NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI08:GRAVITATION
359809
A spaceship is launched into a circular orbit close to earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull? (Radius of earth \( = 6400\;km\) and \(g = 9.8\;m{s^{ - 2}}\) )
1 \(3.28\,\,km{s^{ - 1}}\)
2 \(12\,\,km{s^{ - 1}}\)
3 \(10\,\,km{s^{ - 1}}\)
4 \(40\,\,km{s^{ - 1}}\)
Explanation:
The speed of sattelite very close to earth is nearly \(8\,km{s^{ - 1}}\). The additional velocity required to make it to escape is nearly \(3.2\,km{s^{ - 1}}\).
AIIMS - 2017
PHXI08:GRAVITATION
359810
A planet having mass \(9 M_{e}\) and radius \(4 R_{e}\), where \(M_{e}\) and \(R_{e}\) are mass and radius of earth respectively, has escape velocity in \(km/s\) given by (Given escape velocity on earth \({V_e} = 11.2 \times {10^3}\;m/s\))
1 67.2
2 16.8
3 11.2
4 33.6
Explanation:
Given, \(M_{p}=9 M_{e} ; R_{p}=4 R_{e}\) Escape velocity is given by \(v_{e}=\sqrt{\dfrac{2 G M_{e}}{R_{e}}}\) \(v_{e}=\sqrt{\dfrac{2 G M_{p}}{R_{p}}}=\sqrt{\dfrac{2 G \times 9 M_{e}}{4 R_{e}}}\) \(=\sqrt{\dfrac{G M_{e}}{R_{e}} \times \dfrac{9}{2}}=\sqrt{\dfrac{9}{4}} \times \sqrt{\dfrac{2 G M_{e}}{R_{e}}}\) \( = \frac{3}{2} \times {v_e} = \left( {\frac{3}{2} \times 11.2} \right)km/s\) \({v_e} = (3 \times 5.6)km/s = 16.8\;km/s\)
JEE - 2023
PHXI08:GRAVITATION
359811
The ratio of escape velocities of two planets if \(g\) value on the two planets are \(9.9\;m{\rm{/}}{s^2}\) and \(3.3\;m{\rm{/}}{s^2}\) and their radii are \(6400\;km\) and \(3200\;km\) respectively is
359812
The escape velocities of two planets \(A\) and \(B\) are in the ratio \(1: 2\). If the ratio of their radii respectively is \(1: 3\), then the ratio of acceleration due to gravity of planet \(A\) to the acceleration of gravity of planet \(B\) will be
1 \(\dfrac{2}{3}\)
2 \(\dfrac{3}{2}\)
3 \(\dfrac{3}{4}\)
4 \(\dfrac{4}{3}\)
Explanation:
Given, \(\dfrac{v_{A}}{v_{B}}=\dfrac{1}{2}, \dfrac{R_{A}}{R_{B}}=\dfrac{1}{3}, \dfrac{g_{A}}{g_{B}}=\)? Escape velocity is given by, \(v_{e}=\sqrt{2 g R}\) \(v_{e} \propto \sqrt{g R}\) \(\Rightarrow g \propto \dfrac{V e^{2}}{R}\) So, \(\frac{{{v_A}}}{{{v_B}}} = \sqrt {\frac{{{g_A}{R_A}}}{{{g_B}{R_B}}}} \) \( \Rightarrow {\left( {\frac{1}{2}} \right)^2} = \frac{{{g_A}}}{{{g_B}}} \times \frac{1}{3} \Rightarrow \frac{{{g_A}}}{{{g_B}}} = \frac{3}{4}\)
359809
A spaceship is launched into a circular orbit close to earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull? (Radius of earth \( = 6400\;km\) and \(g = 9.8\;m{s^{ - 2}}\) )
1 \(3.28\,\,km{s^{ - 1}}\)
2 \(12\,\,km{s^{ - 1}}\)
3 \(10\,\,km{s^{ - 1}}\)
4 \(40\,\,km{s^{ - 1}}\)
Explanation:
The speed of sattelite very close to earth is nearly \(8\,km{s^{ - 1}}\). The additional velocity required to make it to escape is nearly \(3.2\,km{s^{ - 1}}\).
AIIMS - 2017
PHXI08:GRAVITATION
359810
A planet having mass \(9 M_{e}\) and radius \(4 R_{e}\), where \(M_{e}\) and \(R_{e}\) are mass and radius of earth respectively, has escape velocity in \(km/s\) given by (Given escape velocity on earth \({V_e} = 11.2 \times {10^3}\;m/s\))
1 67.2
2 16.8
3 11.2
4 33.6
Explanation:
Given, \(M_{p}=9 M_{e} ; R_{p}=4 R_{e}\) Escape velocity is given by \(v_{e}=\sqrt{\dfrac{2 G M_{e}}{R_{e}}}\) \(v_{e}=\sqrt{\dfrac{2 G M_{p}}{R_{p}}}=\sqrt{\dfrac{2 G \times 9 M_{e}}{4 R_{e}}}\) \(=\sqrt{\dfrac{G M_{e}}{R_{e}} \times \dfrac{9}{2}}=\sqrt{\dfrac{9}{4}} \times \sqrt{\dfrac{2 G M_{e}}{R_{e}}}\) \( = \frac{3}{2} \times {v_e} = \left( {\frac{3}{2} \times 11.2} \right)km/s\) \({v_e} = (3 \times 5.6)km/s = 16.8\;km/s\)
JEE - 2023
PHXI08:GRAVITATION
359811
The ratio of escape velocities of two planets if \(g\) value on the two planets are \(9.9\;m{\rm{/}}{s^2}\) and \(3.3\;m{\rm{/}}{s^2}\) and their radii are \(6400\;km\) and \(3200\;km\) respectively is
359812
The escape velocities of two planets \(A\) and \(B\) are in the ratio \(1: 2\). If the ratio of their radii respectively is \(1: 3\), then the ratio of acceleration due to gravity of planet \(A\) to the acceleration of gravity of planet \(B\) will be
1 \(\dfrac{2}{3}\)
2 \(\dfrac{3}{2}\)
3 \(\dfrac{3}{4}\)
4 \(\dfrac{4}{3}\)
Explanation:
Given, \(\dfrac{v_{A}}{v_{B}}=\dfrac{1}{2}, \dfrac{R_{A}}{R_{B}}=\dfrac{1}{3}, \dfrac{g_{A}}{g_{B}}=\)? Escape velocity is given by, \(v_{e}=\sqrt{2 g R}\) \(v_{e} \propto \sqrt{g R}\) \(\Rightarrow g \propto \dfrac{V e^{2}}{R}\) So, \(\frac{{{v_A}}}{{{v_B}}} = \sqrt {\frac{{{g_A}{R_A}}}{{{g_B}{R_B}}}} \) \( \Rightarrow {\left( {\frac{1}{2}} \right)^2} = \frac{{{g_A}}}{{{g_B}}} \times \frac{1}{3} \Rightarrow \frac{{{g_A}}}{{{g_B}}} = \frac{3}{4}\)
359809
A spaceship is launched into a circular orbit close to earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull? (Radius of earth \( = 6400\;km\) and \(g = 9.8\;m{s^{ - 2}}\) )
1 \(3.28\,\,km{s^{ - 1}}\)
2 \(12\,\,km{s^{ - 1}}\)
3 \(10\,\,km{s^{ - 1}}\)
4 \(40\,\,km{s^{ - 1}}\)
Explanation:
The speed of sattelite very close to earth is nearly \(8\,km{s^{ - 1}}\). The additional velocity required to make it to escape is nearly \(3.2\,km{s^{ - 1}}\).
AIIMS - 2017
PHXI08:GRAVITATION
359810
A planet having mass \(9 M_{e}\) and radius \(4 R_{e}\), where \(M_{e}\) and \(R_{e}\) are mass and radius of earth respectively, has escape velocity in \(km/s\) given by (Given escape velocity on earth \({V_e} = 11.2 \times {10^3}\;m/s\))
1 67.2
2 16.8
3 11.2
4 33.6
Explanation:
Given, \(M_{p}=9 M_{e} ; R_{p}=4 R_{e}\) Escape velocity is given by \(v_{e}=\sqrt{\dfrac{2 G M_{e}}{R_{e}}}\) \(v_{e}=\sqrt{\dfrac{2 G M_{p}}{R_{p}}}=\sqrt{\dfrac{2 G \times 9 M_{e}}{4 R_{e}}}\) \(=\sqrt{\dfrac{G M_{e}}{R_{e}} \times \dfrac{9}{2}}=\sqrt{\dfrac{9}{4}} \times \sqrt{\dfrac{2 G M_{e}}{R_{e}}}\) \( = \frac{3}{2} \times {v_e} = \left( {\frac{3}{2} \times 11.2} \right)km/s\) \({v_e} = (3 \times 5.6)km/s = 16.8\;km/s\)
JEE - 2023
PHXI08:GRAVITATION
359811
The ratio of escape velocities of two planets if \(g\) value on the two planets are \(9.9\;m{\rm{/}}{s^2}\) and \(3.3\;m{\rm{/}}{s^2}\) and their radii are \(6400\;km\) and \(3200\;km\) respectively is
359812
The escape velocities of two planets \(A\) and \(B\) are in the ratio \(1: 2\). If the ratio of their radii respectively is \(1: 3\), then the ratio of acceleration due to gravity of planet \(A\) to the acceleration of gravity of planet \(B\) will be
1 \(\dfrac{2}{3}\)
2 \(\dfrac{3}{2}\)
3 \(\dfrac{3}{4}\)
4 \(\dfrac{4}{3}\)
Explanation:
Given, \(\dfrac{v_{A}}{v_{B}}=\dfrac{1}{2}, \dfrac{R_{A}}{R_{B}}=\dfrac{1}{3}, \dfrac{g_{A}}{g_{B}}=\)? Escape velocity is given by, \(v_{e}=\sqrt{2 g R}\) \(v_{e} \propto \sqrt{g R}\) \(\Rightarrow g \propto \dfrac{V e^{2}}{R}\) So, \(\frac{{{v_A}}}{{{v_B}}} = \sqrt {\frac{{{g_A}{R_A}}}{{{g_B}{R_B}}}} \) \( \Rightarrow {\left( {\frac{1}{2}} \right)^2} = \frac{{{g_A}}}{{{g_B}}} \times \frac{1}{3} \Rightarrow \frac{{{g_A}}}{{{g_B}}} = \frac{3}{4}\)
359809
A spaceship is launched into a circular orbit close to earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull? (Radius of earth \( = 6400\;km\) and \(g = 9.8\;m{s^{ - 2}}\) )
1 \(3.28\,\,km{s^{ - 1}}\)
2 \(12\,\,km{s^{ - 1}}\)
3 \(10\,\,km{s^{ - 1}}\)
4 \(40\,\,km{s^{ - 1}}\)
Explanation:
The speed of sattelite very close to earth is nearly \(8\,km{s^{ - 1}}\). The additional velocity required to make it to escape is nearly \(3.2\,km{s^{ - 1}}\).
AIIMS - 2017
PHXI08:GRAVITATION
359810
A planet having mass \(9 M_{e}\) and radius \(4 R_{e}\), where \(M_{e}\) and \(R_{e}\) are mass and radius of earth respectively, has escape velocity in \(km/s\) given by (Given escape velocity on earth \({V_e} = 11.2 \times {10^3}\;m/s\))
1 67.2
2 16.8
3 11.2
4 33.6
Explanation:
Given, \(M_{p}=9 M_{e} ; R_{p}=4 R_{e}\) Escape velocity is given by \(v_{e}=\sqrt{\dfrac{2 G M_{e}}{R_{e}}}\) \(v_{e}=\sqrt{\dfrac{2 G M_{p}}{R_{p}}}=\sqrt{\dfrac{2 G \times 9 M_{e}}{4 R_{e}}}\) \(=\sqrt{\dfrac{G M_{e}}{R_{e}} \times \dfrac{9}{2}}=\sqrt{\dfrac{9}{4}} \times \sqrt{\dfrac{2 G M_{e}}{R_{e}}}\) \( = \frac{3}{2} \times {v_e} = \left( {\frac{3}{2} \times 11.2} \right)km/s\) \({v_e} = (3 \times 5.6)km/s = 16.8\;km/s\)
JEE - 2023
PHXI08:GRAVITATION
359811
The ratio of escape velocities of two planets if \(g\) value on the two planets are \(9.9\;m{\rm{/}}{s^2}\) and \(3.3\;m{\rm{/}}{s^2}\) and their radii are \(6400\;km\) and \(3200\;km\) respectively is
359812
The escape velocities of two planets \(A\) and \(B\) are in the ratio \(1: 2\). If the ratio of their radii respectively is \(1: 3\), then the ratio of acceleration due to gravity of planet \(A\) to the acceleration of gravity of planet \(B\) will be
1 \(\dfrac{2}{3}\)
2 \(\dfrac{3}{2}\)
3 \(\dfrac{3}{4}\)
4 \(\dfrac{4}{3}\)
Explanation:
Given, \(\dfrac{v_{A}}{v_{B}}=\dfrac{1}{2}, \dfrac{R_{A}}{R_{B}}=\dfrac{1}{3}, \dfrac{g_{A}}{g_{B}}=\)? Escape velocity is given by, \(v_{e}=\sqrt{2 g R}\) \(v_{e} \propto \sqrt{g R}\) \(\Rightarrow g \propto \dfrac{V e^{2}}{R}\) So, \(\frac{{{v_A}}}{{{v_B}}} = \sqrt {\frac{{{g_A}{R_A}}}{{{g_B}{R_B}}}} \) \( \Rightarrow {\left( {\frac{1}{2}} \right)^2} = \frac{{{g_A}}}{{{g_B}}} \times \frac{1}{3} \Rightarrow \frac{{{g_A}}}{{{g_B}}} = \frac{3}{4}\)
