NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
150306
A wheel is rotating freely at some angular speed. A second wheel initially at rest and with thrice the rotational inertia of the first, is suddenly coupled to the first wheel. The fraction of the original rotational kinetic energy lost is
1 0.50
2 0.25
3 0.66
4 0.75
Explanation:
D According to law of conservation of angular momentum, \(\mathrm{I} \omega+0=\mathrm{I} \omega_{\mathrm{c}}+3 \mathrm{I} \omega_{\mathrm{c}}\) \(\mathrm{I} \omega=4 \mathrm{I} \omega_{\mathrm{c}}\) \(\omega_{\mathrm{c}}=\frac{\omega}{4}\) Loss of rotational kinetic energy, \(\Delta \mathrm{K} . \mathrm{E} .)_{\mathrm{loss}} =\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{2}(\mathrm{I}+3 \mathrm{I}) \omega_{\mathrm{c}}{ }^{2}\) \(=\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{2} \times 4 \mathrm{I} \times \frac{\omega^{2}}{4 \times 4}\) \(=\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{8} \mathrm{I} \omega^{2}\) \((\Delta \mathrm{KE})_{\text {loss }} =\frac{3}{8} \mathrm{I} \omega^{2}\) Resultant fractional loss in kinetic energy, \(\frac{(\Delta \mathrm{KE})_{\text {loss }}}{(\mathrm{KE})_{\text {initial }}} =\frac{\frac{3}{8} \mathrm{I} \omega^{2}}{\frac{1}{2} \mathrm{I} \omega^{2}}\) \(=\frac{3}{4}=0.75\)
AP EAMCET-07.07.2022
Rotational Motion
150307
A flywheel having moment of inertia \(10 \mathrm{~kg} . \mathrm{m}^{2}\) is rotating at \(50 \mathrm{rad}^{-1}\). What amount of work needs to be done in order to bring this flywheel to rest in 10 seconds?
150308
When no external torque acts on a rotating system,
1 Angular momentum of the system is not conserved
2 Its rotational kinetic energy is conserved
3 Its rotational kinetic energy is independent of moment of inertia
4 Its rotational kinetic energy is directly proportional to moment of inertia
5 Its rotational kinetic energy is inversely proportional to moment of inertia
Explanation:
E If external torque is zero. Then, angular momentum of system is conserved. We know, Kinetic energy (K.E.) \(=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \frac{\mathrm{I} \times \omega^{2} \times \mathrm{I}}{\mathrm{I}}\) \(=\frac{1}{2} \frac{(\mathrm{I} \omega)^{2}}{\mathrm{I}}\) \(=\frac{1}{2} \frac{\mathrm{L}^{2}}{\mathrm{I}}\) K.E. \(\propto \frac{1}{\mathrm{I}}\) However, K.E. is inversely proportional to moment of inertia.
Kerala CEE 2021
Rotational Motion
150309
If ' \(R\) ' is the radius of orbit of a satellite, then the kinetic energy of the satellite is
1 \(\propto \frac{1}{\mathrm{R}}\)
2 \(\propto \frac{1}{\sqrt{\mathrm{R}}}\)
3 \(\propto \mathrm{R}\)
4 \(\propto \frac{1}{\mathrm{R}^{3 / 2}}\)
Explanation:
A Given, radius of orbit of satellite \(=\mathrm{R}\) We know, Kinetic Energy (K.E.) \(=\frac{1}{2} \mathrm{mv}_{0}{ }^{2}\) Where, \(\mathrm{v}_{0}=\) Orbital velocity \(\because \quad \mathrm{v}_{0}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\) Then, K.E. \(=\frac{1}{2} \mathrm{mv}_{0}^{2}\) K.E. \(=\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right)^{2}\) \(=\frac{\mathrm{GMm}}{2 \mathrm{R}}\) So, \(\quad\) K.E. \(\propto \frac{1}{\mathrm{R}}\)
150306
A wheel is rotating freely at some angular speed. A second wheel initially at rest and with thrice the rotational inertia of the first, is suddenly coupled to the first wheel. The fraction of the original rotational kinetic energy lost is
1 0.50
2 0.25
3 0.66
4 0.75
Explanation:
D According to law of conservation of angular momentum, \(\mathrm{I} \omega+0=\mathrm{I} \omega_{\mathrm{c}}+3 \mathrm{I} \omega_{\mathrm{c}}\) \(\mathrm{I} \omega=4 \mathrm{I} \omega_{\mathrm{c}}\) \(\omega_{\mathrm{c}}=\frac{\omega}{4}\) Loss of rotational kinetic energy, \(\Delta \mathrm{K} . \mathrm{E} .)_{\mathrm{loss}} =\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{2}(\mathrm{I}+3 \mathrm{I}) \omega_{\mathrm{c}}{ }^{2}\) \(=\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{2} \times 4 \mathrm{I} \times \frac{\omega^{2}}{4 \times 4}\) \(=\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{8} \mathrm{I} \omega^{2}\) \((\Delta \mathrm{KE})_{\text {loss }} =\frac{3}{8} \mathrm{I} \omega^{2}\) Resultant fractional loss in kinetic energy, \(\frac{(\Delta \mathrm{KE})_{\text {loss }}}{(\mathrm{KE})_{\text {initial }}} =\frac{\frac{3}{8} \mathrm{I} \omega^{2}}{\frac{1}{2} \mathrm{I} \omega^{2}}\) \(=\frac{3}{4}=0.75\)
AP EAMCET-07.07.2022
Rotational Motion
150307
A flywheel having moment of inertia \(10 \mathrm{~kg} . \mathrm{m}^{2}\) is rotating at \(50 \mathrm{rad}^{-1}\). What amount of work needs to be done in order to bring this flywheel to rest in 10 seconds?
150308
When no external torque acts on a rotating system,
1 Angular momentum of the system is not conserved
2 Its rotational kinetic energy is conserved
3 Its rotational kinetic energy is independent of moment of inertia
4 Its rotational kinetic energy is directly proportional to moment of inertia
5 Its rotational kinetic energy is inversely proportional to moment of inertia
Explanation:
E If external torque is zero. Then, angular momentum of system is conserved. We know, Kinetic energy (K.E.) \(=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \frac{\mathrm{I} \times \omega^{2} \times \mathrm{I}}{\mathrm{I}}\) \(=\frac{1}{2} \frac{(\mathrm{I} \omega)^{2}}{\mathrm{I}}\) \(=\frac{1}{2} \frac{\mathrm{L}^{2}}{\mathrm{I}}\) K.E. \(\propto \frac{1}{\mathrm{I}}\) However, K.E. is inversely proportional to moment of inertia.
Kerala CEE 2021
Rotational Motion
150309
If ' \(R\) ' is the radius of orbit of a satellite, then the kinetic energy of the satellite is
1 \(\propto \frac{1}{\mathrm{R}}\)
2 \(\propto \frac{1}{\sqrt{\mathrm{R}}}\)
3 \(\propto \mathrm{R}\)
4 \(\propto \frac{1}{\mathrm{R}^{3 / 2}}\)
Explanation:
A Given, radius of orbit of satellite \(=\mathrm{R}\) We know, Kinetic Energy (K.E.) \(=\frac{1}{2} \mathrm{mv}_{0}{ }^{2}\) Where, \(\mathrm{v}_{0}=\) Orbital velocity \(\because \quad \mathrm{v}_{0}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\) Then, K.E. \(=\frac{1}{2} \mathrm{mv}_{0}^{2}\) K.E. \(=\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right)^{2}\) \(=\frac{\mathrm{GMm}}{2 \mathrm{R}}\) So, \(\quad\) K.E. \(\propto \frac{1}{\mathrm{R}}\)
150306
A wheel is rotating freely at some angular speed. A second wheel initially at rest and with thrice the rotational inertia of the first, is suddenly coupled to the first wheel. The fraction of the original rotational kinetic energy lost is
1 0.50
2 0.25
3 0.66
4 0.75
Explanation:
D According to law of conservation of angular momentum, \(\mathrm{I} \omega+0=\mathrm{I} \omega_{\mathrm{c}}+3 \mathrm{I} \omega_{\mathrm{c}}\) \(\mathrm{I} \omega=4 \mathrm{I} \omega_{\mathrm{c}}\) \(\omega_{\mathrm{c}}=\frac{\omega}{4}\) Loss of rotational kinetic energy, \(\Delta \mathrm{K} . \mathrm{E} .)_{\mathrm{loss}} =\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{2}(\mathrm{I}+3 \mathrm{I}) \omega_{\mathrm{c}}{ }^{2}\) \(=\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{2} \times 4 \mathrm{I} \times \frac{\omega^{2}}{4 \times 4}\) \(=\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{8} \mathrm{I} \omega^{2}\) \((\Delta \mathrm{KE})_{\text {loss }} =\frac{3}{8} \mathrm{I} \omega^{2}\) Resultant fractional loss in kinetic energy, \(\frac{(\Delta \mathrm{KE})_{\text {loss }}}{(\mathrm{KE})_{\text {initial }}} =\frac{\frac{3}{8} \mathrm{I} \omega^{2}}{\frac{1}{2} \mathrm{I} \omega^{2}}\) \(=\frac{3}{4}=0.75\)
AP EAMCET-07.07.2022
Rotational Motion
150307
A flywheel having moment of inertia \(10 \mathrm{~kg} . \mathrm{m}^{2}\) is rotating at \(50 \mathrm{rad}^{-1}\). What amount of work needs to be done in order to bring this flywheel to rest in 10 seconds?
150308
When no external torque acts on a rotating system,
1 Angular momentum of the system is not conserved
2 Its rotational kinetic energy is conserved
3 Its rotational kinetic energy is independent of moment of inertia
4 Its rotational kinetic energy is directly proportional to moment of inertia
5 Its rotational kinetic energy is inversely proportional to moment of inertia
Explanation:
E If external torque is zero. Then, angular momentum of system is conserved. We know, Kinetic energy (K.E.) \(=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \frac{\mathrm{I} \times \omega^{2} \times \mathrm{I}}{\mathrm{I}}\) \(=\frac{1}{2} \frac{(\mathrm{I} \omega)^{2}}{\mathrm{I}}\) \(=\frac{1}{2} \frac{\mathrm{L}^{2}}{\mathrm{I}}\) K.E. \(\propto \frac{1}{\mathrm{I}}\) However, K.E. is inversely proportional to moment of inertia.
Kerala CEE 2021
Rotational Motion
150309
If ' \(R\) ' is the radius of orbit of a satellite, then the kinetic energy of the satellite is
1 \(\propto \frac{1}{\mathrm{R}}\)
2 \(\propto \frac{1}{\sqrt{\mathrm{R}}}\)
3 \(\propto \mathrm{R}\)
4 \(\propto \frac{1}{\mathrm{R}^{3 / 2}}\)
Explanation:
A Given, radius of orbit of satellite \(=\mathrm{R}\) We know, Kinetic Energy (K.E.) \(=\frac{1}{2} \mathrm{mv}_{0}{ }^{2}\) Where, \(\mathrm{v}_{0}=\) Orbital velocity \(\because \quad \mathrm{v}_{0}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\) Then, K.E. \(=\frac{1}{2} \mathrm{mv}_{0}^{2}\) K.E. \(=\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right)^{2}\) \(=\frac{\mathrm{GMm}}{2 \mathrm{R}}\) So, \(\quad\) K.E. \(\propto \frac{1}{\mathrm{R}}\)
150306
A wheel is rotating freely at some angular speed. A second wheel initially at rest and with thrice the rotational inertia of the first, is suddenly coupled to the first wheel. The fraction of the original rotational kinetic energy lost is
1 0.50
2 0.25
3 0.66
4 0.75
Explanation:
D According to law of conservation of angular momentum, \(\mathrm{I} \omega+0=\mathrm{I} \omega_{\mathrm{c}}+3 \mathrm{I} \omega_{\mathrm{c}}\) \(\mathrm{I} \omega=4 \mathrm{I} \omega_{\mathrm{c}}\) \(\omega_{\mathrm{c}}=\frac{\omega}{4}\) Loss of rotational kinetic energy, \(\Delta \mathrm{K} . \mathrm{E} .)_{\mathrm{loss}} =\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{2}(\mathrm{I}+3 \mathrm{I}) \omega_{\mathrm{c}}{ }^{2}\) \(=\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{2} \times 4 \mathrm{I} \times \frac{\omega^{2}}{4 \times 4}\) \(=\frac{1}{2} \mathrm{I} \omega^{2}-\frac{1}{8} \mathrm{I} \omega^{2}\) \((\Delta \mathrm{KE})_{\text {loss }} =\frac{3}{8} \mathrm{I} \omega^{2}\) Resultant fractional loss in kinetic energy, \(\frac{(\Delta \mathrm{KE})_{\text {loss }}}{(\mathrm{KE})_{\text {initial }}} =\frac{\frac{3}{8} \mathrm{I} \omega^{2}}{\frac{1}{2} \mathrm{I} \omega^{2}}\) \(=\frac{3}{4}=0.75\)
AP EAMCET-07.07.2022
Rotational Motion
150307
A flywheel having moment of inertia \(10 \mathrm{~kg} . \mathrm{m}^{2}\) is rotating at \(50 \mathrm{rad}^{-1}\). What amount of work needs to be done in order to bring this flywheel to rest in 10 seconds?
150308
When no external torque acts on a rotating system,
1 Angular momentum of the system is not conserved
2 Its rotational kinetic energy is conserved
3 Its rotational kinetic energy is independent of moment of inertia
4 Its rotational kinetic energy is directly proportional to moment of inertia
5 Its rotational kinetic energy is inversely proportional to moment of inertia
Explanation:
E If external torque is zero. Then, angular momentum of system is conserved. We know, Kinetic energy (K.E.) \(=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \frac{\mathrm{I} \times \omega^{2} \times \mathrm{I}}{\mathrm{I}}\) \(=\frac{1}{2} \frac{(\mathrm{I} \omega)^{2}}{\mathrm{I}}\) \(=\frac{1}{2} \frac{\mathrm{L}^{2}}{\mathrm{I}}\) K.E. \(\propto \frac{1}{\mathrm{I}}\) However, K.E. is inversely proportional to moment of inertia.
Kerala CEE 2021
Rotational Motion
150309
If ' \(R\) ' is the radius of orbit of a satellite, then the kinetic energy of the satellite is
1 \(\propto \frac{1}{\mathrm{R}}\)
2 \(\propto \frac{1}{\sqrt{\mathrm{R}}}\)
3 \(\propto \mathrm{R}\)
4 \(\propto \frac{1}{\mathrm{R}^{3 / 2}}\)
Explanation:
A Given, radius of orbit of satellite \(=\mathrm{R}\) We know, Kinetic Energy (K.E.) \(=\frac{1}{2} \mathrm{mv}_{0}{ }^{2}\) Where, \(\mathrm{v}_{0}=\) Orbital velocity \(\because \quad \mathrm{v}_{0}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\) Then, K.E. \(=\frac{1}{2} \mathrm{mv}_{0}^{2}\) K.E. \(=\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right)^{2}\) \(=\frac{\mathrm{GMm}}{2 \mathrm{R}}\) So, \(\quad\) K.E. \(\propto \frac{1}{\mathrm{R}}\)
