

3 A scalar as well as vector

4 An axial vector
SRM JEE-2018]

Explanation:

D Axial vectors are those vectors which represent rotational effect and act along the axis of rotation.
Eg.: Angular velocity, torque, angular momentum etc are Axial vectors.
Angular momentum- Property characterizing the rotary inertia of an object or system of objects in motion about an axis that may or not pass through the object or system.
Formula-
$L = mvr in vector form \vec{L} = \vec{r} \times \vec{p}$
$where -$
$L = angular momentum.$
$m = mass$
$v = velocity$
$r = radius.$
$SI Unit - kg . m^{2} s^{- 1}$
$Dimension - [{ML}^{2} T^{- 1}]$

Rotational Motion

149969 Two particles of masses $m_{1}$ and $m_{2}$ have equal kinetic energies. The ratio of their momentum is:

1

m_{1} : m_{2}

2

m_{2} : m_{1}

3

\sqrt{m_{1}} : \sqrt{m_{2}}

4

m_{1}^{2} : m_{2}^{2}

Explanation:

C We know,
$Kinetic energies (K.E.) = \frac{1}{2} {mv}^{2}$
$K.E. = \frac{p^{2}}{2 m}$
Where $m =$ mass of the particle
For particle (i),
$K.E.)_{1} = \frac{p_{1}^{2}}{2 m_{1}}$
and for particle (ii),
Given,
$K.E.)_{2} = \frac{p_{2}^{2}}{2 m_{2}}$
$K.E.)_{1} = (K.E.)_{2}$
$\frac{p_{1}^{2}}{2 m_{1}} = \frac{p_{2}^{2}}{2 m_{2}}$
Now,
$p_{1}, p_{2}$ are momentum of particles and $m_{1}$ and $m_{2}$ are mass of particles respectively.
$∴ \vec{p} = 2 q l$
$l e f t (\frac{p_{1}}{p_{2}} r i g h t)^{2} = l e f t (\frac{m_{1}}{m_{2}} r i g h t$
$\frac{p_{1}}{p_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$
$p_{1} : p_{2} = \sqrt{m_{1}} : \sqrt{m_{2}}$

Karnataka CET-2018

Rotational Motion

149971 Assertion: Torque on a body can be zero even if there is a net force on it.
Reason: Torque and force on a body are always perpendicular.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

B Torque on a body $τ = \vec{r} \times \vec{F}$
If $\vec{F} = 0$ then $\vec{τ} = 0$
Hence, assertion is correct
And torque is always perpendicular to radius from vector but it is not reason behind torque to be zero.

AIIMS-26.05.2018(M)

Rotational Motion

149965 A thin circular ring of mass $m$ and radius $R$ is rotating about its axis perpendicular to the plane of the ring with a constant angular velocity $ω$ . Two point particles each of mass $M$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $ω / 2$ . Then, the ratio $m / M$ is

1 1

2 2

3

\frac{1}{2}

4

\sqrt{2}

5

\frac{1}{\sqrt{2}}

Explanation:

B
$τ_{external} = 0$
$∴$ Angular momentum is conserved.
$L_{i} = L_{f}$
$I_{ring} ω = I_{ring} ω^{'} + 2 {MR}^{2} ω^{'}$
${mR}^{2} ω = {mR}^{2} \frac{ω}{2} + 2 {MR}^{2} \frac{ω}{2}$
$m - \frac{m}{2} = M$
$\frac{m}{2} = M$
$\frac{m}{M} = 2$

Kerala CEE -2018

Rotational Motion

149966 A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?

1 Rotational kinetic energy

2 Moment of inertia

3 Angular velocity

4 Angular momentum

Explanation:

D In free space,
$τ_{external} = 0$
$\frac{dL}{dt} = 0$
$L =$ constant
Taking equal masses sphere, if radius is increased. Then moment of inertia, rotational kinetic energy and angular velocity will change but due to the law of conservation of momentum, angular momentum will not change.
So,
$L_{1} = L_{2}$

NEET- 2018

Rotational Motion

149968 Angular momentum is

1 A scalar

2 A polar vector

3 A scalar as well as vector

4 An axial vector
SRM JEE-2018]

Explanation:

D Axial vectors are those vectors which represent rotational effect and act along the axis of rotation.
Eg.: Angular velocity, torque, angular momentum etc are Axial vectors.
Angular momentum- Property characterizing the rotary inertia of an object or system of objects in motion about an axis that may or not pass through the object or system.
Formula-
$L = mvr in vector form \vec{L} = \vec{r} \times \vec{p}$
$where -$
$L = angular momentum.$
$m = mass$
$v = velocity$
$r = radius.$
$SI Unit - kg . m^{2} s^{- 1}$
$Dimension - [{ML}^{2} T^{- 1}]$

Rotational Motion

149969 Two particles of masses $m_{1}$ and $m_{2}$ have equal kinetic energies. The ratio of their momentum is:

1

m_{1} : m_{2}

2

m_{2} : m_{1}

3

\sqrt{m_{1}} : \sqrt{m_{2}}

4

m_{1}^{2} : m_{2}^{2}

Explanation:

C We know,
$Kinetic energies (K.E.) = \frac{1}{2} {mv}^{2}$
$K.E. = \frac{p^{2}}{2 m}$
Where $m =$ mass of the particle
For particle (i),
$K.E.)_{1} = \frac{p_{1}^{2}}{2 m_{1}}$
and for particle (ii),
Given,
$K.E.)_{2} = \frac{p_{2}^{2}}{2 m_{2}}$
$K.E.)_{1} = (K.E.)_{2}$
$\frac{p_{1}^{2}}{2 m_{1}} = \frac{p_{2}^{2}}{2 m_{2}}$
Now,
$p_{1}, p_{2}$ are momentum of particles and $m_{1}$ and $m_{2}$ are mass of particles respectively.
$∴ \vec{p} = 2 q l$
$l e f t (\frac{p_{1}}{p_{2}} r i g h t)^{2} = l e f t (\frac{m_{1}}{m_{2}} r i g h t$
$\frac{p_{1}}{p_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$
$p_{1} : p_{2} = \sqrt{m_{1}} : \sqrt{m_{2}}$

Karnataka CET-2018

Rotational Motion

149971 Assertion: Torque on a body can be zero even if there is a net force on it.
Reason: Torque and force on a body are always perpendicular.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

B Torque on a body $τ = \vec{r} \times \vec{F}$
If $\vec{F} = 0$ then $\vec{τ} = 0$
Hence, assertion is correct
And torque is always perpendicular to radius from vector but it is not reason behind torque to be zero.

AIIMS-26.05.2018(M)

Rotational Motion

149965 A thin circular ring of mass $m$ and radius $R$ is rotating about its axis perpendicular to the plane of the ring with a constant angular velocity $ω$ . Two point particles each of mass $M$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $ω / 2$ . Then, the ratio $m / M$ is

1 1

2 2

3

\frac{1}{2}

4

\sqrt{2}

5

\frac{1}{\sqrt{2}}

Explanation:

B
$τ_{external} = 0$
$∴$ Angular momentum is conserved.
$L_{i} = L_{f}$
$I_{ring} ω = I_{ring} ω^{'} + 2 {MR}^{2} ω^{'}$
${mR}^{2} ω = {mR}^{2} \frac{ω}{2} + 2 {MR}^{2} \frac{ω}{2}$
$m - \frac{m}{2} = M$
$\frac{m}{2} = M$
$\frac{m}{M} = 2$

Kerala CEE -2018

Rotational Motion

149966 A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?

1 Rotational kinetic energy

2 Moment of inertia

3 Angular velocity

4 Angular momentum

Explanation:

D In free space,
$τ_{external} = 0$
$\frac{dL}{dt} = 0$
$L =$ constant
Taking equal masses sphere, if radius is increased. Then moment of inertia, rotational kinetic energy and angular velocity will change but due to the law of conservation of momentum, angular momentum will not change.
So,
$L_{1} = L_{2}$

NEET- 2018

Rotational Motion

149968 Angular momentum is

1 A scalar

2 A polar vector

3 A scalar as well as vector

4 An axial vector
SRM JEE-2018]

Explanation:

D Axial vectors are those vectors which represent rotational effect and act along the axis of rotation.
Eg.: Angular velocity, torque, angular momentum etc are Axial vectors.
Angular momentum- Property characterizing the rotary inertia of an object or system of objects in motion about an axis that may or not pass through the object or system.
Formula-
$L = mvr in vector form \vec{L} = \vec{r} \times \vec{p}$
$where -$
$L = angular momentum.$
$m = mass$
$v = velocity$
$r = radius.$
$SI Unit - kg . m^{2} s^{- 1}$
$Dimension - [{ML}^{2} T^{- 1}]$

Rotational Motion

149969 Two particles of masses $m_{1}$ and $m_{2}$ have equal kinetic energies. The ratio of their momentum is:

1

m_{1} : m_{2}

2

m_{2} : m_{1}

3

\sqrt{m_{1}} : \sqrt{m_{2}}

4

m_{1}^{2} : m_{2}^{2}

Explanation:

C We know,
$Kinetic energies (K.E.) = \frac{1}{2} {mv}^{2}$
$K.E. = \frac{p^{2}}{2 m}$
Where $m =$ mass of the particle
For particle (i),
$K.E.)_{1} = \frac{p_{1}^{2}}{2 m_{1}}$
and for particle (ii),
Given,
$K.E.)_{2} = \frac{p_{2}^{2}}{2 m_{2}}$
$K.E.)_{1} = (K.E.)_{2}$
$\frac{p_{1}^{2}}{2 m_{1}} = \frac{p_{2}^{2}}{2 m_{2}}$
Now,
$p_{1}, p_{2}$ are momentum of particles and $m_{1}$ and $m_{2}$ are mass of particles respectively.
$∴ \vec{p} = 2 q l$
$l e f t (\frac{p_{1}}{p_{2}} r i g h t)^{2} = l e f t (\frac{m_{1}}{m_{2}} r i g h t$
$\frac{p_{1}}{p_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$
$p_{1} : p_{2} = \sqrt{m_{1}} : \sqrt{m_{2}}$

Karnataka CET-2018

Rotational Motion

149971 Assertion: Torque on a body can be zero even if there is a net force on it.
Reason: Torque and force on a body are always perpendicular.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

B Torque on a body $τ = \vec{r} \times \vec{F}$
If $\vec{F} = 0$ then $\vec{τ} = 0$
Hence, assertion is correct
And torque is always perpendicular to radius from vector but it is not reason behind torque to be zero.

AIIMS-26.05.2018(M)