NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI08:GRAVITATION
359688
The approximate height from the surface of earth at which the weight of body becomes \(\dfrac{1}{3}\) of its weight on the surface of the earth is : [ Radius of earth \(R = 6400\,km\) and \(\sqrt{3}=1.732\) ]
359689
A \(90\,kg\) body placed at \(2 R\) distance from surface of earth experiences gravitational pull of ( \(R=\) Radius of earth, \(g = 10\;m{s^{ - 2}}\) )
1 \(225\,N\)
2 \(100\,N\)
3 \(300\,N\)
4 \(120\,N\)
Explanation:
Given, \(m = 90\;kg,\) \(h = 2R,\) \(g = 10\;m/{s^2}\) Acceleration due to gravity at height \(2 R\) is given by \(g^{\prime}=\dfrac{G M}{(R+h)^{2}}\) \(\frac{{g'}}{g} = {\left( {\frac{R}{{h + R}}} \right)^2} = {\left( {\frac{R}{{2R + R}}} \right)^2} = \frac{1}{9};g' = \frac{g}{9}\) \(mg' = \frac{{mg}}{9} = \frac{{90 \times 10}}{9} = 100\;N\)
JEE - 2024
PHXI08:GRAVITATION
359690
A body weighs 72 \(N\) on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?
359691
At what height the gravitational field reduces by \(75 \%\) the gravitational field at the surface of earth?
1 \(R\)
2 \(2R\)
3 \(3R\)
4 \(4R\)
Explanation:
Reduces by \(75 \%\) means \(25 \%\) is left \(\frac{{GM}}{{{r^2}}} = \frac{1}{4}\left( {\frac{{GM}}{{{R^2}}}} \right)\) or \(r = 2R\) \(\therefore h = r - R\) \( = 2R - R = R\)
359688
The approximate height from the surface of earth at which the weight of body becomes \(\dfrac{1}{3}\) of its weight on the surface of the earth is : [ Radius of earth \(R = 6400\,km\) and \(\sqrt{3}=1.732\) ]
359689
A \(90\,kg\) body placed at \(2 R\) distance from surface of earth experiences gravitational pull of ( \(R=\) Radius of earth, \(g = 10\;m{s^{ - 2}}\) )
1 \(225\,N\)
2 \(100\,N\)
3 \(300\,N\)
4 \(120\,N\)
Explanation:
Given, \(m = 90\;kg,\) \(h = 2R,\) \(g = 10\;m/{s^2}\) Acceleration due to gravity at height \(2 R\) is given by \(g^{\prime}=\dfrac{G M}{(R+h)^{2}}\) \(\frac{{g'}}{g} = {\left( {\frac{R}{{h + R}}} \right)^2} = {\left( {\frac{R}{{2R + R}}} \right)^2} = \frac{1}{9};g' = \frac{g}{9}\) \(mg' = \frac{{mg}}{9} = \frac{{90 \times 10}}{9} = 100\;N\)
JEE - 2024
PHXI08:GRAVITATION
359690
A body weighs 72 \(N\) on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?
359691
At what height the gravitational field reduces by \(75 \%\) the gravitational field at the surface of earth?
1 \(R\)
2 \(2R\)
3 \(3R\)
4 \(4R\)
Explanation:
Reduces by \(75 \%\) means \(25 \%\) is left \(\frac{{GM}}{{{r^2}}} = \frac{1}{4}\left( {\frac{{GM}}{{{R^2}}}} \right)\) or \(r = 2R\) \(\therefore h = r - R\) \( = 2R - R = R\)
359688
The approximate height from the surface of earth at which the weight of body becomes \(\dfrac{1}{3}\) of its weight on the surface of the earth is : [ Radius of earth \(R = 6400\,km\) and \(\sqrt{3}=1.732\) ]
359689
A \(90\,kg\) body placed at \(2 R\) distance from surface of earth experiences gravitational pull of ( \(R=\) Radius of earth, \(g = 10\;m{s^{ - 2}}\) )
1 \(225\,N\)
2 \(100\,N\)
3 \(300\,N\)
4 \(120\,N\)
Explanation:
Given, \(m = 90\;kg,\) \(h = 2R,\) \(g = 10\;m/{s^2}\) Acceleration due to gravity at height \(2 R\) is given by \(g^{\prime}=\dfrac{G M}{(R+h)^{2}}\) \(\frac{{g'}}{g} = {\left( {\frac{R}{{h + R}}} \right)^2} = {\left( {\frac{R}{{2R + R}}} \right)^2} = \frac{1}{9};g' = \frac{g}{9}\) \(mg' = \frac{{mg}}{9} = \frac{{90 \times 10}}{9} = 100\;N\)
JEE - 2024
PHXI08:GRAVITATION
359690
A body weighs 72 \(N\) on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?
359691
At what height the gravitational field reduces by \(75 \%\) the gravitational field at the surface of earth?
1 \(R\)
2 \(2R\)
3 \(3R\)
4 \(4R\)
Explanation:
Reduces by \(75 \%\) means \(25 \%\) is left \(\frac{{GM}}{{{r^2}}} = \frac{1}{4}\left( {\frac{{GM}}{{{R^2}}}} \right)\) or \(r = 2R\) \(\therefore h = r - R\) \( = 2R - R = R\)
359688
The approximate height from the surface of earth at which the weight of body becomes \(\dfrac{1}{3}\) of its weight on the surface of the earth is : [ Radius of earth \(R = 6400\,km\) and \(\sqrt{3}=1.732\) ]
359689
A \(90\,kg\) body placed at \(2 R\) distance from surface of earth experiences gravitational pull of ( \(R=\) Radius of earth, \(g = 10\;m{s^{ - 2}}\) )
1 \(225\,N\)
2 \(100\,N\)
3 \(300\,N\)
4 \(120\,N\)
Explanation:
Given, \(m = 90\;kg,\) \(h = 2R,\) \(g = 10\;m/{s^2}\) Acceleration due to gravity at height \(2 R\) is given by \(g^{\prime}=\dfrac{G M}{(R+h)^{2}}\) \(\frac{{g'}}{g} = {\left( {\frac{R}{{h + R}}} \right)^2} = {\left( {\frac{R}{{2R + R}}} \right)^2} = \frac{1}{9};g' = \frac{g}{9}\) \(mg' = \frac{{mg}}{9} = \frac{{90 \times 10}}{9} = 100\;N\)
JEE - 2024
PHXI08:GRAVITATION
359690
A body weighs 72 \(N\) on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?
359691
At what height the gravitational field reduces by \(75 \%\) the gravitational field at the surface of earth?
1 \(R\)
2 \(2R\)
3 \(3R\)
4 \(4R\)
Explanation:
Reduces by \(75 \%\) means \(25 \%\) is left \(\frac{{GM}}{{{r^2}}} = \frac{1}{4}\left( {\frac{{GM}}{{{R^2}}}} \right)\) or \(r = 2R\) \(\therefore h = r - R\) \( = 2R - R = R\)
