138714
The earth rotates on its axis takes 24 hours to complete one revolution. How much time it takes at sun from earth to have shift of $1^{0}$ ?
1 $4 \mathrm{~min}$.
2 $4 \mathrm{hrs}$.
3 $4 \mathrm{sec}$.
4 $24 \mathrm{hrs}$.
Explanation:
A According to question, Time taken for $360^{\circ}$ shift $=24$ hours $=1440 \mathrm{~min}$. Since for a complete rotation $\left(360^{\circ}\right)$ the earth takes 24 hours, Then, time taken for $1^{\circ}=\frac{24 \mathrm{~h}}{360^{\circ}}=\frac{1440 \mathrm{~min}}{360^{\circ}}$ $=4 \mathrm{~min}$
GUJCET 2020
Gravitation
138715
The radius of a planet is twice the radius of earth. Both have almost equal average massdensities. $V_{P}$ and $V_{E}$ are escape velocities of the planet and the earth, respectively, then
B For earth, $\text { Escape velocity }\left(\mathrm{V}_{\mathrm{E}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}}$ Where, Mass $=$ density $\times$ volume $\mathrm{M}_{\mathrm{E}}=\rho_{\mathrm{E}} \times \frac{4}{3} \pi \mathrm{R}_{\mathrm{E}}^{3}$ So, from equation (i) $\mathrm{V}_{\mathrm{E}}=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}_{\mathrm{E}}} \times\left(\frac{4}{3} \pi \mathrm{R}_{\mathrm{E}}^{3}\right) \times \rho_{\mathrm{E}}}$ $\mathrm{V}_{\mathrm{E}}=\mathrm{R}_{\mathrm{E}} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{E}}}$ For planet, $V_{P}=R_{P} \sqrt{\frac{8}{3} G \pi \rho_{P}}$ From equation (ii) and (iii) $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{R}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{P}}} \sqrt{\frac{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{E}}}{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{P}}}}$ Given, $\quad R_{P}=2 R_{E}, \rho_{E}=\rho_{P}$ $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{R}_{\mathrm{E}}}{2 \mathrm{R}_{\mathrm{E}}} \sqrt{\frac{\rho_{\mathrm{E}}}{\rho_{\mathrm{E}}}}$ $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{1}{2}$ $\mathrm{~V}_{\mathrm{P}}=2 \mathrm{~V}_{\mathrm{E}}$
COMEDK 2013
Gravitation
138716
A satellite can be in a geostationary orbit around a planet if it is at a distance $R$ from the centre of the planet. If the planet starts rotating about its axis with double the angular velocity, then to make the satellite geostationary, its orbital radius should be
1 $2 \mathrm{R}$
2 $\frac{R}{2}$
3 $\frac{\mathrm{R}}{\frac{1}{3}}$
4 $\frac{\mathrm{R}}{4^{\frac{1}{3}}}$
Explanation:
D Given, $\mathrm{R}_{1}=\mathrm{R}$ We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ ${\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=\left[\frac{\mathrm{R}}{\mathrm{R}_{2}}\right]^{3}}$ According to the equation, From equation (i) $\mathrm{T}_{1}=\frac{2 \pi}{\omega} \text { and } \mathrm{T}_{2}=\frac{2 \pi}{2 \omega}$ $\frac{\left(\frac{2 \pi}{\omega}\right)^{2}}{\left(\frac{2 \pi}{2 \omega}\right)^{2}}=\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\frac{4 \pi^{2}}{\omega^{2}}}{\frac{4 \pi^{2}}{4 \omega^{2}}}=\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}$ $\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}=4$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=(4)^{1 / 3}$ $\mathrm{R}_{2}=\frac{\mathrm{R}}{(4)^{1 / 3}}$ $\therefore \quad \mathrm{R}_{2}=\frac{\mathrm{R}}{(4)^{1 / 3}}$
COMEDK 2020
Gravitation
138717
The radius of a planet is $\frac{1}{4}$ th of the earth's radius and its acceleration due to gravity is double that of the earth. What is the value of escape velocity at the planet as compared to its value on the earth?
1 $\frac{1}{\sqrt{2}}$
2 $\sqrt{2}$
3 2
4 $2 \sqrt{2}$
Explanation:
A Given that, $\mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{e}}}{4}, \mathrm{~g}_{\mathrm{P}}=2 \mathrm{~g}_{\mathrm{e}}$ We know that, Escape velocity of earth $v_{e}=\sqrt{2 g_{e} R_{e}}$ Escape velocity of planet $v_{P}=\sqrt{2 g_{P} R_{P}}$ $\therefore \frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{2 \times \mathrm{g}_{\mathrm{p}} \mathrm{R}_{\mathrm{P}}}{2 \times \mathrm{g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}}=\sqrt{\frac{2 \times 2 \mathrm{~g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}{4 \times 2 \mathrm{~g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}}$ $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{2}{4}}$ $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{1}{\sqrt{2}}$
COMEDK 2019
Gravitation
138719
Assertion: If an object is projected from earth surface with escape velocity path of object will be parabola. Reason: When object is projected with velocity less than escape velocity from horizontal surface and greater than orbital velocity path of object will be ellipse.
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 When the velocity of a object is equal to the escape velocity, the path of object will be parabolic. When the velocity of object is less than the escape velocity, the path of object will be elliptical.
138714
The earth rotates on its axis takes 24 hours to complete one revolution. How much time it takes at sun from earth to have shift of $1^{0}$ ?
1 $4 \mathrm{~min}$.
2 $4 \mathrm{hrs}$.
3 $4 \mathrm{sec}$.
4 $24 \mathrm{hrs}$.
Explanation:
A According to question, Time taken for $360^{\circ}$ shift $=24$ hours $=1440 \mathrm{~min}$. Since for a complete rotation $\left(360^{\circ}\right)$ the earth takes 24 hours, Then, time taken for $1^{\circ}=\frac{24 \mathrm{~h}}{360^{\circ}}=\frac{1440 \mathrm{~min}}{360^{\circ}}$ $=4 \mathrm{~min}$
GUJCET 2020
Gravitation
138715
The radius of a planet is twice the radius of earth. Both have almost equal average massdensities. $V_{P}$ and $V_{E}$ are escape velocities of the planet and the earth, respectively, then
B For earth, $\text { Escape velocity }\left(\mathrm{V}_{\mathrm{E}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}}$ Where, Mass $=$ density $\times$ volume $\mathrm{M}_{\mathrm{E}}=\rho_{\mathrm{E}} \times \frac{4}{3} \pi \mathrm{R}_{\mathrm{E}}^{3}$ So, from equation (i) $\mathrm{V}_{\mathrm{E}}=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}_{\mathrm{E}}} \times\left(\frac{4}{3} \pi \mathrm{R}_{\mathrm{E}}^{3}\right) \times \rho_{\mathrm{E}}}$ $\mathrm{V}_{\mathrm{E}}=\mathrm{R}_{\mathrm{E}} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{E}}}$ For planet, $V_{P}=R_{P} \sqrt{\frac{8}{3} G \pi \rho_{P}}$ From equation (ii) and (iii) $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{R}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{P}}} \sqrt{\frac{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{E}}}{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{P}}}}$ Given, $\quad R_{P}=2 R_{E}, \rho_{E}=\rho_{P}$ $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{R}_{\mathrm{E}}}{2 \mathrm{R}_{\mathrm{E}}} \sqrt{\frac{\rho_{\mathrm{E}}}{\rho_{\mathrm{E}}}}$ $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{1}{2}$ $\mathrm{~V}_{\mathrm{P}}=2 \mathrm{~V}_{\mathrm{E}}$
COMEDK 2013
Gravitation
138716
A satellite can be in a geostationary orbit around a planet if it is at a distance $R$ from the centre of the planet. If the planet starts rotating about its axis with double the angular velocity, then to make the satellite geostationary, its orbital radius should be
1 $2 \mathrm{R}$
2 $\frac{R}{2}$
3 $\frac{\mathrm{R}}{\frac{1}{3}}$
4 $\frac{\mathrm{R}}{4^{\frac{1}{3}}}$
Explanation:
D Given, $\mathrm{R}_{1}=\mathrm{R}$ We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ ${\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=\left[\frac{\mathrm{R}}{\mathrm{R}_{2}}\right]^{3}}$ According to the equation, From equation (i) $\mathrm{T}_{1}=\frac{2 \pi}{\omega} \text { and } \mathrm{T}_{2}=\frac{2 \pi}{2 \omega}$ $\frac{\left(\frac{2 \pi}{\omega}\right)^{2}}{\left(\frac{2 \pi}{2 \omega}\right)^{2}}=\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\frac{4 \pi^{2}}{\omega^{2}}}{\frac{4 \pi^{2}}{4 \omega^{2}}}=\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}$ $\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}=4$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=(4)^{1 / 3}$ $\mathrm{R}_{2}=\frac{\mathrm{R}}{(4)^{1 / 3}}$ $\therefore \quad \mathrm{R}_{2}=\frac{\mathrm{R}}{(4)^{1 / 3}}$
COMEDK 2020
Gravitation
138717
The radius of a planet is $\frac{1}{4}$ th of the earth's radius and its acceleration due to gravity is double that of the earth. What is the value of escape velocity at the planet as compared to its value on the earth?
1 $\frac{1}{\sqrt{2}}$
2 $\sqrt{2}$
3 2
4 $2 \sqrt{2}$
Explanation:
A Given that, $\mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{e}}}{4}, \mathrm{~g}_{\mathrm{P}}=2 \mathrm{~g}_{\mathrm{e}}$ We know that, Escape velocity of earth $v_{e}=\sqrt{2 g_{e} R_{e}}$ Escape velocity of planet $v_{P}=\sqrt{2 g_{P} R_{P}}$ $\therefore \frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{2 \times \mathrm{g}_{\mathrm{p}} \mathrm{R}_{\mathrm{P}}}{2 \times \mathrm{g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}}=\sqrt{\frac{2 \times 2 \mathrm{~g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}{4 \times 2 \mathrm{~g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}}$ $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{2}{4}}$ $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{1}{\sqrt{2}}$
COMEDK 2019
Gravitation
138719
Assertion: If an object is projected from earth surface with escape velocity path of object will be parabola. Reason: When object is projected with velocity less than escape velocity from horizontal surface and greater than orbital velocity path of object will be ellipse.
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 When the velocity of a object is equal to the escape velocity, the path of object will be parabolic. When the velocity of object is less than the escape velocity, the path of object will be elliptical.
138714
The earth rotates on its axis takes 24 hours to complete one revolution. How much time it takes at sun from earth to have shift of $1^{0}$ ?
1 $4 \mathrm{~min}$.
2 $4 \mathrm{hrs}$.
3 $4 \mathrm{sec}$.
4 $24 \mathrm{hrs}$.
Explanation:
A According to question, Time taken for $360^{\circ}$ shift $=24$ hours $=1440 \mathrm{~min}$. Since for a complete rotation $\left(360^{\circ}\right)$ the earth takes 24 hours, Then, time taken for $1^{\circ}=\frac{24 \mathrm{~h}}{360^{\circ}}=\frac{1440 \mathrm{~min}}{360^{\circ}}$ $=4 \mathrm{~min}$
GUJCET 2020
Gravitation
138715
The radius of a planet is twice the radius of earth. Both have almost equal average massdensities. $V_{P}$ and $V_{E}$ are escape velocities of the planet and the earth, respectively, then
B For earth, $\text { Escape velocity }\left(\mathrm{V}_{\mathrm{E}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}}$ Where, Mass $=$ density $\times$ volume $\mathrm{M}_{\mathrm{E}}=\rho_{\mathrm{E}} \times \frac{4}{3} \pi \mathrm{R}_{\mathrm{E}}^{3}$ So, from equation (i) $\mathrm{V}_{\mathrm{E}}=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}_{\mathrm{E}}} \times\left(\frac{4}{3} \pi \mathrm{R}_{\mathrm{E}}^{3}\right) \times \rho_{\mathrm{E}}}$ $\mathrm{V}_{\mathrm{E}}=\mathrm{R}_{\mathrm{E}} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{E}}}$ For planet, $V_{P}=R_{P} \sqrt{\frac{8}{3} G \pi \rho_{P}}$ From equation (ii) and (iii) $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{R}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{P}}} \sqrt{\frac{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{E}}}{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{P}}}}$ Given, $\quad R_{P}=2 R_{E}, \rho_{E}=\rho_{P}$ $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{R}_{\mathrm{E}}}{2 \mathrm{R}_{\mathrm{E}}} \sqrt{\frac{\rho_{\mathrm{E}}}{\rho_{\mathrm{E}}}}$ $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{1}{2}$ $\mathrm{~V}_{\mathrm{P}}=2 \mathrm{~V}_{\mathrm{E}}$
COMEDK 2013
Gravitation
138716
A satellite can be in a geostationary orbit around a planet if it is at a distance $R$ from the centre of the planet. If the planet starts rotating about its axis with double the angular velocity, then to make the satellite geostationary, its orbital radius should be
1 $2 \mathrm{R}$
2 $\frac{R}{2}$
3 $\frac{\mathrm{R}}{\frac{1}{3}}$
4 $\frac{\mathrm{R}}{4^{\frac{1}{3}}}$
Explanation:
D Given, $\mathrm{R}_{1}=\mathrm{R}$ We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ ${\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=\left[\frac{\mathrm{R}}{\mathrm{R}_{2}}\right]^{3}}$ According to the equation, From equation (i) $\mathrm{T}_{1}=\frac{2 \pi}{\omega} \text { and } \mathrm{T}_{2}=\frac{2 \pi}{2 \omega}$ $\frac{\left(\frac{2 \pi}{\omega}\right)^{2}}{\left(\frac{2 \pi}{2 \omega}\right)^{2}}=\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\frac{4 \pi^{2}}{\omega^{2}}}{\frac{4 \pi^{2}}{4 \omega^{2}}}=\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}$ $\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}=4$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=(4)^{1 / 3}$ $\mathrm{R}_{2}=\frac{\mathrm{R}}{(4)^{1 / 3}}$ $\therefore \quad \mathrm{R}_{2}=\frac{\mathrm{R}}{(4)^{1 / 3}}$
COMEDK 2020
Gravitation
138717
The radius of a planet is $\frac{1}{4}$ th of the earth's radius and its acceleration due to gravity is double that of the earth. What is the value of escape velocity at the planet as compared to its value on the earth?
1 $\frac{1}{\sqrt{2}}$
2 $\sqrt{2}$
3 2
4 $2 \sqrt{2}$
Explanation:
A Given that, $\mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{e}}}{4}, \mathrm{~g}_{\mathrm{P}}=2 \mathrm{~g}_{\mathrm{e}}$ We know that, Escape velocity of earth $v_{e}=\sqrt{2 g_{e} R_{e}}$ Escape velocity of planet $v_{P}=\sqrt{2 g_{P} R_{P}}$ $\therefore \frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{2 \times \mathrm{g}_{\mathrm{p}} \mathrm{R}_{\mathrm{P}}}{2 \times \mathrm{g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}}=\sqrt{\frac{2 \times 2 \mathrm{~g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}{4 \times 2 \mathrm{~g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}}$ $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{2}{4}}$ $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{1}{\sqrt{2}}$
COMEDK 2019
Gravitation
138719
Assertion: If an object is projected from earth surface with escape velocity path of object will be parabola. Reason: When object is projected with velocity less than escape velocity from horizontal surface and greater than orbital velocity path of object will be ellipse.
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 When the velocity of a object is equal to the escape velocity, the path of object will be parabolic. When the velocity of object is less than the escape velocity, the path of object will be elliptical.
138714
The earth rotates on its axis takes 24 hours to complete one revolution. How much time it takes at sun from earth to have shift of $1^{0}$ ?
1 $4 \mathrm{~min}$.
2 $4 \mathrm{hrs}$.
3 $4 \mathrm{sec}$.
4 $24 \mathrm{hrs}$.
Explanation:
A According to question, Time taken for $360^{\circ}$ shift $=24$ hours $=1440 \mathrm{~min}$. Since for a complete rotation $\left(360^{\circ}\right)$ the earth takes 24 hours, Then, time taken for $1^{\circ}=\frac{24 \mathrm{~h}}{360^{\circ}}=\frac{1440 \mathrm{~min}}{360^{\circ}}$ $=4 \mathrm{~min}$
GUJCET 2020
Gravitation
138715
The radius of a planet is twice the radius of earth. Both have almost equal average massdensities. $V_{P}$ and $V_{E}$ are escape velocities of the planet and the earth, respectively, then
B For earth, $\text { Escape velocity }\left(\mathrm{V}_{\mathrm{E}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}}$ Where, Mass $=$ density $\times$ volume $\mathrm{M}_{\mathrm{E}}=\rho_{\mathrm{E}} \times \frac{4}{3} \pi \mathrm{R}_{\mathrm{E}}^{3}$ So, from equation (i) $\mathrm{V}_{\mathrm{E}}=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}_{\mathrm{E}}} \times\left(\frac{4}{3} \pi \mathrm{R}_{\mathrm{E}}^{3}\right) \times \rho_{\mathrm{E}}}$ $\mathrm{V}_{\mathrm{E}}=\mathrm{R}_{\mathrm{E}} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{E}}}$ For planet, $V_{P}=R_{P} \sqrt{\frac{8}{3} G \pi \rho_{P}}$ From equation (ii) and (iii) $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{R}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{P}}} \sqrt{\frac{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{E}}}{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{P}}}}$ Given, $\quad R_{P}=2 R_{E}, \rho_{E}=\rho_{P}$ $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{R}_{\mathrm{E}}}{2 \mathrm{R}_{\mathrm{E}}} \sqrt{\frac{\rho_{\mathrm{E}}}{\rho_{\mathrm{E}}}}$ $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{1}{2}$ $\mathrm{~V}_{\mathrm{P}}=2 \mathrm{~V}_{\mathrm{E}}$
COMEDK 2013
Gravitation
138716
A satellite can be in a geostationary orbit around a planet if it is at a distance $R$ from the centre of the planet. If the planet starts rotating about its axis with double the angular velocity, then to make the satellite geostationary, its orbital radius should be
1 $2 \mathrm{R}$
2 $\frac{R}{2}$
3 $\frac{\mathrm{R}}{\frac{1}{3}}$
4 $\frac{\mathrm{R}}{4^{\frac{1}{3}}}$
Explanation:
D Given, $\mathrm{R}_{1}=\mathrm{R}$ We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ ${\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=\left[\frac{\mathrm{R}}{\mathrm{R}_{2}}\right]^{3}}$ According to the equation, From equation (i) $\mathrm{T}_{1}=\frac{2 \pi}{\omega} \text { and } \mathrm{T}_{2}=\frac{2 \pi}{2 \omega}$ $\frac{\left(\frac{2 \pi}{\omega}\right)^{2}}{\left(\frac{2 \pi}{2 \omega}\right)^{2}}=\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\frac{4 \pi^{2}}{\omega^{2}}}{\frac{4 \pi^{2}}{4 \omega^{2}}}=\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}$ $\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}=4$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=(4)^{1 / 3}$ $\mathrm{R}_{2}=\frac{\mathrm{R}}{(4)^{1 / 3}}$ $\therefore \quad \mathrm{R}_{2}=\frac{\mathrm{R}}{(4)^{1 / 3}}$
COMEDK 2020
Gravitation
138717
The radius of a planet is $\frac{1}{4}$ th of the earth's radius and its acceleration due to gravity is double that of the earth. What is the value of escape velocity at the planet as compared to its value on the earth?
1 $\frac{1}{\sqrt{2}}$
2 $\sqrt{2}$
3 2
4 $2 \sqrt{2}$
Explanation:
A Given that, $\mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{e}}}{4}, \mathrm{~g}_{\mathrm{P}}=2 \mathrm{~g}_{\mathrm{e}}$ We know that, Escape velocity of earth $v_{e}=\sqrt{2 g_{e} R_{e}}$ Escape velocity of planet $v_{P}=\sqrt{2 g_{P} R_{P}}$ $\therefore \frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{2 \times \mathrm{g}_{\mathrm{p}} \mathrm{R}_{\mathrm{P}}}{2 \times \mathrm{g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}}=\sqrt{\frac{2 \times 2 \mathrm{~g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}{4 \times 2 \mathrm{~g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}}$ $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{2}{4}}$ $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{1}{\sqrt{2}}$
COMEDK 2019
Gravitation
138719
Assertion: If an object is projected from earth surface with escape velocity path of object will be parabola. Reason: When object is projected with velocity less than escape velocity from horizontal surface and greater than orbital velocity path of object will be ellipse.
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 When the velocity of a object is equal to the escape velocity, the path of object will be parabolic. When the velocity of object is less than the escape velocity, the path of object will be elliptical.
138714
The earth rotates on its axis takes 24 hours to complete one revolution. How much time it takes at sun from earth to have shift of $1^{0}$ ?
1 $4 \mathrm{~min}$.
2 $4 \mathrm{hrs}$.
3 $4 \mathrm{sec}$.
4 $24 \mathrm{hrs}$.
Explanation:
A According to question, Time taken for $360^{\circ}$ shift $=24$ hours $=1440 \mathrm{~min}$. Since for a complete rotation $\left(360^{\circ}\right)$ the earth takes 24 hours, Then, time taken for $1^{\circ}=\frac{24 \mathrm{~h}}{360^{\circ}}=\frac{1440 \mathrm{~min}}{360^{\circ}}$ $=4 \mathrm{~min}$
GUJCET 2020
Gravitation
138715
The radius of a planet is twice the radius of earth. Both have almost equal average massdensities. $V_{P}$ and $V_{E}$ are escape velocities of the planet and the earth, respectively, then
B For earth, $\text { Escape velocity }\left(\mathrm{V}_{\mathrm{E}}\right)=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}}$ Where, Mass $=$ density $\times$ volume $\mathrm{M}_{\mathrm{E}}=\rho_{\mathrm{E}} \times \frac{4}{3} \pi \mathrm{R}_{\mathrm{E}}^{3}$ So, from equation (i) $\mathrm{V}_{\mathrm{E}}=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}_{\mathrm{E}}} \times\left(\frac{4}{3} \pi \mathrm{R}_{\mathrm{E}}^{3}\right) \times \rho_{\mathrm{E}}}$ $\mathrm{V}_{\mathrm{E}}=\mathrm{R}_{\mathrm{E}} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{E}}}$ For planet, $V_{P}=R_{P} \sqrt{\frac{8}{3} G \pi \rho_{P}}$ From equation (ii) and (iii) $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{R}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{P}}} \sqrt{\frac{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{E}}}{\frac{8}{3} \pi \mathrm{G} \rho_{\mathrm{P}}}}$ Given, $\quad R_{P}=2 R_{E}, \rho_{E}=\rho_{P}$ $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{R}_{\mathrm{E}}}{2 \mathrm{R}_{\mathrm{E}}} \sqrt{\frac{\rho_{\mathrm{E}}}{\rho_{\mathrm{E}}}}$ $\frac{\mathrm{V}_{\mathrm{E}}}{\mathrm{V}_{\mathrm{P}}}=\frac{1}{2}$ $\mathrm{~V}_{\mathrm{P}}=2 \mathrm{~V}_{\mathrm{E}}$
COMEDK 2013
Gravitation
138716
A satellite can be in a geostationary orbit around a planet if it is at a distance $R$ from the centre of the planet. If the planet starts rotating about its axis with double the angular velocity, then to make the satellite geostationary, its orbital radius should be
1 $2 \mathrm{R}$
2 $\frac{R}{2}$
3 $\frac{\mathrm{R}}{\frac{1}{3}}$
4 $\frac{\mathrm{R}}{4^{\frac{1}{3}}}$
Explanation:
D Given, $\mathrm{R}_{1}=\mathrm{R}$ We know that, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ ${\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=\left[\frac{\mathrm{R}}{\mathrm{R}_{2}}\right]^{3}}$ According to the equation, From equation (i) $\mathrm{T}_{1}=\frac{2 \pi}{\omega} \text { and } \mathrm{T}_{2}=\frac{2 \pi}{2 \omega}$ $\frac{\left(\frac{2 \pi}{\omega}\right)^{2}}{\left(\frac{2 \pi}{2 \omega}\right)^{2}}=\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\frac{4 \pi^{2}}{\omega^{2}}}{\frac{4 \pi^{2}}{4 \omega^{2}}}=\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}$ $\left(\frac{\mathrm{R}}{\mathrm{R}_{2}}\right)^{3}=4$ $\frac{\mathrm{R}}{\mathrm{R}_{2}}=(4)^{1 / 3}$ $\mathrm{R}_{2}=\frac{\mathrm{R}}{(4)^{1 / 3}}$ $\therefore \quad \mathrm{R}_{2}=\frac{\mathrm{R}}{(4)^{1 / 3}}$
COMEDK 2020
Gravitation
138717
The radius of a planet is $\frac{1}{4}$ th of the earth's radius and its acceleration due to gravity is double that of the earth. What is the value of escape velocity at the planet as compared to its value on the earth?
1 $\frac{1}{\sqrt{2}}$
2 $\sqrt{2}$
3 2
4 $2 \sqrt{2}$
Explanation:
A Given that, $\mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{e}}}{4}, \mathrm{~g}_{\mathrm{P}}=2 \mathrm{~g}_{\mathrm{e}}$ We know that, Escape velocity of earth $v_{e}=\sqrt{2 g_{e} R_{e}}$ Escape velocity of planet $v_{P}=\sqrt{2 g_{P} R_{P}}$ $\therefore \frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{2 \times \mathrm{g}_{\mathrm{p}} \mathrm{R}_{\mathrm{P}}}{2 \times \mathrm{g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}}=\sqrt{\frac{2 \times 2 \mathrm{~g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}{4 \times 2 \mathrm{~g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}}$ $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{2}{4}}$ $\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{e}}}=\frac{1}{\sqrt{2}}$
COMEDK 2019
Gravitation
138719
Assertion: If an object is projected from earth surface with escape velocity path of object will be parabola. Reason: When object is projected with velocity less than escape velocity from horizontal surface and greater than orbital velocity path of object will be ellipse.
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 When the velocity of a object is equal to the escape velocity, the path of object will be parabolic. When the velocity of object is less than the escape velocity, the path of object will be elliptical.
