138665
The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.
1 12 hours
2 6 hours
3 4 hours
4 3 hours
Explanation:
D Given, $\mathrm{T}_{1}=24 \mathrm{hr}, \mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=\mathrm{R} / 4, \mathrm{~T}_{2}=$ ? According to Keplers' third law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{R}}{\mathrm{R} / 4}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=(4)^{3}$ $\mathrm{~T}_{2}^{2}=\frac{\mathrm{T}_{1}^{2}}{64}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{8}=\frac{24}{8}=3 \text { hours }$
JEE Main-29.01.2023
Gravitation
138666
If earth has a mass nine times and radius twice to that of a planet $P$. Then $\frac{v_{e}}{3} \sqrt{x_{m}} s^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$, where $v_{e}$ is escape velocity on earth. The value of $x$ is
1 1
2 18
3 3
4 2
Explanation:
D Given that, mass of planet $\left(m_{P}\right)=\frac{M_{e}}{9}$ Radius of planet $\left(R_{P}\right)=\frac{R_{e}}{2}$, escape velocity of $\operatorname{planet}\left(\mathrm{v}_{\mathrm{P}}\right)=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{\mathrm{x}}$ We know that, escape velocity $(\mathrm{v})=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}}=\sqrt{\frac{2 \times \mathrm{G} \times \mathrm{M}_{\mathrm{e}} / 9}{\mathrm{R}_{\mathrm{e}} / 2}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \times 2 \mathrm{GM}_{\mathrm{e}}}{9 \mathrm{R}_{\mathrm{e}}}}$ Also $\quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From equation (i) and equation (ii), we get - $\mathrm{v}_{\mathrm{P}}=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{2}$ We compare with escape velocity of planet, $\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{\mathrm{x}}=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{2}$ So, $\quad \mathrm{x}=2$
JEE Main-01.02.2023
Gravitation
138667
The escape velocities of two planets $A$ and $B$ are in the ratio 1:2. if the ration 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 $\frac{3}{2}$
2 $\frac{2}{3}$
3 $\frac{3}{4}$
4 $\frac{4}{3}$
Explanation:
C Given that, Ratio of escape velocities, $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{1}{2}$ Ratio of radius of planets, $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1}{3}$ We know that, escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gR}}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}} \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}}$ $\frac{1}{2}=\sqrt{\frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}}\left(\frac{1}{3}\right)}$ Squares on both side, we get - $\frac{1}{4}=\frac{g_{A}}{g_{B}} \times \frac{1}{3}$ $\frac{g_{A}}{g_{B}}=\frac{3}{4}$
JEE Main-01.02.2023
Gravitation
138668
A satellite is orbiting the Earth in a circular orbit of radius $R$. Which one of the following statements is true?
1 Angular momentum varies as $\frac{1}{\sqrt{\mathrm{R}}}$
2 Linear momentum varies as $\sqrt{\mathrm{R}}$
3 Frequency of revolution varies as $\frac{1}{\mathrm{R}^{2}}$
4 Kinetic energy varies as $\frac{1}{\mathrm{R}}$
5 Potential energy varies as $R$
Explanation:
D For option (a), We know that - Velocity, $v=\sqrt{\frac{\mathrm{G} \cdot \mathrm{M}}{\mathrm{R}}}$ Angular momentum $(\mathrm{L})=\mathrm{mvr}$ $\mathrm{L}=\mathrm{m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right) \mathrm{R}=\mathrm{m} \sqrt{\frac{\mathrm{GMR}^{2}}{\mathrm{R}}}$ $\mathrm{L}=\mathrm{m} \sqrt{\mathrm{GMR}}$ $\therefore \quad \mathrm{L} \propto \sqrt{\mathrm{R}}$ So, statement first is wrong For option (b), We know that - Linear momentum $(\mathrm{L})=\mathrm{mv}$ $\mathrm{L}=\mathrm{m} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\mathrm{L} \propto \frac{1}{\sqrt{\mathrm{R}}}$ So, Second statement is wrong. For option (c), We know that - Frequency, $\mathrm{f}=\frac{1}{\mathrm{~T}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{V}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ $\mathrm{T} \propto \sqrt{\mathrm{R}^{3}}$ So, third statement are also wrong. For option (d), We know that - Kinetic Energy of satellite is, $\text { K.E. } =\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right)^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}}{\mathrm{R}}$ K.E. $\propto \frac{1}{\mathrm{R}}$, So statement (d) is true.
Kerala CEE -2018
Gravitation
138671
A body is projected with a velocity of $2 \times 11.2$ $\mathrm{kms}^{-1}$ from the surface of earth. The velocity of the body when it escapes the gravitational pull of earth is
1 $\sqrt{3} \times 11.2 \mathrm{kms}^{-1}$
2 $11.2 \mathrm{kms}^{-1}$
3 $\sqrt{2} \times 11.2 \mathrm{kms}^{-1}$
4 $6.5 \times 11.2 \mathrm{kms}^{-1}$
5 $2 \times 11.2 \mathrm{kms}^{-1}$
Explanation:
A Given, $\mathrm{v}=2 \times 11.2 \mathrm{kms}^{-1}$ Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{f}}=\text { ? }$ Now we using the formula of kinetic energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}=\frac{1}{2} \mathrm{~m}\left(2 \mathrm{v}_{\mathrm{e}}\right)^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ $\mathrm{v}_{\mathrm{f}}^{2}=\left(2 \times \mathrm{v}_{\mathrm{e}}\right)^{2}-\mathrm{v}_{\mathrm{e}}^{2}$ $\mathrm{v}_{\mathrm{f}}^{2}=\mathrm{v}_{\mathrm{e}}^{2}\left(2^{2}-1\right)$ $\mathrm{v}_{\mathrm{f}}^{2}=\mathrm{v}_{\mathrm{e}}^{2} \times 3$ $\mathrm{v}_{\mathrm{f}}=\sqrt{3} \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{3} \times 11.2 \mathrm{~km} / \mathrm{s}$
138665
The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.
1 12 hours
2 6 hours
3 4 hours
4 3 hours
Explanation:
D Given, $\mathrm{T}_{1}=24 \mathrm{hr}, \mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=\mathrm{R} / 4, \mathrm{~T}_{2}=$ ? According to Keplers' third law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{R}}{\mathrm{R} / 4}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=(4)^{3}$ $\mathrm{~T}_{2}^{2}=\frac{\mathrm{T}_{1}^{2}}{64}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{8}=\frac{24}{8}=3 \text { hours }$
JEE Main-29.01.2023
Gravitation
138666
If earth has a mass nine times and radius twice to that of a planet $P$. Then $\frac{v_{e}}{3} \sqrt{x_{m}} s^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$, where $v_{e}$ is escape velocity on earth. The value of $x$ is
1 1
2 18
3 3
4 2
Explanation:
D Given that, mass of planet $\left(m_{P}\right)=\frac{M_{e}}{9}$ Radius of planet $\left(R_{P}\right)=\frac{R_{e}}{2}$, escape velocity of $\operatorname{planet}\left(\mathrm{v}_{\mathrm{P}}\right)=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{\mathrm{x}}$ We know that, escape velocity $(\mathrm{v})=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}}=\sqrt{\frac{2 \times \mathrm{G} \times \mathrm{M}_{\mathrm{e}} / 9}{\mathrm{R}_{\mathrm{e}} / 2}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \times 2 \mathrm{GM}_{\mathrm{e}}}{9 \mathrm{R}_{\mathrm{e}}}}$ Also $\quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From equation (i) and equation (ii), we get - $\mathrm{v}_{\mathrm{P}}=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{2}$ We compare with escape velocity of planet, $\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{\mathrm{x}}=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{2}$ So, $\quad \mathrm{x}=2$
JEE Main-01.02.2023
Gravitation
138667
The escape velocities of two planets $A$ and $B$ are in the ratio 1:2. if the ration 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 $\frac{3}{2}$
2 $\frac{2}{3}$
3 $\frac{3}{4}$
4 $\frac{4}{3}$
Explanation:
C Given that, Ratio of escape velocities, $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{1}{2}$ Ratio of radius of planets, $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1}{3}$ We know that, escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gR}}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}} \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}}$ $\frac{1}{2}=\sqrt{\frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}}\left(\frac{1}{3}\right)}$ Squares on both side, we get - $\frac{1}{4}=\frac{g_{A}}{g_{B}} \times \frac{1}{3}$ $\frac{g_{A}}{g_{B}}=\frac{3}{4}$
JEE Main-01.02.2023
Gravitation
138668
A satellite is orbiting the Earth in a circular orbit of radius $R$. Which one of the following statements is true?
1 Angular momentum varies as $\frac{1}{\sqrt{\mathrm{R}}}$
2 Linear momentum varies as $\sqrt{\mathrm{R}}$
3 Frequency of revolution varies as $\frac{1}{\mathrm{R}^{2}}$
4 Kinetic energy varies as $\frac{1}{\mathrm{R}}$
5 Potential energy varies as $R$
Explanation:
D For option (a), We know that - Velocity, $v=\sqrt{\frac{\mathrm{G} \cdot \mathrm{M}}{\mathrm{R}}}$ Angular momentum $(\mathrm{L})=\mathrm{mvr}$ $\mathrm{L}=\mathrm{m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right) \mathrm{R}=\mathrm{m} \sqrt{\frac{\mathrm{GMR}^{2}}{\mathrm{R}}}$ $\mathrm{L}=\mathrm{m} \sqrt{\mathrm{GMR}}$ $\therefore \quad \mathrm{L} \propto \sqrt{\mathrm{R}}$ So, statement first is wrong For option (b), We know that - Linear momentum $(\mathrm{L})=\mathrm{mv}$ $\mathrm{L}=\mathrm{m} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\mathrm{L} \propto \frac{1}{\sqrt{\mathrm{R}}}$ So, Second statement is wrong. For option (c), We know that - Frequency, $\mathrm{f}=\frac{1}{\mathrm{~T}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{V}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ $\mathrm{T} \propto \sqrt{\mathrm{R}^{3}}$ So, third statement are also wrong. For option (d), We know that - Kinetic Energy of satellite is, $\text { K.E. } =\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right)^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}}{\mathrm{R}}$ K.E. $\propto \frac{1}{\mathrm{R}}$, So statement (d) is true.
Kerala CEE -2018
Gravitation
138671
A body is projected with a velocity of $2 \times 11.2$ $\mathrm{kms}^{-1}$ from the surface of earth. The velocity of the body when it escapes the gravitational pull of earth is
1 $\sqrt{3} \times 11.2 \mathrm{kms}^{-1}$
2 $11.2 \mathrm{kms}^{-1}$
3 $\sqrt{2} \times 11.2 \mathrm{kms}^{-1}$
4 $6.5 \times 11.2 \mathrm{kms}^{-1}$
5 $2 \times 11.2 \mathrm{kms}^{-1}$
Explanation:
A Given, $\mathrm{v}=2 \times 11.2 \mathrm{kms}^{-1}$ Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{f}}=\text { ? }$ Now we using the formula of kinetic energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}=\frac{1}{2} \mathrm{~m}\left(2 \mathrm{v}_{\mathrm{e}}\right)^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ $\mathrm{v}_{\mathrm{f}}^{2}=\left(2 \times \mathrm{v}_{\mathrm{e}}\right)^{2}-\mathrm{v}_{\mathrm{e}}^{2}$ $\mathrm{v}_{\mathrm{f}}^{2}=\mathrm{v}_{\mathrm{e}}^{2}\left(2^{2}-1\right)$ $\mathrm{v}_{\mathrm{f}}^{2}=\mathrm{v}_{\mathrm{e}}^{2} \times 3$ $\mathrm{v}_{\mathrm{f}}=\sqrt{3} \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{3} \times 11.2 \mathrm{~km} / \mathrm{s}$
138665
The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.
1 12 hours
2 6 hours
3 4 hours
4 3 hours
Explanation:
D Given, $\mathrm{T}_{1}=24 \mathrm{hr}, \mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=\mathrm{R} / 4, \mathrm{~T}_{2}=$ ? According to Keplers' third law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{R}}{\mathrm{R} / 4}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=(4)^{3}$ $\mathrm{~T}_{2}^{2}=\frac{\mathrm{T}_{1}^{2}}{64}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{8}=\frac{24}{8}=3 \text { hours }$
JEE Main-29.01.2023
Gravitation
138666
If earth has a mass nine times and radius twice to that of a planet $P$. Then $\frac{v_{e}}{3} \sqrt{x_{m}} s^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$, where $v_{e}$ is escape velocity on earth. The value of $x$ is
1 1
2 18
3 3
4 2
Explanation:
D Given that, mass of planet $\left(m_{P}\right)=\frac{M_{e}}{9}$ Radius of planet $\left(R_{P}\right)=\frac{R_{e}}{2}$, escape velocity of $\operatorname{planet}\left(\mathrm{v}_{\mathrm{P}}\right)=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{\mathrm{x}}$ We know that, escape velocity $(\mathrm{v})=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}}=\sqrt{\frac{2 \times \mathrm{G} \times \mathrm{M}_{\mathrm{e}} / 9}{\mathrm{R}_{\mathrm{e}} / 2}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \times 2 \mathrm{GM}_{\mathrm{e}}}{9 \mathrm{R}_{\mathrm{e}}}}$ Also $\quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From equation (i) and equation (ii), we get - $\mathrm{v}_{\mathrm{P}}=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{2}$ We compare with escape velocity of planet, $\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{\mathrm{x}}=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{2}$ So, $\quad \mathrm{x}=2$
JEE Main-01.02.2023
Gravitation
138667
The escape velocities of two planets $A$ and $B$ are in the ratio 1:2. if the ration 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 $\frac{3}{2}$
2 $\frac{2}{3}$
3 $\frac{3}{4}$
4 $\frac{4}{3}$
Explanation:
C Given that, Ratio of escape velocities, $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{1}{2}$ Ratio of radius of planets, $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1}{3}$ We know that, escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gR}}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}} \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}}$ $\frac{1}{2}=\sqrt{\frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}}\left(\frac{1}{3}\right)}$ Squares on both side, we get - $\frac{1}{4}=\frac{g_{A}}{g_{B}} \times \frac{1}{3}$ $\frac{g_{A}}{g_{B}}=\frac{3}{4}$
JEE Main-01.02.2023
Gravitation
138668
A satellite is orbiting the Earth in a circular orbit of radius $R$. Which one of the following statements is true?
1 Angular momentum varies as $\frac{1}{\sqrt{\mathrm{R}}}$
2 Linear momentum varies as $\sqrt{\mathrm{R}}$
3 Frequency of revolution varies as $\frac{1}{\mathrm{R}^{2}}$
4 Kinetic energy varies as $\frac{1}{\mathrm{R}}$
5 Potential energy varies as $R$
Explanation:
D For option (a), We know that - Velocity, $v=\sqrt{\frac{\mathrm{G} \cdot \mathrm{M}}{\mathrm{R}}}$ Angular momentum $(\mathrm{L})=\mathrm{mvr}$ $\mathrm{L}=\mathrm{m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right) \mathrm{R}=\mathrm{m} \sqrt{\frac{\mathrm{GMR}^{2}}{\mathrm{R}}}$ $\mathrm{L}=\mathrm{m} \sqrt{\mathrm{GMR}}$ $\therefore \quad \mathrm{L} \propto \sqrt{\mathrm{R}}$ So, statement first is wrong For option (b), We know that - Linear momentum $(\mathrm{L})=\mathrm{mv}$ $\mathrm{L}=\mathrm{m} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\mathrm{L} \propto \frac{1}{\sqrt{\mathrm{R}}}$ So, Second statement is wrong. For option (c), We know that - Frequency, $\mathrm{f}=\frac{1}{\mathrm{~T}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{V}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ $\mathrm{T} \propto \sqrt{\mathrm{R}^{3}}$ So, third statement are also wrong. For option (d), We know that - Kinetic Energy of satellite is, $\text { K.E. } =\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right)^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}}{\mathrm{R}}$ K.E. $\propto \frac{1}{\mathrm{R}}$, So statement (d) is true.
Kerala CEE -2018
Gravitation
138671
A body is projected with a velocity of $2 \times 11.2$ $\mathrm{kms}^{-1}$ from the surface of earth. The velocity of the body when it escapes the gravitational pull of earth is
1 $\sqrt{3} \times 11.2 \mathrm{kms}^{-1}$
2 $11.2 \mathrm{kms}^{-1}$
3 $\sqrt{2} \times 11.2 \mathrm{kms}^{-1}$
4 $6.5 \times 11.2 \mathrm{kms}^{-1}$
5 $2 \times 11.2 \mathrm{kms}^{-1}$
Explanation:
A Given, $\mathrm{v}=2 \times 11.2 \mathrm{kms}^{-1}$ Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{f}}=\text { ? }$ Now we using the formula of kinetic energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}=\frac{1}{2} \mathrm{~m}\left(2 \mathrm{v}_{\mathrm{e}}\right)^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ $\mathrm{v}_{\mathrm{f}}^{2}=\left(2 \times \mathrm{v}_{\mathrm{e}}\right)^{2}-\mathrm{v}_{\mathrm{e}}^{2}$ $\mathrm{v}_{\mathrm{f}}^{2}=\mathrm{v}_{\mathrm{e}}^{2}\left(2^{2}-1\right)$ $\mathrm{v}_{\mathrm{f}}^{2}=\mathrm{v}_{\mathrm{e}}^{2} \times 3$ $\mathrm{v}_{\mathrm{f}}=\sqrt{3} \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{3} \times 11.2 \mathrm{~km} / \mathrm{s}$
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Gravitation
138665
The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.
1 12 hours
2 6 hours
3 4 hours
4 3 hours
Explanation:
D Given, $\mathrm{T}_{1}=24 \mathrm{hr}, \mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=\mathrm{R} / 4, \mathrm{~T}_{2}=$ ? According to Keplers' third law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{R}}{\mathrm{R} / 4}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=(4)^{3}$ $\mathrm{~T}_{2}^{2}=\frac{\mathrm{T}_{1}^{2}}{64}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{8}=\frac{24}{8}=3 \text { hours }$
JEE Main-29.01.2023
Gravitation
138666
If earth has a mass nine times and radius twice to that of a planet $P$. Then $\frac{v_{e}}{3} \sqrt{x_{m}} s^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$, where $v_{e}$ is escape velocity on earth. The value of $x$ is
1 1
2 18
3 3
4 2
Explanation:
D Given that, mass of planet $\left(m_{P}\right)=\frac{M_{e}}{9}$ Radius of planet $\left(R_{P}\right)=\frac{R_{e}}{2}$, escape velocity of $\operatorname{planet}\left(\mathrm{v}_{\mathrm{P}}\right)=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{\mathrm{x}}$ We know that, escape velocity $(\mathrm{v})=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}}=\sqrt{\frac{2 \times \mathrm{G} \times \mathrm{M}_{\mathrm{e}} / 9}{\mathrm{R}_{\mathrm{e}} / 2}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \times 2 \mathrm{GM}_{\mathrm{e}}}{9 \mathrm{R}_{\mathrm{e}}}}$ Also $\quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From equation (i) and equation (ii), we get - $\mathrm{v}_{\mathrm{P}}=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{2}$ We compare with escape velocity of planet, $\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{\mathrm{x}}=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{2}$ So, $\quad \mathrm{x}=2$
JEE Main-01.02.2023
Gravitation
138667
The escape velocities of two planets $A$ and $B$ are in the ratio 1:2. if the ration 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 $\frac{3}{2}$
2 $\frac{2}{3}$
3 $\frac{3}{4}$
4 $\frac{4}{3}$
Explanation:
C Given that, Ratio of escape velocities, $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{1}{2}$ Ratio of radius of planets, $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1}{3}$ We know that, escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gR}}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}} \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}}$ $\frac{1}{2}=\sqrt{\frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}}\left(\frac{1}{3}\right)}$ Squares on both side, we get - $\frac{1}{4}=\frac{g_{A}}{g_{B}} \times \frac{1}{3}$ $\frac{g_{A}}{g_{B}}=\frac{3}{4}$
JEE Main-01.02.2023
Gravitation
138668
A satellite is orbiting the Earth in a circular orbit of radius $R$. Which one of the following statements is true?
1 Angular momentum varies as $\frac{1}{\sqrt{\mathrm{R}}}$
2 Linear momentum varies as $\sqrt{\mathrm{R}}$
3 Frequency of revolution varies as $\frac{1}{\mathrm{R}^{2}}$
4 Kinetic energy varies as $\frac{1}{\mathrm{R}}$
5 Potential energy varies as $R$
Explanation:
D For option (a), We know that - Velocity, $v=\sqrt{\frac{\mathrm{G} \cdot \mathrm{M}}{\mathrm{R}}}$ Angular momentum $(\mathrm{L})=\mathrm{mvr}$ $\mathrm{L}=\mathrm{m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right) \mathrm{R}=\mathrm{m} \sqrt{\frac{\mathrm{GMR}^{2}}{\mathrm{R}}}$ $\mathrm{L}=\mathrm{m} \sqrt{\mathrm{GMR}}$ $\therefore \quad \mathrm{L} \propto \sqrt{\mathrm{R}}$ So, statement first is wrong For option (b), We know that - Linear momentum $(\mathrm{L})=\mathrm{mv}$ $\mathrm{L}=\mathrm{m} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\mathrm{L} \propto \frac{1}{\sqrt{\mathrm{R}}}$ So, Second statement is wrong. For option (c), We know that - Frequency, $\mathrm{f}=\frac{1}{\mathrm{~T}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{V}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ $\mathrm{T} \propto \sqrt{\mathrm{R}^{3}}$ So, third statement are also wrong. For option (d), We know that - Kinetic Energy of satellite is, $\text { K.E. } =\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right)^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}}{\mathrm{R}}$ K.E. $\propto \frac{1}{\mathrm{R}}$, So statement (d) is true.
Kerala CEE -2018
Gravitation
138671
A body is projected with a velocity of $2 \times 11.2$ $\mathrm{kms}^{-1}$ from the surface of earth. The velocity of the body when it escapes the gravitational pull of earth is
1 $\sqrt{3} \times 11.2 \mathrm{kms}^{-1}$
2 $11.2 \mathrm{kms}^{-1}$
3 $\sqrt{2} \times 11.2 \mathrm{kms}^{-1}$
4 $6.5 \times 11.2 \mathrm{kms}^{-1}$
5 $2 \times 11.2 \mathrm{kms}^{-1}$
Explanation:
A Given, $\mathrm{v}=2 \times 11.2 \mathrm{kms}^{-1}$ Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{f}}=\text { ? }$ Now we using the formula of kinetic energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}=\frac{1}{2} \mathrm{~m}\left(2 \mathrm{v}_{\mathrm{e}}\right)^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ $\mathrm{v}_{\mathrm{f}}^{2}=\left(2 \times \mathrm{v}_{\mathrm{e}}\right)^{2}-\mathrm{v}_{\mathrm{e}}^{2}$ $\mathrm{v}_{\mathrm{f}}^{2}=\mathrm{v}_{\mathrm{e}}^{2}\left(2^{2}-1\right)$ $\mathrm{v}_{\mathrm{f}}^{2}=\mathrm{v}_{\mathrm{e}}^{2} \times 3$ $\mathrm{v}_{\mathrm{f}}=\sqrt{3} \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{3} \times 11.2 \mathrm{~km} / \mathrm{s}$
138665
The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.
1 12 hours
2 6 hours
3 4 hours
4 3 hours
Explanation:
D Given, $\mathrm{T}_{1}=24 \mathrm{hr}, \mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=\mathrm{R} / 4, \mathrm{~T}_{2}=$ ? According to Keplers' third law, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{2}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\left(\frac{\mathrm{R}}{\mathrm{R} / 4}\right)^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=(4)^{3}$ $\mathrm{~T}_{2}^{2}=\frac{\mathrm{T}_{1}^{2}}{64}$ $\mathrm{~T}_{2}=\frac{\mathrm{T}_{1}}{8}=\frac{24}{8}=3 \text { hours }$
JEE Main-29.01.2023
Gravitation
138666
If earth has a mass nine times and radius twice to that of a planet $P$. Then $\frac{v_{e}}{3} \sqrt{x_{m}} s^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$, where $v_{e}$ is escape velocity on earth. The value of $x$ is
1 1
2 18
3 3
4 2
Explanation:
D Given that, mass of planet $\left(m_{P}\right)=\frac{M_{e}}{9}$ Radius of planet $\left(R_{P}\right)=\frac{R_{e}}{2}$, escape velocity of $\operatorname{planet}\left(\mathrm{v}_{\mathrm{P}}\right)=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{\mathrm{x}}$ We know that, escape velocity $(\mathrm{v})=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}}=\sqrt{\frac{2 \times \mathrm{G} \times \mathrm{M}_{\mathrm{e}} / 9}{\mathrm{R}_{\mathrm{e}} / 2}}$ $\mathrm{v}_{\mathrm{P}}=\sqrt{\frac{2 \times 2 \mathrm{GM}_{\mathrm{e}}}{9 \mathrm{R}_{\mathrm{e}}}}$ Also $\quad \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ From equation (i) and equation (ii), we get - $\mathrm{v}_{\mathrm{P}}=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{2}$ We compare with escape velocity of planet, $\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{\mathrm{x}}=\frac{\mathrm{v}_{\mathrm{e}}}{3} \sqrt{2}$ So, $\quad \mathrm{x}=2$
JEE Main-01.02.2023
Gravitation
138667
The escape velocities of two planets $A$ and $B$ are in the ratio 1:2. if the ration 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 $\frac{3}{2}$
2 $\frac{2}{3}$
3 $\frac{3}{4}$
4 $\frac{4}{3}$
Explanation:
C Given that, Ratio of escape velocities, $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{1}{2}$ Ratio of radius of planets, $\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1}{3}$ We know that, escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{gR}}$ $\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}} \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}}$ $\frac{1}{2}=\sqrt{\frac{\mathrm{g}_{\mathrm{A}}}{\mathrm{g}_{\mathrm{B}}}\left(\frac{1}{3}\right)}$ Squares on both side, we get - $\frac{1}{4}=\frac{g_{A}}{g_{B}} \times \frac{1}{3}$ $\frac{g_{A}}{g_{B}}=\frac{3}{4}$
JEE Main-01.02.2023
Gravitation
138668
A satellite is orbiting the Earth in a circular orbit of radius $R$. Which one of the following statements is true?
1 Angular momentum varies as $\frac{1}{\sqrt{\mathrm{R}}}$
2 Linear momentum varies as $\sqrt{\mathrm{R}}$
3 Frequency of revolution varies as $\frac{1}{\mathrm{R}^{2}}$
4 Kinetic energy varies as $\frac{1}{\mathrm{R}}$
5 Potential energy varies as $R$
Explanation:
D For option (a), We know that - Velocity, $v=\sqrt{\frac{\mathrm{G} \cdot \mathrm{M}}{\mathrm{R}}}$ Angular momentum $(\mathrm{L})=\mathrm{mvr}$ $\mathrm{L}=\mathrm{m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right) \mathrm{R}=\mathrm{m} \sqrt{\frac{\mathrm{GMR}^{2}}{\mathrm{R}}}$ $\mathrm{L}=\mathrm{m} \sqrt{\mathrm{GMR}}$ $\therefore \quad \mathrm{L} \propto \sqrt{\mathrm{R}}$ So, statement first is wrong For option (b), We know that - Linear momentum $(\mathrm{L})=\mathrm{mv}$ $\mathrm{L}=\mathrm{m} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ $\mathrm{L} \propto \frac{1}{\sqrt{\mathrm{R}}}$ So, Second statement is wrong. For option (c), We know that - Frequency, $\mathrm{f}=\frac{1}{\mathrm{~T}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{V}}$ $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}$ $\mathrm{T} \propto \sqrt{\mathrm{R}^{3}}$ So, third statement are also wrong. For option (d), We know that - Kinetic Energy of satellite is, $\text { K.E. } =\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\right)^{2}$ $=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}}{\mathrm{R}}$ K.E. $\propto \frac{1}{\mathrm{R}}$, So statement (d) is true.
Kerala CEE -2018
Gravitation
138671
A body is projected with a velocity of $2 \times 11.2$ $\mathrm{kms}^{-1}$ from the surface of earth. The velocity of the body when it escapes the gravitational pull of earth is
1 $\sqrt{3} \times 11.2 \mathrm{kms}^{-1}$
2 $11.2 \mathrm{kms}^{-1}$
3 $\sqrt{2} \times 11.2 \mathrm{kms}^{-1}$
4 $6.5 \times 11.2 \mathrm{kms}^{-1}$
5 $2 \times 11.2 \mathrm{kms}^{-1}$
Explanation:
A Given, $\mathrm{v}=2 \times 11.2 \mathrm{kms}^{-1}$ Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=11.2 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{f}}=\text { ? }$ Now we using the formula of kinetic energy, $\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}=\frac{1}{2} \mathrm{~m}\left(2 \mathrm{v}_{\mathrm{e}}\right)^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^{2}$ $\mathrm{v}_{\mathrm{f}}^{2}=\left(2 \times \mathrm{v}_{\mathrm{e}}\right)^{2}-\mathrm{v}_{\mathrm{e}}^{2}$ $\mathrm{v}_{\mathrm{f}}^{2}=\mathrm{v}_{\mathrm{e}}^{2}\left(2^{2}-1\right)$ $\mathrm{v}_{\mathrm{f}}^{2}=\mathrm{v}_{\mathrm{e}}^{2} \times 3$ $\mathrm{v}_{\mathrm{f}}=\sqrt{3} \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{3} \times 11.2 \mathrm{~km} / \mathrm{s}$
