NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138616
A satellite of mass ' $m$ ' is revolving around the earth of mass ' $M$ ' in an orbit of radius ' $r$ ' with constant angular velocity ' $\omega$ '. The angular momentum of the satellite is ( $G$ = gravitational constant)
B A satellite of mass $\mathrm{m}$ is revolving around the earth of mass $\mathrm{M}$ in an orbit of radius $\mathrm{r}$ with constant angular velocity $\omega$. The centripetal force of acting on the satellite is balanced by the gravitational force due to earth. $\therefore$ Centripetal force $=$ Gravitational force $\frac{\mathrm{mv}^{2}}{\mathrm{r}} =\frac{\mathrm{GMm}}{\mathrm{r}^{2}}$ $\mathrm{v}^{2} =\frac{\mathrm{GM}}{\mathrm{r}}$ $\mathrm{v} =\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$ Angular momentum of satellite about the centre of orbit $\mathrm{L}=\mathrm{mvr}$ $=\mathrm{m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\right) \mathrm{r}$ $=\mathrm{m} \sqrt{\mathrm{GM}} \times \mathrm{r} \times \mathrm{r}^{-1 / 2}$ $=\mathrm{m}(\mathrm{GM})^{\frac{1}{2}} \times \mathrm{r}^{1-\frac{1}{2}}$ $=\mathrm{m}(\mathrm{GM})^{\frac{1}{2}} \times \mathrm{r}^{\frac{1}{2}}$ $=\mathrm{m}(\mathrm{GMr})^{\frac{1}{2}}$
MHT-CET 2020
Gravitation
138617
Earth revolves round the sun in a circular orbit of radius ' $R$ '. The angular momentum of the revolving earth is directly proportional to
1 $\mathrm{R}$
2 $\sqrt{\mathrm{R}}$
3 $\mathrm{R}^{2}$
4 $\mathrm{R}^{3}$
Explanation:
B The angular momentum of the earth. Angular momentum $=\mathrm{mvR}$ $=\mathrm{m} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}} \times \mathrm{R} \quad\left\{\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right\}$ $=\mathrm{m} \times \sqrt{\mathrm{GM}} \times \mathrm{R}^{1-\frac{1}{2}}$ $=\mathrm{m} \times \sqrt{\mathrm{GM}} \times \mathrm{R}^{\frac{1}{2}}$ $\mathrm{~L}=\mathrm{m} \sqrt{\mathrm{GMR}}$ $\mathrm{L} \propto \sqrt{\mathrm{R}} .$ So, Angular momentum of the earth is directly proportional to $\sqrt{\mathrm{R}}$.
MHT-CET 2020
Gravitation
138618
A geostationary satellite is orbiting the earth at a height $6 R$ above the surface of the earth, where $R$ is the radius of the earth. This time period of another satellite at a height $(2.5 \mathrm{R})$ from the surface of the earth is
1 10 hours
2 $6 \sqrt{2}$ hours
3 $6 \sqrt{3}$ hours
4 6 hours
Explanation:
B Given that, A geostationary satellite far away from the earth $=6 R$ Radius of first satellite $\left(\mathrm{R}_{1}\right)=\mathrm{R}+6 \mathrm{R}=7 \mathrm{R}$ Radius of second satellite $\left(\mathrm{R}_{2}\right)=\mathrm{R}+2.5 \mathrm{R}=3.5 \mathrm{R}$ From the Kepler's third law- We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{3.5 \mathrm{R}}{7 \mathrm{R}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\sqrt{\frac{1}{8}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{2 \sqrt{2}}$ $\mathrm{~T}_{2}=\frac{24}{2 \sqrt{2}}$ $\mathrm{~T}_{2}=6 \sqrt{2} \text { hours. }$
UPSEE 2002
Gravitation
138619
The time period of an earth satellite in circular orbit is independent of
1 radius of the orbit.
2 neither the mass of the satellite nor the radius of the orbit.
3 the mass of the satellite.
4 both the mass of satellite and radius of the orbit.
Explanation:
C We know that, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}_{0}}$ $\mathrm{~T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ Hence, it is clear that time period of an earth satellite in circular orbit is independent of mass of the satellite.
138616
A satellite of mass ' $m$ ' is revolving around the earth of mass ' $M$ ' in an orbit of radius ' $r$ ' with constant angular velocity ' $\omega$ '. The angular momentum of the satellite is ( $G$ = gravitational constant)
B A satellite of mass $\mathrm{m}$ is revolving around the earth of mass $\mathrm{M}$ in an orbit of radius $\mathrm{r}$ with constant angular velocity $\omega$. The centripetal force of acting on the satellite is balanced by the gravitational force due to earth. $\therefore$ Centripetal force $=$ Gravitational force $\frac{\mathrm{mv}^{2}}{\mathrm{r}} =\frac{\mathrm{GMm}}{\mathrm{r}^{2}}$ $\mathrm{v}^{2} =\frac{\mathrm{GM}}{\mathrm{r}}$ $\mathrm{v} =\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$ Angular momentum of satellite about the centre of orbit $\mathrm{L}=\mathrm{mvr}$ $=\mathrm{m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\right) \mathrm{r}$ $=\mathrm{m} \sqrt{\mathrm{GM}} \times \mathrm{r} \times \mathrm{r}^{-1 / 2}$ $=\mathrm{m}(\mathrm{GM})^{\frac{1}{2}} \times \mathrm{r}^{1-\frac{1}{2}}$ $=\mathrm{m}(\mathrm{GM})^{\frac{1}{2}} \times \mathrm{r}^{\frac{1}{2}}$ $=\mathrm{m}(\mathrm{GMr})^{\frac{1}{2}}$
MHT-CET 2020
Gravitation
138617
Earth revolves round the sun in a circular orbit of radius ' $R$ '. The angular momentum of the revolving earth is directly proportional to
1 $\mathrm{R}$
2 $\sqrt{\mathrm{R}}$
3 $\mathrm{R}^{2}$
4 $\mathrm{R}^{3}$
Explanation:
B The angular momentum of the earth. Angular momentum $=\mathrm{mvR}$ $=\mathrm{m} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}} \times \mathrm{R} \quad\left\{\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right\}$ $=\mathrm{m} \times \sqrt{\mathrm{GM}} \times \mathrm{R}^{1-\frac{1}{2}}$ $=\mathrm{m} \times \sqrt{\mathrm{GM}} \times \mathrm{R}^{\frac{1}{2}}$ $\mathrm{~L}=\mathrm{m} \sqrt{\mathrm{GMR}}$ $\mathrm{L} \propto \sqrt{\mathrm{R}} .$ So, Angular momentum of the earth is directly proportional to $\sqrt{\mathrm{R}}$.
MHT-CET 2020
Gravitation
138618
A geostationary satellite is orbiting the earth at a height $6 R$ above the surface of the earth, where $R$ is the radius of the earth. This time period of another satellite at a height $(2.5 \mathrm{R})$ from the surface of the earth is
1 10 hours
2 $6 \sqrt{2}$ hours
3 $6 \sqrt{3}$ hours
4 6 hours
Explanation:
B Given that, A geostationary satellite far away from the earth $=6 R$ Radius of first satellite $\left(\mathrm{R}_{1}\right)=\mathrm{R}+6 \mathrm{R}=7 \mathrm{R}$ Radius of second satellite $\left(\mathrm{R}_{2}\right)=\mathrm{R}+2.5 \mathrm{R}=3.5 \mathrm{R}$ From the Kepler's third law- We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{3.5 \mathrm{R}}{7 \mathrm{R}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\sqrt{\frac{1}{8}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{2 \sqrt{2}}$ $\mathrm{~T}_{2}=\frac{24}{2 \sqrt{2}}$ $\mathrm{~T}_{2}=6 \sqrt{2} \text { hours. }$
UPSEE 2002
Gravitation
138619
The time period of an earth satellite in circular orbit is independent of
1 radius of the orbit.
2 neither the mass of the satellite nor the radius of the orbit.
3 the mass of the satellite.
4 both the mass of satellite and radius of the orbit.
Explanation:
C We know that, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}_{0}}$ $\mathrm{~T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ Hence, it is clear that time period of an earth satellite in circular orbit is independent of mass of the satellite.
138616
A satellite of mass ' $m$ ' is revolving around the earth of mass ' $M$ ' in an orbit of radius ' $r$ ' with constant angular velocity ' $\omega$ '. The angular momentum of the satellite is ( $G$ = gravitational constant)
B A satellite of mass $\mathrm{m}$ is revolving around the earth of mass $\mathrm{M}$ in an orbit of radius $\mathrm{r}$ with constant angular velocity $\omega$. The centripetal force of acting on the satellite is balanced by the gravitational force due to earth. $\therefore$ Centripetal force $=$ Gravitational force $\frac{\mathrm{mv}^{2}}{\mathrm{r}} =\frac{\mathrm{GMm}}{\mathrm{r}^{2}}$ $\mathrm{v}^{2} =\frac{\mathrm{GM}}{\mathrm{r}}$ $\mathrm{v} =\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$ Angular momentum of satellite about the centre of orbit $\mathrm{L}=\mathrm{mvr}$ $=\mathrm{m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\right) \mathrm{r}$ $=\mathrm{m} \sqrt{\mathrm{GM}} \times \mathrm{r} \times \mathrm{r}^{-1 / 2}$ $=\mathrm{m}(\mathrm{GM})^{\frac{1}{2}} \times \mathrm{r}^{1-\frac{1}{2}}$ $=\mathrm{m}(\mathrm{GM})^{\frac{1}{2}} \times \mathrm{r}^{\frac{1}{2}}$ $=\mathrm{m}(\mathrm{GMr})^{\frac{1}{2}}$
MHT-CET 2020
Gravitation
138617
Earth revolves round the sun in a circular orbit of radius ' $R$ '. The angular momentum of the revolving earth is directly proportional to
1 $\mathrm{R}$
2 $\sqrt{\mathrm{R}}$
3 $\mathrm{R}^{2}$
4 $\mathrm{R}^{3}$
Explanation:
B The angular momentum of the earth. Angular momentum $=\mathrm{mvR}$ $=\mathrm{m} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}} \times \mathrm{R} \quad\left\{\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right\}$ $=\mathrm{m} \times \sqrt{\mathrm{GM}} \times \mathrm{R}^{1-\frac{1}{2}}$ $=\mathrm{m} \times \sqrt{\mathrm{GM}} \times \mathrm{R}^{\frac{1}{2}}$ $\mathrm{~L}=\mathrm{m} \sqrt{\mathrm{GMR}}$ $\mathrm{L} \propto \sqrt{\mathrm{R}} .$ So, Angular momentum of the earth is directly proportional to $\sqrt{\mathrm{R}}$.
MHT-CET 2020
Gravitation
138618
A geostationary satellite is orbiting the earth at a height $6 R$ above the surface of the earth, where $R$ is the radius of the earth. This time period of another satellite at a height $(2.5 \mathrm{R})$ from the surface of the earth is
1 10 hours
2 $6 \sqrt{2}$ hours
3 $6 \sqrt{3}$ hours
4 6 hours
Explanation:
B Given that, A geostationary satellite far away from the earth $=6 R$ Radius of first satellite $\left(\mathrm{R}_{1}\right)=\mathrm{R}+6 \mathrm{R}=7 \mathrm{R}$ Radius of second satellite $\left(\mathrm{R}_{2}\right)=\mathrm{R}+2.5 \mathrm{R}=3.5 \mathrm{R}$ From the Kepler's third law- We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{3.5 \mathrm{R}}{7 \mathrm{R}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\sqrt{\frac{1}{8}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{2 \sqrt{2}}$ $\mathrm{~T}_{2}=\frac{24}{2 \sqrt{2}}$ $\mathrm{~T}_{2}=6 \sqrt{2} \text { hours. }$
UPSEE 2002
Gravitation
138619
The time period of an earth satellite in circular orbit is independent of
1 radius of the orbit.
2 neither the mass of the satellite nor the radius of the orbit.
3 the mass of the satellite.
4 both the mass of satellite and radius of the orbit.
Explanation:
C We know that, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}_{0}}$ $\mathrm{~T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ Hence, it is clear that time period of an earth satellite in circular orbit is independent of mass of the satellite.
138616
A satellite of mass ' $m$ ' is revolving around the earth of mass ' $M$ ' in an orbit of radius ' $r$ ' with constant angular velocity ' $\omega$ '. The angular momentum of the satellite is ( $G$ = gravitational constant)
B A satellite of mass $\mathrm{m}$ is revolving around the earth of mass $\mathrm{M}$ in an orbit of radius $\mathrm{r}$ with constant angular velocity $\omega$. The centripetal force of acting on the satellite is balanced by the gravitational force due to earth. $\therefore$ Centripetal force $=$ Gravitational force $\frac{\mathrm{mv}^{2}}{\mathrm{r}} =\frac{\mathrm{GMm}}{\mathrm{r}^{2}}$ $\mathrm{v}^{2} =\frac{\mathrm{GM}}{\mathrm{r}}$ $\mathrm{v} =\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}$ Angular momentum of satellite about the centre of orbit $\mathrm{L}=\mathrm{mvr}$ $=\mathrm{m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}}\right) \mathrm{r}$ $=\mathrm{m} \sqrt{\mathrm{GM}} \times \mathrm{r} \times \mathrm{r}^{-1 / 2}$ $=\mathrm{m}(\mathrm{GM})^{\frac{1}{2}} \times \mathrm{r}^{1-\frac{1}{2}}$ $=\mathrm{m}(\mathrm{GM})^{\frac{1}{2}} \times \mathrm{r}^{\frac{1}{2}}$ $=\mathrm{m}(\mathrm{GMr})^{\frac{1}{2}}$
MHT-CET 2020
Gravitation
138617
Earth revolves round the sun in a circular orbit of radius ' $R$ '. The angular momentum of the revolving earth is directly proportional to
1 $\mathrm{R}$
2 $\sqrt{\mathrm{R}}$
3 $\mathrm{R}^{2}$
4 $\mathrm{R}^{3}$
Explanation:
B The angular momentum of the earth. Angular momentum $=\mathrm{mvR}$ $=\mathrm{m} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}} \times \mathrm{R} \quad\left\{\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right\}$ $=\mathrm{m} \times \sqrt{\mathrm{GM}} \times \mathrm{R}^{1-\frac{1}{2}}$ $=\mathrm{m} \times \sqrt{\mathrm{GM}} \times \mathrm{R}^{\frac{1}{2}}$ $\mathrm{~L}=\mathrm{m} \sqrt{\mathrm{GMR}}$ $\mathrm{L} \propto \sqrt{\mathrm{R}} .$ So, Angular momentum of the earth is directly proportional to $\sqrt{\mathrm{R}}$.
MHT-CET 2020
Gravitation
138618
A geostationary satellite is orbiting the earth at a height $6 R$ above the surface of the earth, where $R$ is the radius of the earth. This time period of another satellite at a height $(2.5 \mathrm{R})$ from the surface of the earth is
1 10 hours
2 $6 \sqrt{2}$ hours
3 $6 \sqrt{3}$ hours
4 6 hours
Explanation:
B Given that, A geostationary satellite far away from the earth $=6 R$ Radius of first satellite $\left(\mathrm{R}_{1}\right)=\mathrm{R}+6 \mathrm{R}=7 \mathrm{R}$ Radius of second satellite $\left(\mathrm{R}_{2}\right)=\mathrm{R}+2.5 \mathrm{R}=3.5 \mathrm{R}$ From the Kepler's third law- We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{3.5 \mathrm{R}}{7 \mathrm{R}}\right)^{3}$ $\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}=\left(\frac{1}{2}\right)^{3}$ $\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\sqrt{\frac{1}{8}}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{2 \sqrt{2}}$ $\mathrm{~T}_{2}=\frac{24}{2 \sqrt{2}}$ $\mathrm{~T}_{2}=6 \sqrt{2} \text { hours. }$
UPSEE 2002
Gravitation
138619
The time period of an earth satellite in circular orbit is independent of
1 radius of the orbit.
2 neither the mass of the satellite nor the radius of the orbit.
3 the mass of the satellite.
4 both the mass of satellite and radius of the orbit.
Explanation:
C We know that, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}_{0}}$ $\mathrm{~T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ Hence, it is clear that time period of an earth satellite in circular orbit is independent of mass of the satellite.
