149984
If a body of moment of inertia \(2 \mathrm{~kg} \mathrm{~m}^{2}\) revolves about its own axis making 2 rotations per second, then its angular momentum (in Js) is
1 \(2 \pi\)
2 \(4 \pi\)
3 \(6 \pi\)
4 \(8 \pi\)
5 \(10 \pi\)
Explanation:
D Given: Moment of inertia (I) \(=2 \mathrm{~kg}-\mathrm{m}^{2}\) Axis rotation \(\quad(\mathrm{f})=2 \mathrm{rps}\) We know that, Angular momentum, \(\mathrm{L}=\mathrm{I} \omega\) \(\mathrm{L}=2 \times 2 \pi \mathrm{f}\) \(\mathrm{L}=2 \times 2 \pi \times 2\) \(\mathrm{~L}=8 \pi \mathrm{Js}\)
Kerala CEE - 2016
Rotational Motion
149985
Angular momentum of the Earth revolving around the Sun in a circular orbit of radius \(R\) is proportional to
1 \(\sqrt{\mathrm{R}}\)
2 \(\mathrm{R}\)
3 \(\mathrm{R}^{2}\)
4 \(R^{1 / 3}\)
5 \(R^{3 / 2}\)
Explanation:
A We know that, Angular momentum of Earth, \(\mathrm{L}=\mathrm{mvR}\) Now, \(\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\) \(\mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{R}}\) \(\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\) So, \(L=m \sqrt{\frac{G M}{R}} \times R=m \sqrt{\text { GMR }}\) \(\therefore \quad \mathrm{L} \propto \sqrt{\mathrm{R}}\)
Kerala CEE - 2015
Rotational Motion
149986
The angular momentum of a particle describing uniform circular motion is \(L\). If its kinetic energy is halved and angular velocity doubled, its new angular momentum is
1 \(4 \mathrm{~L}\)
2 \(\frac{L}{4}\)
3 \(\frac{\mathrm{L}}{2}\)
4 \(2 \mathrm{~L}\)
5 \(\frac{L}{8}\)
Explanation:
B Kinetic energy \((K)=\frac{1}{2} I \omega^{2}\) \(\mathrm{K}=\frac{1}{2}(\mathrm{I} \omega) \omega\) Angular momentum \(\mathrm{L})=\mathrm{I} \omega\) \(\therefore \quad \mathrm{K} =\frac{1}{2} \mathrm{~L} \omega\) \(\mathrm{L} =\frac{2 \mathrm{~K}}{\omega}\) Second condition \(\omega^{\prime}=2 \omega, K^{\prime}=\frac{K}{2}\) New angular momentum \(\mathrm{L}^{\prime} =\frac{2 \mathrm{~K}^{\prime}}{\omega^{\prime}}\) \(\mathrm{L}^{\prime} =\frac{2\left(\frac{\mathrm{K}}{2}\right)}{(2 \omega)}=\frac{\mathrm{K}}{2 \omega}\) \(\mathrm{L}^{\prime} =\frac{1}{4}\left(\frac{2 \mathrm{~K}}{\omega}\right)\) \(\mathrm{L}^{\prime} =\frac{\mathrm{L}}{4}\) By comparing from equation (i) and (ii) \(\mathrm{L}^{\prime}=\frac{\mathrm{L}}{4}\)
UPSEE-2005
Rotational Motion
149987
If the earth were to contract such that its radius becomes one quarter, without change in its mass, the duration of one full day would be
1 \(3 \mathrm{~h}\)
2 \(1.5 \mathrm{~h}\)
3 \(6 \mathrm{~h}\)
4 \(4 \mathrm{~h}\)
5 \(2 \mathrm{~h}\)
Explanation:
B \(M_{1}=M_{2}=M\) [ Mass of Earth] \(\mathrm{R}_{2}=\frac{1}{4} \mathrm{R}_{1}\) \(\mathrm{~T}_{1}=24 \mathrm{hrs}\) By the law of conservation of angular momentum \(\mathrm{I} \omega=\) Constant \(\mathrm{I}_{1} \omega_{1}=\mathrm{I}_{2} \omega_{2}\) \(\frac{2}{5} \mathrm{MR}_{1}^{2} \omega_{1}=\frac{2}{5} \mathrm{MR}_{2}^{2} \omega_{2}\) \(\frac{2}{5} \mathrm{M}\left(\mathrm{R}_{1}^{2} \omega_{1}\right)=\frac{2}{5} \mathrm{M}\left(\mathrm{R}_{2}^{2} \omega_{2}\right)\) \(\left(\because \omega=\frac{2 \pi}{\mathrm{T}}\right)\) \(\mathrm{R}_{1}^{2} \times \frac{2 \pi}{24}=\frac{\mathrm{R}_{1}^{2}}{16} \times \frac{2 \pi}{\mathrm{T}_{2}}\) \(2 \pi \mathrm{R}_{1}^{2}\left(\frac{1}{24}\right))=\frac{2 \pi \mathrm{R}_{1}^{2}}{16 \times \mathrm{T}_{2}}\) \(\mathrm{T}_{2}=\frac{24}{16}\) \(\mathrm{T}_{2}=\frac{3}{2}\) \(\mathrm{T}_{2}=1.5 \mathrm{~h}\)
Kerala CEE 2007
Rotational Motion
149989
A disc of mass \(2 \mathrm{~kg}\) and radius \(0.2 \mathrm{~m}\) is rotating with angular velocity 30 rads \(^{-1}\) what is angular velocity, if a mass of \(0.25 \mathrm{~kg}\) is put on periphery of the disc?
1 \(24 \mathrm{rad} \mathrm{s}^{-1}\)
2 \(36 \mathrm{rad} \mathrm{s}^{-1}\)
3 \(15 \mathrm{rad} \mathrm{s}^{-1}\)
4 \(26 \mathrm{rad} \mathrm{s}^{-1}\)
Explanation:
A Given, \(\mathrm{m}_{1}=2 \mathrm{~kg}\) \(\mathrm{~m}_{2}=0.25 \mathrm{~kg}\) \(\mathrm{r}=0.2 \mathrm{~m}\) \(\omega_{1}=30 \mathrm{rad} / \mathrm{sec}\) According to law of conservation of angular momentum \(\mathrm{L}=\mathrm{I} \omega=\) Constant \(\therefore \quad \mathrm{I}_{1} \omega_{1}=\mathrm{I}_{2} \omega_{2}\) \(\frac{1}{2} \mathrm{~m}_{1} \mathrm{r}^{2} \omega_{1}=\frac{1}{2}\left(\mathrm{~m}_{1}+2 \mathrm{~m}_{2}\right) \mathrm{r}^{2} \times \omega_{2}\) \(\frac{1}{2} \times 2 \times(0.2)^{2} \times 30=\frac{1}{2} \times(2+2 \times 0.25) \times(0.2)^{2} \times \omega_{2}\) \(\therefore \quad 1.2=0.05 \omega_{2}\) \(\therefore \quad \omega_{2}=24 \mathrm{rad} / \mathrm{sec}\)
149984
If a body of moment of inertia \(2 \mathrm{~kg} \mathrm{~m}^{2}\) revolves about its own axis making 2 rotations per second, then its angular momentum (in Js) is
1 \(2 \pi\)
2 \(4 \pi\)
3 \(6 \pi\)
4 \(8 \pi\)
5 \(10 \pi\)
Explanation:
D Given: Moment of inertia (I) \(=2 \mathrm{~kg}-\mathrm{m}^{2}\) Axis rotation \(\quad(\mathrm{f})=2 \mathrm{rps}\) We know that, Angular momentum, \(\mathrm{L}=\mathrm{I} \omega\) \(\mathrm{L}=2 \times 2 \pi \mathrm{f}\) \(\mathrm{L}=2 \times 2 \pi \times 2\) \(\mathrm{~L}=8 \pi \mathrm{Js}\)
Kerala CEE - 2016
Rotational Motion
149985
Angular momentum of the Earth revolving around the Sun in a circular orbit of radius \(R\) is proportional to
1 \(\sqrt{\mathrm{R}}\)
2 \(\mathrm{R}\)
3 \(\mathrm{R}^{2}\)
4 \(R^{1 / 3}\)
5 \(R^{3 / 2}\)
Explanation:
A We know that, Angular momentum of Earth, \(\mathrm{L}=\mathrm{mvR}\) Now, \(\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\) \(\mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{R}}\) \(\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\) So, \(L=m \sqrt{\frac{G M}{R}} \times R=m \sqrt{\text { GMR }}\) \(\therefore \quad \mathrm{L} \propto \sqrt{\mathrm{R}}\)
Kerala CEE - 2015
Rotational Motion
149986
The angular momentum of a particle describing uniform circular motion is \(L\). If its kinetic energy is halved and angular velocity doubled, its new angular momentum is
1 \(4 \mathrm{~L}\)
2 \(\frac{L}{4}\)
3 \(\frac{\mathrm{L}}{2}\)
4 \(2 \mathrm{~L}\)
5 \(\frac{L}{8}\)
Explanation:
B Kinetic energy \((K)=\frac{1}{2} I \omega^{2}\) \(\mathrm{K}=\frac{1}{2}(\mathrm{I} \omega) \omega\) Angular momentum \(\mathrm{L})=\mathrm{I} \omega\) \(\therefore \quad \mathrm{K} =\frac{1}{2} \mathrm{~L} \omega\) \(\mathrm{L} =\frac{2 \mathrm{~K}}{\omega}\) Second condition \(\omega^{\prime}=2 \omega, K^{\prime}=\frac{K}{2}\) New angular momentum \(\mathrm{L}^{\prime} =\frac{2 \mathrm{~K}^{\prime}}{\omega^{\prime}}\) \(\mathrm{L}^{\prime} =\frac{2\left(\frac{\mathrm{K}}{2}\right)}{(2 \omega)}=\frac{\mathrm{K}}{2 \omega}\) \(\mathrm{L}^{\prime} =\frac{1}{4}\left(\frac{2 \mathrm{~K}}{\omega}\right)\) \(\mathrm{L}^{\prime} =\frac{\mathrm{L}}{4}\) By comparing from equation (i) and (ii) \(\mathrm{L}^{\prime}=\frac{\mathrm{L}}{4}\)
UPSEE-2005
Rotational Motion
149987
If the earth were to contract such that its radius becomes one quarter, without change in its mass, the duration of one full day would be
1 \(3 \mathrm{~h}\)
2 \(1.5 \mathrm{~h}\)
3 \(6 \mathrm{~h}\)
4 \(4 \mathrm{~h}\)
5 \(2 \mathrm{~h}\)
Explanation:
B \(M_{1}=M_{2}=M\) [ Mass of Earth] \(\mathrm{R}_{2}=\frac{1}{4} \mathrm{R}_{1}\) \(\mathrm{~T}_{1}=24 \mathrm{hrs}\) By the law of conservation of angular momentum \(\mathrm{I} \omega=\) Constant \(\mathrm{I}_{1} \omega_{1}=\mathrm{I}_{2} \omega_{2}\) \(\frac{2}{5} \mathrm{MR}_{1}^{2} \omega_{1}=\frac{2}{5} \mathrm{MR}_{2}^{2} \omega_{2}\) \(\frac{2}{5} \mathrm{M}\left(\mathrm{R}_{1}^{2} \omega_{1}\right)=\frac{2}{5} \mathrm{M}\left(\mathrm{R}_{2}^{2} \omega_{2}\right)\) \(\left(\because \omega=\frac{2 \pi}{\mathrm{T}}\right)\) \(\mathrm{R}_{1}^{2} \times \frac{2 \pi}{24}=\frac{\mathrm{R}_{1}^{2}}{16} \times \frac{2 \pi}{\mathrm{T}_{2}}\) \(2 \pi \mathrm{R}_{1}^{2}\left(\frac{1}{24}\right))=\frac{2 \pi \mathrm{R}_{1}^{2}}{16 \times \mathrm{T}_{2}}\) \(\mathrm{T}_{2}=\frac{24}{16}\) \(\mathrm{T}_{2}=\frac{3}{2}\) \(\mathrm{T}_{2}=1.5 \mathrm{~h}\)
Kerala CEE 2007
Rotational Motion
149989
A disc of mass \(2 \mathrm{~kg}\) and radius \(0.2 \mathrm{~m}\) is rotating with angular velocity 30 rads \(^{-1}\) what is angular velocity, if a mass of \(0.25 \mathrm{~kg}\) is put on periphery of the disc?
1 \(24 \mathrm{rad} \mathrm{s}^{-1}\)
2 \(36 \mathrm{rad} \mathrm{s}^{-1}\)
3 \(15 \mathrm{rad} \mathrm{s}^{-1}\)
4 \(26 \mathrm{rad} \mathrm{s}^{-1}\)
Explanation:
A Given, \(\mathrm{m}_{1}=2 \mathrm{~kg}\) \(\mathrm{~m}_{2}=0.25 \mathrm{~kg}\) \(\mathrm{r}=0.2 \mathrm{~m}\) \(\omega_{1}=30 \mathrm{rad} / \mathrm{sec}\) According to law of conservation of angular momentum \(\mathrm{L}=\mathrm{I} \omega=\) Constant \(\therefore \quad \mathrm{I}_{1} \omega_{1}=\mathrm{I}_{2} \omega_{2}\) \(\frac{1}{2} \mathrm{~m}_{1} \mathrm{r}^{2} \omega_{1}=\frac{1}{2}\left(\mathrm{~m}_{1}+2 \mathrm{~m}_{2}\right) \mathrm{r}^{2} \times \omega_{2}\) \(\frac{1}{2} \times 2 \times(0.2)^{2} \times 30=\frac{1}{2} \times(2+2 \times 0.25) \times(0.2)^{2} \times \omega_{2}\) \(\therefore \quad 1.2=0.05 \omega_{2}\) \(\therefore \quad \omega_{2}=24 \mathrm{rad} / \mathrm{sec}\)
149984
If a body of moment of inertia \(2 \mathrm{~kg} \mathrm{~m}^{2}\) revolves about its own axis making 2 rotations per second, then its angular momentum (in Js) is
1 \(2 \pi\)
2 \(4 \pi\)
3 \(6 \pi\)
4 \(8 \pi\)
5 \(10 \pi\)
Explanation:
D Given: Moment of inertia (I) \(=2 \mathrm{~kg}-\mathrm{m}^{2}\) Axis rotation \(\quad(\mathrm{f})=2 \mathrm{rps}\) We know that, Angular momentum, \(\mathrm{L}=\mathrm{I} \omega\) \(\mathrm{L}=2 \times 2 \pi \mathrm{f}\) \(\mathrm{L}=2 \times 2 \pi \times 2\) \(\mathrm{~L}=8 \pi \mathrm{Js}\)
Kerala CEE - 2016
Rotational Motion
149985
Angular momentum of the Earth revolving around the Sun in a circular orbit of radius \(R\) is proportional to
1 \(\sqrt{\mathrm{R}}\)
2 \(\mathrm{R}\)
3 \(\mathrm{R}^{2}\)
4 \(R^{1 / 3}\)
5 \(R^{3 / 2}\)
Explanation:
A We know that, Angular momentum of Earth, \(\mathrm{L}=\mathrm{mvR}\) Now, \(\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\) \(\mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{R}}\) \(\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\) So, \(L=m \sqrt{\frac{G M}{R}} \times R=m \sqrt{\text { GMR }}\) \(\therefore \quad \mathrm{L} \propto \sqrt{\mathrm{R}}\)
Kerala CEE - 2015
Rotational Motion
149986
The angular momentum of a particle describing uniform circular motion is \(L\). If its kinetic energy is halved and angular velocity doubled, its new angular momentum is
1 \(4 \mathrm{~L}\)
2 \(\frac{L}{4}\)
3 \(\frac{\mathrm{L}}{2}\)
4 \(2 \mathrm{~L}\)
5 \(\frac{L}{8}\)
Explanation:
B Kinetic energy \((K)=\frac{1}{2} I \omega^{2}\) \(\mathrm{K}=\frac{1}{2}(\mathrm{I} \omega) \omega\) Angular momentum \(\mathrm{L})=\mathrm{I} \omega\) \(\therefore \quad \mathrm{K} =\frac{1}{2} \mathrm{~L} \omega\) \(\mathrm{L} =\frac{2 \mathrm{~K}}{\omega}\) Second condition \(\omega^{\prime}=2 \omega, K^{\prime}=\frac{K}{2}\) New angular momentum \(\mathrm{L}^{\prime} =\frac{2 \mathrm{~K}^{\prime}}{\omega^{\prime}}\) \(\mathrm{L}^{\prime} =\frac{2\left(\frac{\mathrm{K}}{2}\right)}{(2 \omega)}=\frac{\mathrm{K}}{2 \omega}\) \(\mathrm{L}^{\prime} =\frac{1}{4}\left(\frac{2 \mathrm{~K}}{\omega}\right)\) \(\mathrm{L}^{\prime} =\frac{\mathrm{L}}{4}\) By comparing from equation (i) and (ii) \(\mathrm{L}^{\prime}=\frac{\mathrm{L}}{4}\)
UPSEE-2005
Rotational Motion
149987
If the earth were to contract such that its radius becomes one quarter, without change in its mass, the duration of one full day would be
1 \(3 \mathrm{~h}\)
2 \(1.5 \mathrm{~h}\)
3 \(6 \mathrm{~h}\)
4 \(4 \mathrm{~h}\)
5 \(2 \mathrm{~h}\)
Explanation:
B \(M_{1}=M_{2}=M\) [ Mass of Earth] \(\mathrm{R}_{2}=\frac{1}{4} \mathrm{R}_{1}\) \(\mathrm{~T}_{1}=24 \mathrm{hrs}\) By the law of conservation of angular momentum \(\mathrm{I} \omega=\) Constant \(\mathrm{I}_{1} \omega_{1}=\mathrm{I}_{2} \omega_{2}\) \(\frac{2}{5} \mathrm{MR}_{1}^{2} \omega_{1}=\frac{2}{5} \mathrm{MR}_{2}^{2} \omega_{2}\) \(\frac{2}{5} \mathrm{M}\left(\mathrm{R}_{1}^{2} \omega_{1}\right)=\frac{2}{5} \mathrm{M}\left(\mathrm{R}_{2}^{2} \omega_{2}\right)\) \(\left(\because \omega=\frac{2 \pi}{\mathrm{T}}\right)\) \(\mathrm{R}_{1}^{2} \times \frac{2 \pi}{24}=\frac{\mathrm{R}_{1}^{2}}{16} \times \frac{2 \pi}{\mathrm{T}_{2}}\) \(2 \pi \mathrm{R}_{1}^{2}\left(\frac{1}{24}\right))=\frac{2 \pi \mathrm{R}_{1}^{2}}{16 \times \mathrm{T}_{2}}\) \(\mathrm{T}_{2}=\frac{24}{16}\) \(\mathrm{T}_{2}=\frac{3}{2}\) \(\mathrm{T}_{2}=1.5 \mathrm{~h}\)
Kerala CEE 2007
Rotational Motion
149989
A disc of mass \(2 \mathrm{~kg}\) and radius \(0.2 \mathrm{~m}\) is rotating with angular velocity 30 rads \(^{-1}\) what is angular velocity, if a mass of \(0.25 \mathrm{~kg}\) is put on periphery of the disc?
1 \(24 \mathrm{rad} \mathrm{s}^{-1}\)
2 \(36 \mathrm{rad} \mathrm{s}^{-1}\)
3 \(15 \mathrm{rad} \mathrm{s}^{-1}\)
4 \(26 \mathrm{rad} \mathrm{s}^{-1}\)
Explanation:
A Given, \(\mathrm{m}_{1}=2 \mathrm{~kg}\) \(\mathrm{~m}_{2}=0.25 \mathrm{~kg}\) \(\mathrm{r}=0.2 \mathrm{~m}\) \(\omega_{1}=30 \mathrm{rad} / \mathrm{sec}\) According to law of conservation of angular momentum \(\mathrm{L}=\mathrm{I} \omega=\) Constant \(\therefore \quad \mathrm{I}_{1} \omega_{1}=\mathrm{I}_{2} \omega_{2}\) \(\frac{1}{2} \mathrm{~m}_{1} \mathrm{r}^{2} \omega_{1}=\frac{1}{2}\left(\mathrm{~m}_{1}+2 \mathrm{~m}_{2}\right) \mathrm{r}^{2} \times \omega_{2}\) \(\frac{1}{2} \times 2 \times(0.2)^{2} \times 30=\frac{1}{2} \times(2+2 \times 0.25) \times(0.2)^{2} \times \omega_{2}\) \(\therefore \quad 1.2=0.05 \omega_{2}\) \(\therefore \quad \omega_{2}=24 \mathrm{rad} / \mathrm{sec}\)
149984
If a body of moment of inertia \(2 \mathrm{~kg} \mathrm{~m}^{2}\) revolves about its own axis making 2 rotations per second, then its angular momentum (in Js) is
1 \(2 \pi\)
2 \(4 \pi\)
3 \(6 \pi\)
4 \(8 \pi\)
5 \(10 \pi\)
Explanation:
D Given: Moment of inertia (I) \(=2 \mathrm{~kg}-\mathrm{m}^{2}\) Axis rotation \(\quad(\mathrm{f})=2 \mathrm{rps}\) We know that, Angular momentum, \(\mathrm{L}=\mathrm{I} \omega\) \(\mathrm{L}=2 \times 2 \pi \mathrm{f}\) \(\mathrm{L}=2 \times 2 \pi \times 2\) \(\mathrm{~L}=8 \pi \mathrm{Js}\)
Kerala CEE - 2016
Rotational Motion
149985
Angular momentum of the Earth revolving around the Sun in a circular orbit of radius \(R\) is proportional to
1 \(\sqrt{\mathrm{R}}\)
2 \(\mathrm{R}\)
3 \(\mathrm{R}^{2}\)
4 \(R^{1 / 3}\)
5 \(R^{3 / 2}\)
Explanation:
A We know that, Angular momentum of Earth, \(\mathrm{L}=\mathrm{mvR}\) Now, \(\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\) \(\mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{R}}\) \(\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\) So, \(L=m \sqrt{\frac{G M}{R}} \times R=m \sqrt{\text { GMR }}\) \(\therefore \quad \mathrm{L} \propto \sqrt{\mathrm{R}}\)
Kerala CEE - 2015
Rotational Motion
149986
The angular momentum of a particle describing uniform circular motion is \(L\). If its kinetic energy is halved and angular velocity doubled, its new angular momentum is
1 \(4 \mathrm{~L}\)
2 \(\frac{L}{4}\)
3 \(\frac{\mathrm{L}}{2}\)
4 \(2 \mathrm{~L}\)
5 \(\frac{L}{8}\)
Explanation:
B Kinetic energy \((K)=\frac{1}{2} I \omega^{2}\) \(\mathrm{K}=\frac{1}{2}(\mathrm{I} \omega) \omega\) Angular momentum \(\mathrm{L})=\mathrm{I} \omega\) \(\therefore \quad \mathrm{K} =\frac{1}{2} \mathrm{~L} \omega\) \(\mathrm{L} =\frac{2 \mathrm{~K}}{\omega}\) Second condition \(\omega^{\prime}=2 \omega, K^{\prime}=\frac{K}{2}\) New angular momentum \(\mathrm{L}^{\prime} =\frac{2 \mathrm{~K}^{\prime}}{\omega^{\prime}}\) \(\mathrm{L}^{\prime} =\frac{2\left(\frac{\mathrm{K}}{2}\right)}{(2 \omega)}=\frac{\mathrm{K}}{2 \omega}\) \(\mathrm{L}^{\prime} =\frac{1}{4}\left(\frac{2 \mathrm{~K}}{\omega}\right)\) \(\mathrm{L}^{\prime} =\frac{\mathrm{L}}{4}\) By comparing from equation (i) and (ii) \(\mathrm{L}^{\prime}=\frac{\mathrm{L}}{4}\)
UPSEE-2005
Rotational Motion
149987
If the earth were to contract such that its radius becomes one quarter, without change in its mass, the duration of one full day would be
1 \(3 \mathrm{~h}\)
2 \(1.5 \mathrm{~h}\)
3 \(6 \mathrm{~h}\)
4 \(4 \mathrm{~h}\)
5 \(2 \mathrm{~h}\)
Explanation:
B \(M_{1}=M_{2}=M\) [ Mass of Earth] \(\mathrm{R}_{2}=\frac{1}{4} \mathrm{R}_{1}\) \(\mathrm{~T}_{1}=24 \mathrm{hrs}\) By the law of conservation of angular momentum \(\mathrm{I} \omega=\) Constant \(\mathrm{I}_{1} \omega_{1}=\mathrm{I}_{2} \omega_{2}\) \(\frac{2}{5} \mathrm{MR}_{1}^{2} \omega_{1}=\frac{2}{5} \mathrm{MR}_{2}^{2} \omega_{2}\) \(\frac{2}{5} \mathrm{M}\left(\mathrm{R}_{1}^{2} \omega_{1}\right)=\frac{2}{5} \mathrm{M}\left(\mathrm{R}_{2}^{2} \omega_{2}\right)\) \(\left(\because \omega=\frac{2 \pi}{\mathrm{T}}\right)\) \(\mathrm{R}_{1}^{2} \times \frac{2 \pi}{24}=\frac{\mathrm{R}_{1}^{2}}{16} \times \frac{2 \pi}{\mathrm{T}_{2}}\) \(2 \pi \mathrm{R}_{1}^{2}\left(\frac{1}{24}\right))=\frac{2 \pi \mathrm{R}_{1}^{2}}{16 \times \mathrm{T}_{2}}\) \(\mathrm{T}_{2}=\frac{24}{16}\) \(\mathrm{T}_{2}=\frac{3}{2}\) \(\mathrm{T}_{2}=1.5 \mathrm{~h}\)
Kerala CEE 2007
Rotational Motion
149989
A disc of mass \(2 \mathrm{~kg}\) and radius \(0.2 \mathrm{~m}\) is rotating with angular velocity 30 rads \(^{-1}\) what is angular velocity, if a mass of \(0.25 \mathrm{~kg}\) is put on periphery of the disc?
1 \(24 \mathrm{rad} \mathrm{s}^{-1}\)
2 \(36 \mathrm{rad} \mathrm{s}^{-1}\)
3 \(15 \mathrm{rad} \mathrm{s}^{-1}\)
4 \(26 \mathrm{rad} \mathrm{s}^{-1}\)
Explanation:
A Given, \(\mathrm{m}_{1}=2 \mathrm{~kg}\) \(\mathrm{~m}_{2}=0.25 \mathrm{~kg}\) \(\mathrm{r}=0.2 \mathrm{~m}\) \(\omega_{1}=30 \mathrm{rad} / \mathrm{sec}\) According to law of conservation of angular momentum \(\mathrm{L}=\mathrm{I} \omega=\) Constant \(\therefore \quad \mathrm{I}_{1} \omega_{1}=\mathrm{I}_{2} \omega_{2}\) \(\frac{1}{2} \mathrm{~m}_{1} \mathrm{r}^{2} \omega_{1}=\frac{1}{2}\left(\mathrm{~m}_{1}+2 \mathrm{~m}_{2}\right) \mathrm{r}^{2} \times \omega_{2}\) \(\frac{1}{2} \times 2 \times(0.2)^{2} \times 30=\frac{1}{2} \times(2+2 \times 0.25) \times(0.2)^{2} \times \omega_{2}\) \(\therefore \quad 1.2=0.05 \omega_{2}\) \(\therefore \quad \omega_{2}=24 \mathrm{rad} / \mathrm{sec}\)
149984
If a body of moment of inertia \(2 \mathrm{~kg} \mathrm{~m}^{2}\) revolves about its own axis making 2 rotations per second, then its angular momentum (in Js) is
1 \(2 \pi\)
2 \(4 \pi\)
3 \(6 \pi\)
4 \(8 \pi\)
5 \(10 \pi\)
Explanation:
D Given: Moment of inertia (I) \(=2 \mathrm{~kg}-\mathrm{m}^{2}\) Axis rotation \(\quad(\mathrm{f})=2 \mathrm{rps}\) We know that, Angular momentum, \(\mathrm{L}=\mathrm{I} \omega\) \(\mathrm{L}=2 \times 2 \pi \mathrm{f}\) \(\mathrm{L}=2 \times 2 \pi \times 2\) \(\mathrm{~L}=8 \pi \mathrm{Js}\)
Kerala CEE - 2016
Rotational Motion
149985
Angular momentum of the Earth revolving around the Sun in a circular orbit of radius \(R\) is proportional to
1 \(\sqrt{\mathrm{R}}\)
2 \(\mathrm{R}\)
3 \(\mathrm{R}^{2}\)
4 \(R^{1 / 3}\)
5 \(R^{3 / 2}\)
Explanation:
A We know that, Angular momentum of Earth, \(\mathrm{L}=\mathrm{mvR}\) Now, \(\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\) \(\mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{R}}\) \(\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\) So, \(L=m \sqrt{\frac{G M}{R}} \times R=m \sqrt{\text { GMR }}\) \(\therefore \quad \mathrm{L} \propto \sqrt{\mathrm{R}}\)
Kerala CEE - 2015
Rotational Motion
149986
The angular momentum of a particle describing uniform circular motion is \(L\). If its kinetic energy is halved and angular velocity doubled, its new angular momentum is
1 \(4 \mathrm{~L}\)
2 \(\frac{L}{4}\)
3 \(\frac{\mathrm{L}}{2}\)
4 \(2 \mathrm{~L}\)
5 \(\frac{L}{8}\)
Explanation:
B Kinetic energy \((K)=\frac{1}{2} I \omega^{2}\) \(\mathrm{K}=\frac{1}{2}(\mathrm{I} \omega) \omega\) Angular momentum \(\mathrm{L})=\mathrm{I} \omega\) \(\therefore \quad \mathrm{K} =\frac{1}{2} \mathrm{~L} \omega\) \(\mathrm{L} =\frac{2 \mathrm{~K}}{\omega}\) Second condition \(\omega^{\prime}=2 \omega, K^{\prime}=\frac{K}{2}\) New angular momentum \(\mathrm{L}^{\prime} =\frac{2 \mathrm{~K}^{\prime}}{\omega^{\prime}}\) \(\mathrm{L}^{\prime} =\frac{2\left(\frac{\mathrm{K}}{2}\right)}{(2 \omega)}=\frac{\mathrm{K}}{2 \omega}\) \(\mathrm{L}^{\prime} =\frac{1}{4}\left(\frac{2 \mathrm{~K}}{\omega}\right)\) \(\mathrm{L}^{\prime} =\frac{\mathrm{L}}{4}\) By comparing from equation (i) and (ii) \(\mathrm{L}^{\prime}=\frac{\mathrm{L}}{4}\)
UPSEE-2005
Rotational Motion
149987
If the earth were to contract such that its radius becomes one quarter, without change in its mass, the duration of one full day would be
1 \(3 \mathrm{~h}\)
2 \(1.5 \mathrm{~h}\)
3 \(6 \mathrm{~h}\)
4 \(4 \mathrm{~h}\)
5 \(2 \mathrm{~h}\)
Explanation:
B \(M_{1}=M_{2}=M\) [ Mass of Earth] \(\mathrm{R}_{2}=\frac{1}{4} \mathrm{R}_{1}\) \(\mathrm{~T}_{1}=24 \mathrm{hrs}\) By the law of conservation of angular momentum \(\mathrm{I} \omega=\) Constant \(\mathrm{I}_{1} \omega_{1}=\mathrm{I}_{2} \omega_{2}\) \(\frac{2}{5} \mathrm{MR}_{1}^{2} \omega_{1}=\frac{2}{5} \mathrm{MR}_{2}^{2} \omega_{2}\) \(\frac{2}{5} \mathrm{M}\left(\mathrm{R}_{1}^{2} \omega_{1}\right)=\frac{2}{5} \mathrm{M}\left(\mathrm{R}_{2}^{2} \omega_{2}\right)\) \(\left(\because \omega=\frac{2 \pi}{\mathrm{T}}\right)\) \(\mathrm{R}_{1}^{2} \times \frac{2 \pi}{24}=\frac{\mathrm{R}_{1}^{2}}{16} \times \frac{2 \pi}{\mathrm{T}_{2}}\) \(2 \pi \mathrm{R}_{1}^{2}\left(\frac{1}{24}\right))=\frac{2 \pi \mathrm{R}_{1}^{2}}{16 \times \mathrm{T}_{2}}\) \(\mathrm{T}_{2}=\frac{24}{16}\) \(\mathrm{T}_{2}=\frac{3}{2}\) \(\mathrm{T}_{2}=1.5 \mathrm{~h}\)
Kerala CEE 2007
Rotational Motion
149989
A disc of mass \(2 \mathrm{~kg}\) and radius \(0.2 \mathrm{~m}\) is rotating with angular velocity 30 rads \(^{-1}\) what is angular velocity, if a mass of \(0.25 \mathrm{~kg}\) is put on periphery of the disc?
1 \(24 \mathrm{rad} \mathrm{s}^{-1}\)
2 \(36 \mathrm{rad} \mathrm{s}^{-1}\)
3 \(15 \mathrm{rad} \mathrm{s}^{-1}\)
4 \(26 \mathrm{rad} \mathrm{s}^{-1}\)
Explanation:
A Given, \(\mathrm{m}_{1}=2 \mathrm{~kg}\) \(\mathrm{~m}_{2}=0.25 \mathrm{~kg}\) \(\mathrm{r}=0.2 \mathrm{~m}\) \(\omega_{1}=30 \mathrm{rad} / \mathrm{sec}\) According to law of conservation of angular momentum \(\mathrm{L}=\mathrm{I} \omega=\) Constant \(\therefore \quad \mathrm{I}_{1} \omega_{1}=\mathrm{I}_{2} \omega_{2}\) \(\frac{1}{2} \mathrm{~m}_{1} \mathrm{r}^{2} \omega_{1}=\frac{1}{2}\left(\mathrm{~m}_{1}+2 \mathrm{~m}_{2}\right) \mathrm{r}^{2} \times \omega_{2}\) \(\frac{1}{2} \times 2 \times(0.2)^{2} \times 30=\frac{1}{2} \times(2+2 \times 0.25) \times(0.2)^{2} \times \omega_{2}\) \(\therefore \quad 1.2=0.05 \omega_{2}\) \(\therefore \quad \omega_{2}=24 \mathrm{rad} / \mathrm{sec}\)
