NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138755
The orbital velocity of a satellite at a height $h$ above the surface of earth is $v$. The value of escape velocity from the same location is given by
1 $\sqrt{2} \mathrm{v}$
2 $\mathrm{v}$
3 $\frac{\mathrm{v}}{\sqrt{2}}$
4 $\frac{\mathrm{v}}{2}$
Explanation:
A We know, Orbital velocity $(\mathrm{v})=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ From equation (i) $\therefore \quad \mathrm{v}_{\mathrm{e}}=\sqrt{2} \mathrm{v}$
J and K CET- 2000
Gravitation
138757
The mass of mars is 0.11 times the mass of the earth and the radius of mars is 0.53 times the radius of the earth. The escape velocity from the earth's surface is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$, the escape velocity of a body from mars's surface is
1 $1.12 \mathrm{~km} \mathrm{~s}^{-1}$
2 $0.51 \mathrm{~km} \mathrm{~s}^{-1}$
3 $5.1 \mathrm{~km} \mathrm{~s}^{-1}$
4 $10.2 \mathrm{~km} \mathrm{~s}^{-1}$
Explanation:
C Suppose that, $\mathrm{v}_{\mathrm{e}}=11.2 \mathrm{~km} / \mathrm{s}$ $\mathrm{M}_{\mathrm{m}}=0.11 \mathrm{M}_{\mathrm{e}}$ $\mathrm{R}_{\mathrm{m}}=0.53 \mathrm{R}_{\mathrm{e}}$ We know, Escape velocity of earth, $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Escape velocity of mars's $\therefore \quad \mathrm{v}_{\mathrm{m}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}}}$ Put the value, $\mathrm{v}_{\mathrm{m}} =\sqrt{\frac{2 \times \mathrm{G} \times 0.11 \mathrm{M}_{\mathrm{e}}}{0.53 \mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{m}} =\sqrt{\frac{0.11}{0.53}} \cdot \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{m}} =0.455 \times 11.2$ $\mathrm{v}_{\mathrm{m}} =5.1 \mathrm{~km} / \mathrm{s}$
J and K CET- 1997
Gravitation
138758
The escape velocity form the surface of earth is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$. What is the escape velocity in a planet whose radius is three times that of earth and on which the acceleration due to gravity is three times of that on earth?
138755
The orbital velocity of a satellite at a height $h$ above the surface of earth is $v$. The value of escape velocity from the same location is given by
1 $\sqrt{2} \mathrm{v}$
2 $\mathrm{v}$
3 $\frac{\mathrm{v}}{\sqrt{2}}$
4 $\frac{\mathrm{v}}{2}$
Explanation:
A We know, Orbital velocity $(\mathrm{v})=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ From equation (i) $\therefore \quad \mathrm{v}_{\mathrm{e}}=\sqrt{2} \mathrm{v}$
J and K CET- 2000
Gravitation
138757
The mass of mars is 0.11 times the mass of the earth and the radius of mars is 0.53 times the radius of the earth. The escape velocity from the earth's surface is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$, the escape velocity of a body from mars's surface is
1 $1.12 \mathrm{~km} \mathrm{~s}^{-1}$
2 $0.51 \mathrm{~km} \mathrm{~s}^{-1}$
3 $5.1 \mathrm{~km} \mathrm{~s}^{-1}$
4 $10.2 \mathrm{~km} \mathrm{~s}^{-1}$
Explanation:
C Suppose that, $\mathrm{v}_{\mathrm{e}}=11.2 \mathrm{~km} / \mathrm{s}$ $\mathrm{M}_{\mathrm{m}}=0.11 \mathrm{M}_{\mathrm{e}}$ $\mathrm{R}_{\mathrm{m}}=0.53 \mathrm{R}_{\mathrm{e}}$ We know, Escape velocity of earth, $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Escape velocity of mars's $\therefore \quad \mathrm{v}_{\mathrm{m}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}}}$ Put the value, $\mathrm{v}_{\mathrm{m}} =\sqrt{\frac{2 \times \mathrm{G} \times 0.11 \mathrm{M}_{\mathrm{e}}}{0.53 \mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{m}} =\sqrt{\frac{0.11}{0.53}} \cdot \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{m}} =0.455 \times 11.2$ $\mathrm{v}_{\mathrm{m}} =5.1 \mathrm{~km} / \mathrm{s}$
J and K CET- 1997
Gravitation
138758
The escape velocity form the surface of earth is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$. What is the escape velocity in a planet whose radius is three times that of earth and on which the acceleration due to gravity is three times of that on earth?
138755
The orbital velocity of a satellite at a height $h$ above the surface of earth is $v$. The value of escape velocity from the same location is given by
1 $\sqrt{2} \mathrm{v}$
2 $\mathrm{v}$
3 $\frac{\mathrm{v}}{\sqrt{2}}$
4 $\frac{\mathrm{v}}{2}$
Explanation:
A We know, Orbital velocity $(\mathrm{v})=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ From equation (i) $\therefore \quad \mathrm{v}_{\mathrm{e}}=\sqrt{2} \mathrm{v}$
J and K CET- 2000
Gravitation
138757
The mass of mars is 0.11 times the mass of the earth and the radius of mars is 0.53 times the radius of the earth. The escape velocity from the earth's surface is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$, the escape velocity of a body from mars's surface is
1 $1.12 \mathrm{~km} \mathrm{~s}^{-1}$
2 $0.51 \mathrm{~km} \mathrm{~s}^{-1}$
3 $5.1 \mathrm{~km} \mathrm{~s}^{-1}$
4 $10.2 \mathrm{~km} \mathrm{~s}^{-1}$
Explanation:
C Suppose that, $\mathrm{v}_{\mathrm{e}}=11.2 \mathrm{~km} / \mathrm{s}$ $\mathrm{M}_{\mathrm{m}}=0.11 \mathrm{M}_{\mathrm{e}}$ $\mathrm{R}_{\mathrm{m}}=0.53 \mathrm{R}_{\mathrm{e}}$ We know, Escape velocity of earth, $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Escape velocity of mars's $\therefore \quad \mathrm{v}_{\mathrm{m}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}}}$ Put the value, $\mathrm{v}_{\mathrm{m}} =\sqrt{\frac{2 \times \mathrm{G} \times 0.11 \mathrm{M}_{\mathrm{e}}}{0.53 \mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{m}} =\sqrt{\frac{0.11}{0.53}} \cdot \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{m}} =0.455 \times 11.2$ $\mathrm{v}_{\mathrm{m}} =5.1 \mathrm{~km} / \mathrm{s}$
J and K CET- 1997
Gravitation
138758
The escape velocity form the surface of earth is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$. What is the escape velocity in a planet whose radius is three times that of earth and on which the acceleration due to gravity is three times of that on earth?
138755
The orbital velocity of a satellite at a height $h$ above the surface of earth is $v$. The value of escape velocity from the same location is given by
1 $\sqrt{2} \mathrm{v}$
2 $\mathrm{v}$
3 $\frac{\mathrm{v}}{\sqrt{2}}$
4 $\frac{\mathrm{v}}{2}$
Explanation:
A We know, Orbital velocity $(\mathrm{v})=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\left(\mathrm{v}_{\mathrm{e}}\right)=\sqrt{2} \times \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ From equation (i) $\therefore \quad \mathrm{v}_{\mathrm{e}}=\sqrt{2} \mathrm{v}$
J and K CET- 2000
Gravitation
138757
The mass of mars is 0.11 times the mass of the earth and the radius of mars is 0.53 times the radius of the earth. The escape velocity from the earth's surface is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$, the escape velocity of a body from mars's surface is
1 $1.12 \mathrm{~km} \mathrm{~s}^{-1}$
2 $0.51 \mathrm{~km} \mathrm{~s}^{-1}$
3 $5.1 \mathrm{~km} \mathrm{~s}^{-1}$
4 $10.2 \mathrm{~km} \mathrm{~s}^{-1}$
Explanation:
C Suppose that, $\mathrm{v}_{\mathrm{e}}=11.2 \mathrm{~km} / \mathrm{s}$ $\mathrm{M}_{\mathrm{m}}=0.11 \mathrm{M}_{\mathrm{e}}$ $\mathrm{R}_{\mathrm{m}}=0.53 \mathrm{R}_{\mathrm{e}}$ We know, Escape velocity of earth, $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}}$ Escape velocity of mars's $\therefore \quad \mathrm{v}_{\mathrm{m}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}}}$ Put the value, $\mathrm{v}_{\mathrm{m}} =\sqrt{\frac{2 \times \mathrm{G} \times 0.11 \mathrm{M}_{\mathrm{e}}}{0.53 \mathrm{R}_{\mathrm{e}}}}$ $\mathrm{v}_{\mathrm{m}} =\sqrt{\frac{0.11}{0.53}} \cdot \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{m}} =0.455 \times 11.2$ $\mathrm{v}_{\mathrm{m}} =5.1 \mathrm{~km} / \mathrm{s}$
J and K CET- 1997
Gravitation
138758
The escape velocity form the surface of earth is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$. What is the escape velocity in a planet whose radius is three times that of earth and on which the acceleration due to gravity is three times of that on earth?
