150301
A solid cylinder of radius \(R\) is at rest at a height \(h\) on an inclined plane. If it rolls down then its velocity on reaching the ground is
1 \(\sqrt{\frac{5 \mathrm{gh}}{3}}\)
2 \(\sqrt{\frac{2 h}{3 g}}\)
3 \(\sqrt{\frac{2 \mathrm{gh}}{3}}\)
4 \(\sqrt{\frac{4 \mathrm{gh}}{3}}\)
Explanation:
D Given that, Radius \(=\mathrm{R}\) Height \(=\mathrm{h}\) According to Work Energy theorem, \(M g h=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) Moment of inertia of solid cylinder is \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}\) \(\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{MR}^{2} \omega^{2} \quad(\because \mathrm{v}=\mathrm{R} \omega\) \(\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{4} \mathrm{Mv}^{2}\) \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}^{2}\) \(\mathrm{v}^{2}=\frac{4}{3} \mathrm{gh}\) \(\mathrm{v}=\sqrt{\frac{4 \mathrm{gh}}{3}}\)
AP EAMCET-05.07.2022
Rotational Motion
150302
A hydrogen atom is in ground state absorbs \(10.2 \mathrm{eV}\) energy. The angular momentum of electron of the hydrogen atom will increase by the value of : (Given, Planck's constant \(=6.6 \times 10^{-34} \mathrm{Js}\)
1 \(2.10 \times 10^{-34} \mathrm{Js}\)
2 \(1.05 \times 10^{-34} \mathrm{Js}\)
3 \(3.15 \times 10^{-34} \mathrm{Js}\)
4 \(4.2 \times 10^{-34} \mathrm{Js}\)
Explanation:
B Given that, Energy absorbs in ground \(=10.2 \mathrm{eV}\) Plank's constant \(=6.6 \times 10^{-34} \mathrm{Js}\) So, after absorbing, the energy of electron become \(-13.6+10.2=\frac{-13.6}{\mathrm{n}^{2}}\) \(-3.4=\frac{-13.6}{\mathrm{n}^{2}}\) \(\frac{13.6}{\mathrm{n}^{2}}=3.4\) \(\mathrm{n}^{2}=\frac{13.6}{3.4}\) \(\mathrm{n}^{2}=4\) \(\mathrm{n}=2\) Then, Increase in angular momentum given by \(\Delta \mathrm{L} =2 \times \frac{\mathrm{h}}{2 \pi}-1 \times \frac{\mathrm{h}}{2 \pi}\) \(=\frac{6.6 \times 10^{-34}}{2 \times 3.14}\) \(\Delta \mathrm{L} \cong 1.05 \times 10^{-34} \mathrm{Js} .\)
JEE Main-27.06.2022
Rotational Motion
150303
A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is:-
1 \(\frac{2}{5}\)
2 \(\frac{2}{7}\)
3 \(\frac{1}{5}\)
4 \(\frac{7}{10}\)
Explanation:
B When a solid spherical ball is rolling on a frictionless horizontal plane surface. Then, \(\mathrm{K} \cdot \mathrm{E}_{\text {Total }}=\mathrm{K} \cdot \mathrm{E}_{\text {Translation }}+\mathrm{K} \cdot \mathrm{E}_{\text {rotational }}\) \(\mathrm{K} . \mathrm{E}_{\text {Total }}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{v}_{\mathrm{cm}}=\mathrm{R} \omega\), for pure rolling \(\mathrm{I}_{\mathrm{cm}}=\frac{2}{5} \mathrm{mR}^{2} \quad \text { for solid sphere }\) \(\text { K. } E_{\text {Rotational }}=\frac{1}{2} I_{\mathrm{cm}} \omega^{2}\) \(=\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^{2} \times \frac{\mathrm{v}_{\mathrm{cm}}^{2}}{\mathrm{R}^{2}}\) \(=\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(\mathrm{~K} \cdot \mathrm{E}_{\text {Total }}=\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}+\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(=\frac{7}{10} \mathrm{mv}_{\mathrm{cm}}^{2}\) Hence ratio of K.E. rotational and K.E. total is \(\frac{\mathrm{K} \cdot \mathrm{E}_{\text {Rotational }}}{\mathrm{K} \cdot \mathrm{E}_{\text {Total }}}=\frac{\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}}{\frac{7}{10} \mathrm{mv}_{\mathrm{cm}}^{2}}=\frac{2}{7}\)
JEE Main-26.06.2022
Rotational Motion
150304
A fly wheel is accelerated uniformly from rest and rotates through \(5 \mathrm{rad}\) in the first second. The angle rotated by the fly wheel in the next second, will be:
1 \(7.5 \mathrm{rad}\)
2 \(15 \mathrm{rad}\)
3 \(20 \mathrm{rad}\)
4 \(30 \mathrm{rad}\)
Explanation:
B Case-I :- Given that, Angular displacement, \(\theta=5 \mathrm{rad}\) Time, \(\mathrm{t}=1\) sec. \(\omega_{0}=0\) As the angular acceleration is constant, we have \(\theta=\omega_{0} \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(\theta=0+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(5=\frac{1}{2} \alpha(1)^{2}\) Angular acceleration, \(\alpha=10 \mathrm{rad} \mathrm{s}^{-2}\) Case- II :- The angular displacement in first two second is given by \(\theta^{\prime}=\frac{1}{2} \times 10 \times(2)^{2}\) \(\theta^{\prime}=20 \mathrm{rad}\) Thus, the angle rotated by flywheel during the \(2^{\text {nd }}\) second \(=\theta^{\prime}-\theta\) \(=20-5\) \(=15 \mathrm{rad}\)
150301
A solid cylinder of radius \(R\) is at rest at a height \(h\) on an inclined plane. If it rolls down then its velocity on reaching the ground is
1 \(\sqrt{\frac{5 \mathrm{gh}}{3}}\)
2 \(\sqrt{\frac{2 h}{3 g}}\)
3 \(\sqrt{\frac{2 \mathrm{gh}}{3}}\)
4 \(\sqrt{\frac{4 \mathrm{gh}}{3}}\)
Explanation:
D Given that, Radius \(=\mathrm{R}\) Height \(=\mathrm{h}\) According to Work Energy theorem, \(M g h=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) Moment of inertia of solid cylinder is \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}\) \(\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{MR}^{2} \omega^{2} \quad(\because \mathrm{v}=\mathrm{R} \omega\) \(\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{4} \mathrm{Mv}^{2}\) \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}^{2}\) \(\mathrm{v}^{2}=\frac{4}{3} \mathrm{gh}\) \(\mathrm{v}=\sqrt{\frac{4 \mathrm{gh}}{3}}\)
AP EAMCET-05.07.2022
Rotational Motion
150302
A hydrogen atom is in ground state absorbs \(10.2 \mathrm{eV}\) energy. The angular momentum of electron of the hydrogen atom will increase by the value of : (Given, Planck's constant \(=6.6 \times 10^{-34} \mathrm{Js}\)
1 \(2.10 \times 10^{-34} \mathrm{Js}\)
2 \(1.05 \times 10^{-34} \mathrm{Js}\)
3 \(3.15 \times 10^{-34} \mathrm{Js}\)
4 \(4.2 \times 10^{-34} \mathrm{Js}\)
Explanation:
B Given that, Energy absorbs in ground \(=10.2 \mathrm{eV}\) Plank's constant \(=6.6 \times 10^{-34} \mathrm{Js}\) So, after absorbing, the energy of electron become \(-13.6+10.2=\frac{-13.6}{\mathrm{n}^{2}}\) \(-3.4=\frac{-13.6}{\mathrm{n}^{2}}\) \(\frac{13.6}{\mathrm{n}^{2}}=3.4\) \(\mathrm{n}^{2}=\frac{13.6}{3.4}\) \(\mathrm{n}^{2}=4\) \(\mathrm{n}=2\) Then, Increase in angular momentum given by \(\Delta \mathrm{L} =2 \times \frac{\mathrm{h}}{2 \pi}-1 \times \frac{\mathrm{h}}{2 \pi}\) \(=\frac{6.6 \times 10^{-34}}{2 \times 3.14}\) \(\Delta \mathrm{L} \cong 1.05 \times 10^{-34} \mathrm{Js} .\)
JEE Main-27.06.2022
Rotational Motion
150303
A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is:-
1 \(\frac{2}{5}\)
2 \(\frac{2}{7}\)
3 \(\frac{1}{5}\)
4 \(\frac{7}{10}\)
Explanation:
B When a solid spherical ball is rolling on a frictionless horizontal plane surface. Then, \(\mathrm{K} \cdot \mathrm{E}_{\text {Total }}=\mathrm{K} \cdot \mathrm{E}_{\text {Translation }}+\mathrm{K} \cdot \mathrm{E}_{\text {rotational }}\) \(\mathrm{K} . \mathrm{E}_{\text {Total }}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{v}_{\mathrm{cm}}=\mathrm{R} \omega\), for pure rolling \(\mathrm{I}_{\mathrm{cm}}=\frac{2}{5} \mathrm{mR}^{2} \quad \text { for solid sphere }\) \(\text { K. } E_{\text {Rotational }}=\frac{1}{2} I_{\mathrm{cm}} \omega^{2}\) \(=\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^{2} \times \frac{\mathrm{v}_{\mathrm{cm}}^{2}}{\mathrm{R}^{2}}\) \(=\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(\mathrm{~K} \cdot \mathrm{E}_{\text {Total }}=\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}+\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(=\frac{7}{10} \mathrm{mv}_{\mathrm{cm}}^{2}\) Hence ratio of K.E. rotational and K.E. total is \(\frac{\mathrm{K} \cdot \mathrm{E}_{\text {Rotational }}}{\mathrm{K} \cdot \mathrm{E}_{\text {Total }}}=\frac{\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}}{\frac{7}{10} \mathrm{mv}_{\mathrm{cm}}^{2}}=\frac{2}{7}\)
JEE Main-26.06.2022
Rotational Motion
150304
A fly wheel is accelerated uniformly from rest and rotates through \(5 \mathrm{rad}\) in the first second. The angle rotated by the fly wheel in the next second, will be:
1 \(7.5 \mathrm{rad}\)
2 \(15 \mathrm{rad}\)
3 \(20 \mathrm{rad}\)
4 \(30 \mathrm{rad}\)
Explanation:
B Case-I :- Given that, Angular displacement, \(\theta=5 \mathrm{rad}\) Time, \(\mathrm{t}=1\) sec. \(\omega_{0}=0\) As the angular acceleration is constant, we have \(\theta=\omega_{0} \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(\theta=0+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(5=\frac{1}{2} \alpha(1)^{2}\) Angular acceleration, \(\alpha=10 \mathrm{rad} \mathrm{s}^{-2}\) Case- II :- The angular displacement in first two second is given by \(\theta^{\prime}=\frac{1}{2} \times 10 \times(2)^{2}\) \(\theta^{\prime}=20 \mathrm{rad}\) Thus, the angle rotated by flywheel during the \(2^{\text {nd }}\) second \(=\theta^{\prime}-\theta\) \(=20-5\) \(=15 \mathrm{rad}\)
150301
A solid cylinder of radius \(R\) is at rest at a height \(h\) on an inclined plane. If it rolls down then its velocity on reaching the ground is
1 \(\sqrt{\frac{5 \mathrm{gh}}{3}}\)
2 \(\sqrt{\frac{2 h}{3 g}}\)
3 \(\sqrt{\frac{2 \mathrm{gh}}{3}}\)
4 \(\sqrt{\frac{4 \mathrm{gh}}{3}}\)
Explanation:
D Given that, Radius \(=\mathrm{R}\) Height \(=\mathrm{h}\) According to Work Energy theorem, \(M g h=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) Moment of inertia of solid cylinder is \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}\) \(\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{MR}^{2} \omega^{2} \quad(\because \mathrm{v}=\mathrm{R} \omega\) \(\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{4} \mathrm{Mv}^{2}\) \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}^{2}\) \(\mathrm{v}^{2}=\frac{4}{3} \mathrm{gh}\) \(\mathrm{v}=\sqrt{\frac{4 \mathrm{gh}}{3}}\)
AP EAMCET-05.07.2022
Rotational Motion
150302
A hydrogen atom is in ground state absorbs \(10.2 \mathrm{eV}\) energy. The angular momentum of electron of the hydrogen atom will increase by the value of : (Given, Planck's constant \(=6.6 \times 10^{-34} \mathrm{Js}\)
1 \(2.10 \times 10^{-34} \mathrm{Js}\)
2 \(1.05 \times 10^{-34} \mathrm{Js}\)
3 \(3.15 \times 10^{-34} \mathrm{Js}\)
4 \(4.2 \times 10^{-34} \mathrm{Js}\)
Explanation:
B Given that, Energy absorbs in ground \(=10.2 \mathrm{eV}\) Plank's constant \(=6.6 \times 10^{-34} \mathrm{Js}\) So, after absorbing, the energy of electron become \(-13.6+10.2=\frac{-13.6}{\mathrm{n}^{2}}\) \(-3.4=\frac{-13.6}{\mathrm{n}^{2}}\) \(\frac{13.6}{\mathrm{n}^{2}}=3.4\) \(\mathrm{n}^{2}=\frac{13.6}{3.4}\) \(\mathrm{n}^{2}=4\) \(\mathrm{n}=2\) Then, Increase in angular momentum given by \(\Delta \mathrm{L} =2 \times \frac{\mathrm{h}}{2 \pi}-1 \times \frac{\mathrm{h}}{2 \pi}\) \(=\frac{6.6 \times 10^{-34}}{2 \times 3.14}\) \(\Delta \mathrm{L} \cong 1.05 \times 10^{-34} \mathrm{Js} .\)
JEE Main-27.06.2022
Rotational Motion
150303
A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is:-
1 \(\frac{2}{5}\)
2 \(\frac{2}{7}\)
3 \(\frac{1}{5}\)
4 \(\frac{7}{10}\)
Explanation:
B When a solid spherical ball is rolling on a frictionless horizontal plane surface. Then, \(\mathrm{K} \cdot \mathrm{E}_{\text {Total }}=\mathrm{K} \cdot \mathrm{E}_{\text {Translation }}+\mathrm{K} \cdot \mathrm{E}_{\text {rotational }}\) \(\mathrm{K} . \mathrm{E}_{\text {Total }}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{v}_{\mathrm{cm}}=\mathrm{R} \omega\), for pure rolling \(\mathrm{I}_{\mathrm{cm}}=\frac{2}{5} \mathrm{mR}^{2} \quad \text { for solid sphere }\) \(\text { K. } E_{\text {Rotational }}=\frac{1}{2} I_{\mathrm{cm}} \omega^{2}\) \(=\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^{2} \times \frac{\mathrm{v}_{\mathrm{cm}}^{2}}{\mathrm{R}^{2}}\) \(=\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(\mathrm{~K} \cdot \mathrm{E}_{\text {Total }}=\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}+\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(=\frac{7}{10} \mathrm{mv}_{\mathrm{cm}}^{2}\) Hence ratio of K.E. rotational and K.E. total is \(\frac{\mathrm{K} \cdot \mathrm{E}_{\text {Rotational }}}{\mathrm{K} \cdot \mathrm{E}_{\text {Total }}}=\frac{\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}}{\frac{7}{10} \mathrm{mv}_{\mathrm{cm}}^{2}}=\frac{2}{7}\)
JEE Main-26.06.2022
Rotational Motion
150304
A fly wheel is accelerated uniformly from rest and rotates through \(5 \mathrm{rad}\) in the first second. The angle rotated by the fly wheel in the next second, will be:
1 \(7.5 \mathrm{rad}\)
2 \(15 \mathrm{rad}\)
3 \(20 \mathrm{rad}\)
4 \(30 \mathrm{rad}\)
Explanation:
B Case-I :- Given that, Angular displacement, \(\theta=5 \mathrm{rad}\) Time, \(\mathrm{t}=1\) sec. \(\omega_{0}=0\) As the angular acceleration is constant, we have \(\theta=\omega_{0} \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(\theta=0+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(5=\frac{1}{2} \alpha(1)^{2}\) Angular acceleration, \(\alpha=10 \mathrm{rad} \mathrm{s}^{-2}\) Case- II :- The angular displacement in first two second is given by \(\theta^{\prime}=\frac{1}{2} \times 10 \times(2)^{2}\) \(\theta^{\prime}=20 \mathrm{rad}\) Thus, the angle rotated by flywheel during the \(2^{\text {nd }}\) second \(=\theta^{\prime}-\theta\) \(=20-5\) \(=15 \mathrm{rad}\)
150301
A solid cylinder of radius \(R\) is at rest at a height \(h\) on an inclined plane. If it rolls down then its velocity on reaching the ground is
1 \(\sqrt{\frac{5 \mathrm{gh}}{3}}\)
2 \(\sqrt{\frac{2 h}{3 g}}\)
3 \(\sqrt{\frac{2 \mathrm{gh}}{3}}\)
4 \(\sqrt{\frac{4 \mathrm{gh}}{3}}\)
Explanation:
D Given that, Radius \(=\mathrm{R}\) Height \(=\mathrm{h}\) According to Work Energy theorem, \(M g h=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) Moment of inertia of solid cylinder is \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}\) \(\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{MR}^{2} \omega^{2} \quad(\because \mathrm{v}=\mathrm{R} \omega\) \(\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{4} \mathrm{Mv}^{2}\) \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}^{2}\) \(\mathrm{v}^{2}=\frac{4}{3} \mathrm{gh}\) \(\mathrm{v}=\sqrt{\frac{4 \mathrm{gh}}{3}}\)
AP EAMCET-05.07.2022
Rotational Motion
150302
A hydrogen atom is in ground state absorbs \(10.2 \mathrm{eV}\) energy. The angular momentum of electron of the hydrogen atom will increase by the value of : (Given, Planck's constant \(=6.6 \times 10^{-34} \mathrm{Js}\)
1 \(2.10 \times 10^{-34} \mathrm{Js}\)
2 \(1.05 \times 10^{-34} \mathrm{Js}\)
3 \(3.15 \times 10^{-34} \mathrm{Js}\)
4 \(4.2 \times 10^{-34} \mathrm{Js}\)
Explanation:
B Given that, Energy absorbs in ground \(=10.2 \mathrm{eV}\) Plank's constant \(=6.6 \times 10^{-34} \mathrm{Js}\) So, after absorbing, the energy of electron become \(-13.6+10.2=\frac{-13.6}{\mathrm{n}^{2}}\) \(-3.4=\frac{-13.6}{\mathrm{n}^{2}}\) \(\frac{13.6}{\mathrm{n}^{2}}=3.4\) \(\mathrm{n}^{2}=\frac{13.6}{3.4}\) \(\mathrm{n}^{2}=4\) \(\mathrm{n}=2\) Then, Increase in angular momentum given by \(\Delta \mathrm{L} =2 \times \frac{\mathrm{h}}{2 \pi}-1 \times \frac{\mathrm{h}}{2 \pi}\) \(=\frac{6.6 \times 10^{-34}}{2 \times 3.14}\) \(\Delta \mathrm{L} \cong 1.05 \times 10^{-34} \mathrm{Js} .\)
JEE Main-27.06.2022
Rotational Motion
150303
A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is:-
1 \(\frac{2}{5}\)
2 \(\frac{2}{7}\)
3 \(\frac{1}{5}\)
4 \(\frac{7}{10}\)
Explanation:
B When a solid spherical ball is rolling on a frictionless horizontal plane surface. Then, \(\mathrm{K} \cdot \mathrm{E}_{\text {Total }}=\mathrm{K} \cdot \mathrm{E}_{\text {Translation }}+\mathrm{K} \cdot \mathrm{E}_{\text {rotational }}\) \(\mathrm{K} . \mathrm{E}_{\text {Total }}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) \(\mathrm{v}_{\mathrm{cm}}=\mathrm{R} \omega\), for pure rolling \(\mathrm{I}_{\mathrm{cm}}=\frac{2}{5} \mathrm{mR}^{2} \quad \text { for solid sphere }\) \(\text { K. } E_{\text {Rotational }}=\frac{1}{2} I_{\mathrm{cm}} \omega^{2}\) \(=\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^{2} \times \frac{\mathrm{v}_{\mathrm{cm}}^{2}}{\mathrm{R}^{2}}\) \(=\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(\mathrm{~K} \cdot \mathrm{E}_{\text {Total }}=\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}+\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(=\frac{7}{10} \mathrm{mv}_{\mathrm{cm}}^{2}\) Hence ratio of K.E. rotational and K.E. total is \(\frac{\mathrm{K} \cdot \mathrm{E}_{\text {Rotational }}}{\mathrm{K} \cdot \mathrm{E}_{\text {Total }}}=\frac{\frac{1}{5} \mathrm{mv}_{\mathrm{cm}}^{2}}{\frac{7}{10} \mathrm{mv}_{\mathrm{cm}}^{2}}=\frac{2}{7}\)
JEE Main-26.06.2022
Rotational Motion
150304
A fly wheel is accelerated uniformly from rest and rotates through \(5 \mathrm{rad}\) in the first second. The angle rotated by the fly wheel in the next second, will be:
1 \(7.5 \mathrm{rad}\)
2 \(15 \mathrm{rad}\)
3 \(20 \mathrm{rad}\)
4 \(30 \mathrm{rad}\)
Explanation:
B Case-I :- Given that, Angular displacement, \(\theta=5 \mathrm{rad}\) Time, \(\mathrm{t}=1\) sec. \(\omega_{0}=0\) As the angular acceleration is constant, we have \(\theta=\omega_{0} \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(\theta=0+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(5=\frac{1}{2} \alpha(1)^{2}\) Angular acceleration, \(\alpha=10 \mathrm{rad} \mathrm{s}^{-2}\) Case- II :- The angular displacement in first two second is given by \(\theta^{\prime}=\frac{1}{2} \times 10 \times(2)^{2}\) \(\theta^{\prime}=20 \mathrm{rad}\) Thus, the angle rotated by flywheel during the \(2^{\text {nd }}\) second \(=\theta^{\prime}-\theta\) \(=20-5\) \(=15 \mathrm{rad}\)
