138296
Assuming the mass of Earth to be ten times the mass of Mars, its radius to be twice the radius of mars and the acceleration due to gravity on the surface of Earth is $10 \mathrm{~m} / \mathrm{s}^{2}$. Then the acceleration due to gravity on the surface of Mars is given by
1 $0.2 \mathrm{~m} / \mathrm{s}^{2}$
2 $0.4 \mathrm{~m} / \mathrm{s}^{2}$
3 $2 \mathrm{~m} / \mathrm{s}^{2}$
4 $4 \mathrm{~m} / \mathrm{s}^{2}$
5 $5 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
D Given, $M_{e}=10 \times M_{m} \quad$ or $\quad M_{m}=\frac{M_{e}}{10}$ $\mathrm{R}_{\mathrm{e}}=2 \times \mathrm{R}_{\mathrm{m}} \quad \text { or } \quad \mathrm{R}_{\mathrm{m}}=\frac{\mathrm{R}_{\mathrm{e}}}{2}$ $\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ Acceleration due to gravity on earth $(g)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ Acceleration due to gravity on mars $\left(\mathrm{g}^{\prime}\right)=\frac{\mathrm{GM}_{\mathrm{m}}}{\left(\mathrm{R}_{\mathrm{m}}\right)^{2}}$ Putting the value of these, we get - $\mathrm{g}^{\prime}=\mathrm{G} \frac{\mathrm{M}_{\mathrm{e}} / 10}{\left(\mathrm{R}_{\mathrm{e}} / 2\right)^{2}}$ $\mathrm{~g}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \times 4}{10 \mathrm{R}_{\mathrm{e}}{ }^{2}}$ $\mathrm{~g}^{\prime}=\frac{2}{5} \mathrm{~g}=0.4 \times 10=4 \mathrm{~m} / \mathrm{s}^{2}$
Kerala CEE - 2004
Gravitation
138297
The acceleration due to gravity on the surface of a planet is one-fourth of the value of Earth. When a brass ball is brought to this planet, its
1 mass is halved
2 weight is halved
3 mass becomes one-fourth
4 weight becomes one-fourth
5 mass and weight remain the same
Explanation:
D According to question $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{g}}}=\frac{\mathrm{g}_{\mathrm{p}}}{\mathrm{g}_{\mathrm{e}}}$ $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{g}}}=\frac{\mathrm{g}_{\mathrm{e}} / 4}{\mathrm{~g}_{\mathrm{e}}}$ $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{e}}}=\frac{1}{4}$ $\mathrm{w}_{\mathrm{p}}=\frac{\mathrm{w}_{\mathrm{e}}}{4}$ Thus, when a brass ball is brought to this planet its weight becomes one fourth.
Kerala CEE - 2016
Gravitation
138298
A body hanging from a massless spring stretches is by $3 \mathbf{c m}$ on Earth's surface. At a place $800 \mathrm{~km}$ above the Earth's surface, the same body will stretch the spring by (Radius of Earth $=6400 \mathrm{~km}$ )
1 $\left(\frac{34}{27}\right) \mathrm{cm}$
2 $\left(\frac{64}{27}\right) \mathrm{cm}$
3 $\left(\frac{27}{64}\right) \mathrm{cm}$
4 $\left(\frac{27}{34}\right) \mathrm{cm}$
5 $\left(\frac{35}{81}\right) \mathrm{cm}$
Explanation:
B Given, $\mathrm{x}=3 \mathrm{~cm}$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}=6400 \times 10^{3} \mathrm{~m}$ Height from the earth surface (h) $=800 \mathrm{~km}=800 \times 10^{3} \mathrm{~m}$ When a body hanging from a massless spring on earth surface. So, Acceleration due to gravity, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{~F}=-\mathrm{kx}=\mathrm{mg}$ At a place $800 \mathrm{~km}$ above the earth's surface, the same body will stretch the spring by $x^{\prime}$ and the value of $g^{\prime}$ $\mathrm{F} =-k x^{\prime}=m g^{\prime}$ $g^{\prime} =\frac{G M}{(R+h)^{2}}$ Dividing equation (iii) by equation (ii), we get- $\frac{-\mathrm{kx}^{\prime}}{-\mathrm{kx}}=\frac{\mathrm{mg}}{\mathrm{mg}}$ $\frac{\mathrm{x}^{\prime}}{\mathrm{x}}=\frac{\mathrm{g}^{\prime}}{\mathrm{g}}$ $\frac{x^{\prime}}{x}=\frac{(R+h)^2}{\frac{G M}{R^2}}$ $\frac{x^{\prime}}{x}=\frac{R^2}{(R+h)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{h}{R}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{800 \times 10^3}{6400 \times 10^3}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{1}{8}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{64}{81}$ $x^{\prime}=\frac{64}{81} x$ $x^{\prime}=\frac{64}{81} \times 3$ $x^{\prime}=\left(\frac{64}{27}\right) \mathrm{cm}$
Kerala CEE - 2016
Gravitation
138299
At what depth below the surface of the earth, the value of $g$ is the same as that at a height of 5 km?
1 $1.25 \mathrm{~km}$
2 $2.5 \mathrm{~km}$
3 $5 \mathrm{~km}$
4 $7.5 \mathrm{~km}$
5 $10 \mathrm{~km}$
Explanation:
E Given, Height $(\mathrm{h})=5 \mathrm{~km}$ Variation of $\mathrm{g}$ with depth $g=g_{o}\left(1-\frac{d}{R}\right)$ Variation of $\mathrm{g}$ with height $\mathrm{g}^{\prime}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ On Solving eq ${ }^{\mathrm{n}}$ (i) and (ii) We get $g_{o}\left(1-\frac{2 h}{R}\right)=g_{o}\left(1-\frac{d}{R}\right) \quad\left[\because g=g^{\prime}\right]$ $\frac{2 h}{R}=\frac{d}{R}$ $d=2 h$ $d=2 \times 5$ $d=10 k m .$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138296
Assuming the mass of Earth to be ten times the mass of Mars, its radius to be twice the radius of mars and the acceleration due to gravity on the surface of Earth is $10 \mathrm{~m} / \mathrm{s}^{2}$. Then the acceleration due to gravity on the surface of Mars is given by
1 $0.2 \mathrm{~m} / \mathrm{s}^{2}$
2 $0.4 \mathrm{~m} / \mathrm{s}^{2}$
3 $2 \mathrm{~m} / \mathrm{s}^{2}$
4 $4 \mathrm{~m} / \mathrm{s}^{2}$
5 $5 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
D Given, $M_{e}=10 \times M_{m} \quad$ or $\quad M_{m}=\frac{M_{e}}{10}$ $\mathrm{R}_{\mathrm{e}}=2 \times \mathrm{R}_{\mathrm{m}} \quad \text { or } \quad \mathrm{R}_{\mathrm{m}}=\frac{\mathrm{R}_{\mathrm{e}}}{2}$ $\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ Acceleration due to gravity on earth $(g)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ Acceleration due to gravity on mars $\left(\mathrm{g}^{\prime}\right)=\frac{\mathrm{GM}_{\mathrm{m}}}{\left(\mathrm{R}_{\mathrm{m}}\right)^{2}}$ Putting the value of these, we get - $\mathrm{g}^{\prime}=\mathrm{G} \frac{\mathrm{M}_{\mathrm{e}} / 10}{\left(\mathrm{R}_{\mathrm{e}} / 2\right)^{2}}$ $\mathrm{~g}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \times 4}{10 \mathrm{R}_{\mathrm{e}}{ }^{2}}$ $\mathrm{~g}^{\prime}=\frac{2}{5} \mathrm{~g}=0.4 \times 10=4 \mathrm{~m} / \mathrm{s}^{2}$
Kerala CEE - 2004
Gravitation
138297
The acceleration due to gravity on the surface of a planet is one-fourth of the value of Earth. When a brass ball is brought to this planet, its
1 mass is halved
2 weight is halved
3 mass becomes one-fourth
4 weight becomes one-fourth
5 mass and weight remain the same
Explanation:
D According to question $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{g}}}=\frac{\mathrm{g}_{\mathrm{p}}}{\mathrm{g}_{\mathrm{e}}}$ $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{g}}}=\frac{\mathrm{g}_{\mathrm{e}} / 4}{\mathrm{~g}_{\mathrm{e}}}$ $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{e}}}=\frac{1}{4}$ $\mathrm{w}_{\mathrm{p}}=\frac{\mathrm{w}_{\mathrm{e}}}{4}$ Thus, when a brass ball is brought to this planet its weight becomes one fourth.
Kerala CEE - 2016
Gravitation
138298
A body hanging from a massless spring stretches is by $3 \mathbf{c m}$ on Earth's surface. At a place $800 \mathrm{~km}$ above the Earth's surface, the same body will stretch the spring by (Radius of Earth $=6400 \mathrm{~km}$ )
1 $\left(\frac{34}{27}\right) \mathrm{cm}$
2 $\left(\frac{64}{27}\right) \mathrm{cm}$
3 $\left(\frac{27}{64}\right) \mathrm{cm}$
4 $\left(\frac{27}{34}\right) \mathrm{cm}$
5 $\left(\frac{35}{81}\right) \mathrm{cm}$
Explanation:
B Given, $\mathrm{x}=3 \mathrm{~cm}$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}=6400 \times 10^{3} \mathrm{~m}$ Height from the earth surface (h) $=800 \mathrm{~km}=800 \times 10^{3} \mathrm{~m}$ When a body hanging from a massless spring on earth surface. So, Acceleration due to gravity, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{~F}=-\mathrm{kx}=\mathrm{mg}$ At a place $800 \mathrm{~km}$ above the earth's surface, the same body will stretch the spring by $x^{\prime}$ and the value of $g^{\prime}$ $\mathrm{F} =-k x^{\prime}=m g^{\prime}$ $g^{\prime} =\frac{G M}{(R+h)^{2}}$ Dividing equation (iii) by equation (ii), we get- $\frac{-\mathrm{kx}^{\prime}}{-\mathrm{kx}}=\frac{\mathrm{mg}}{\mathrm{mg}}$ $\frac{\mathrm{x}^{\prime}}{\mathrm{x}}=\frac{\mathrm{g}^{\prime}}{\mathrm{g}}$ $\frac{x^{\prime}}{x}=\frac{(R+h)^2}{\frac{G M}{R^2}}$ $\frac{x^{\prime}}{x}=\frac{R^2}{(R+h)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{h}{R}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{800 \times 10^3}{6400 \times 10^3}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{1}{8}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{64}{81}$ $x^{\prime}=\frac{64}{81} x$ $x^{\prime}=\frac{64}{81} \times 3$ $x^{\prime}=\left(\frac{64}{27}\right) \mathrm{cm}$
Kerala CEE - 2016
Gravitation
138299
At what depth below the surface of the earth, the value of $g$ is the same as that at a height of 5 km?
1 $1.25 \mathrm{~km}$
2 $2.5 \mathrm{~km}$
3 $5 \mathrm{~km}$
4 $7.5 \mathrm{~km}$
5 $10 \mathrm{~km}$
Explanation:
E Given, Height $(\mathrm{h})=5 \mathrm{~km}$ Variation of $\mathrm{g}$ with depth $g=g_{o}\left(1-\frac{d}{R}\right)$ Variation of $\mathrm{g}$ with height $\mathrm{g}^{\prime}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ On Solving eq ${ }^{\mathrm{n}}$ (i) and (ii) We get $g_{o}\left(1-\frac{2 h}{R}\right)=g_{o}\left(1-\frac{d}{R}\right) \quad\left[\because g=g^{\prime}\right]$ $\frac{2 h}{R}=\frac{d}{R}$ $d=2 h$ $d=2 \times 5$ $d=10 k m .$
138296
Assuming the mass of Earth to be ten times the mass of Mars, its radius to be twice the radius of mars and the acceleration due to gravity on the surface of Earth is $10 \mathrm{~m} / \mathrm{s}^{2}$. Then the acceleration due to gravity on the surface of Mars is given by
1 $0.2 \mathrm{~m} / \mathrm{s}^{2}$
2 $0.4 \mathrm{~m} / \mathrm{s}^{2}$
3 $2 \mathrm{~m} / \mathrm{s}^{2}$
4 $4 \mathrm{~m} / \mathrm{s}^{2}$
5 $5 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
D Given, $M_{e}=10 \times M_{m} \quad$ or $\quad M_{m}=\frac{M_{e}}{10}$ $\mathrm{R}_{\mathrm{e}}=2 \times \mathrm{R}_{\mathrm{m}} \quad \text { or } \quad \mathrm{R}_{\mathrm{m}}=\frac{\mathrm{R}_{\mathrm{e}}}{2}$ $\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ Acceleration due to gravity on earth $(g)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ Acceleration due to gravity on mars $\left(\mathrm{g}^{\prime}\right)=\frac{\mathrm{GM}_{\mathrm{m}}}{\left(\mathrm{R}_{\mathrm{m}}\right)^{2}}$ Putting the value of these, we get - $\mathrm{g}^{\prime}=\mathrm{G} \frac{\mathrm{M}_{\mathrm{e}} / 10}{\left(\mathrm{R}_{\mathrm{e}} / 2\right)^{2}}$ $\mathrm{~g}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \times 4}{10 \mathrm{R}_{\mathrm{e}}{ }^{2}}$ $\mathrm{~g}^{\prime}=\frac{2}{5} \mathrm{~g}=0.4 \times 10=4 \mathrm{~m} / \mathrm{s}^{2}$
Kerala CEE - 2004
Gravitation
138297
The acceleration due to gravity on the surface of a planet is one-fourth of the value of Earth. When a brass ball is brought to this planet, its
1 mass is halved
2 weight is halved
3 mass becomes one-fourth
4 weight becomes one-fourth
5 mass and weight remain the same
Explanation:
D According to question $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{g}}}=\frac{\mathrm{g}_{\mathrm{p}}}{\mathrm{g}_{\mathrm{e}}}$ $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{g}}}=\frac{\mathrm{g}_{\mathrm{e}} / 4}{\mathrm{~g}_{\mathrm{e}}}$ $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{e}}}=\frac{1}{4}$ $\mathrm{w}_{\mathrm{p}}=\frac{\mathrm{w}_{\mathrm{e}}}{4}$ Thus, when a brass ball is brought to this planet its weight becomes one fourth.
Kerala CEE - 2016
Gravitation
138298
A body hanging from a massless spring stretches is by $3 \mathbf{c m}$ on Earth's surface. At a place $800 \mathrm{~km}$ above the Earth's surface, the same body will stretch the spring by (Radius of Earth $=6400 \mathrm{~km}$ )
1 $\left(\frac{34}{27}\right) \mathrm{cm}$
2 $\left(\frac{64}{27}\right) \mathrm{cm}$
3 $\left(\frac{27}{64}\right) \mathrm{cm}$
4 $\left(\frac{27}{34}\right) \mathrm{cm}$
5 $\left(\frac{35}{81}\right) \mathrm{cm}$
Explanation:
B Given, $\mathrm{x}=3 \mathrm{~cm}$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}=6400 \times 10^{3} \mathrm{~m}$ Height from the earth surface (h) $=800 \mathrm{~km}=800 \times 10^{3} \mathrm{~m}$ When a body hanging from a massless spring on earth surface. So, Acceleration due to gravity, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{~F}=-\mathrm{kx}=\mathrm{mg}$ At a place $800 \mathrm{~km}$ above the earth's surface, the same body will stretch the spring by $x^{\prime}$ and the value of $g^{\prime}$ $\mathrm{F} =-k x^{\prime}=m g^{\prime}$ $g^{\prime} =\frac{G M}{(R+h)^{2}}$ Dividing equation (iii) by equation (ii), we get- $\frac{-\mathrm{kx}^{\prime}}{-\mathrm{kx}}=\frac{\mathrm{mg}}{\mathrm{mg}}$ $\frac{\mathrm{x}^{\prime}}{\mathrm{x}}=\frac{\mathrm{g}^{\prime}}{\mathrm{g}}$ $\frac{x^{\prime}}{x}=\frac{(R+h)^2}{\frac{G M}{R^2}}$ $\frac{x^{\prime}}{x}=\frac{R^2}{(R+h)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{h}{R}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{800 \times 10^3}{6400 \times 10^3}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{1}{8}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{64}{81}$ $x^{\prime}=\frac{64}{81} x$ $x^{\prime}=\frac{64}{81} \times 3$ $x^{\prime}=\left(\frac{64}{27}\right) \mathrm{cm}$
Kerala CEE - 2016
Gravitation
138299
At what depth below the surface of the earth, the value of $g$ is the same as that at a height of 5 km?
1 $1.25 \mathrm{~km}$
2 $2.5 \mathrm{~km}$
3 $5 \mathrm{~km}$
4 $7.5 \mathrm{~km}$
5 $10 \mathrm{~km}$
Explanation:
E Given, Height $(\mathrm{h})=5 \mathrm{~km}$ Variation of $\mathrm{g}$ with depth $g=g_{o}\left(1-\frac{d}{R}\right)$ Variation of $\mathrm{g}$ with height $\mathrm{g}^{\prime}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ On Solving eq ${ }^{\mathrm{n}}$ (i) and (ii) We get $g_{o}\left(1-\frac{2 h}{R}\right)=g_{o}\left(1-\frac{d}{R}\right) \quad\left[\because g=g^{\prime}\right]$ $\frac{2 h}{R}=\frac{d}{R}$ $d=2 h$ $d=2 \times 5$ $d=10 k m .$
138296
Assuming the mass of Earth to be ten times the mass of Mars, its radius to be twice the radius of mars and the acceleration due to gravity on the surface of Earth is $10 \mathrm{~m} / \mathrm{s}^{2}$. Then the acceleration due to gravity on the surface of Mars is given by
1 $0.2 \mathrm{~m} / \mathrm{s}^{2}$
2 $0.4 \mathrm{~m} / \mathrm{s}^{2}$
3 $2 \mathrm{~m} / \mathrm{s}^{2}$
4 $4 \mathrm{~m} / \mathrm{s}^{2}$
5 $5 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
D Given, $M_{e}=10 \times M_{m} \quad$ or $\quad M_{m}=\frac{M_{e}}{10}$ $\mathrm{R}_{\mathrm{e}}=2 \times \mathrm{R}_{\mathrm{m}} \quad \text { or } \quad \mathrm{R}_{\mathrm{m}}=\frac{\mathrm{R}_{\mathrm{e}}}{2}$ $\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ Acceleration due to gravity on earth $(g)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ Acceleration due to gravity on mars $\left(\mathrm{g}^{\prime}\right)=\frac{\mathrm{GM}_{\mathrm{m}}}{\left(\mathrm{R}_{\mathrm{m}}\right)^{2}}$ Putting the value of these, we get - $\mathrm{g}^{\prime}=\mathrm{G} \frac{\mathrm{M}_{\mathrm{e}} / 10}{\left(\mathrm{R}_{\mathrm{e}} / 2\right)^{2}}$ $\mathrm{~g}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \times 4}{10 \mathrm{R}_{\mathrm{e}}{ }^{2}}$ $\mathrm{~g}^{\prime}=\frac{2}{5} \mathrm{~g}=0.4 \times 10=4 \mathrm{~m} / \mathrm{s}^{2}$
Kerala CEE - 2004
Gravitation
138297
The acceleration due to gravity on the surface of a planet is one-fourth of the value of Earth. When a brass ball is brought to this planet, its
1 mass is halved
2 weight is halved
3 mass becomes one-fourth
4 weight becomes one-fourth
5 mass and weight remain the same
Explanation:
D According to question $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{g}}}=\frac{\mathrm{g}_{\mathrm{p}}}{\mathrm{g}_{\mathrm{e}}}$ $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{g}}}=\frac{\mathrm{g}_{\mathrm{e}} / 4}{\mathrm{~g}_{\mathrm{e}}}$ $\frac{\mathrm{w}_{\mathrm{p}}}{\mathrm{w}_{\mathrm{e}}}=\frac{1}{4}$ $\mathrm{w}_{\mathrm{p}}=\frac{\mathrm{w}_{\mathrm{e}}}{4}$ Thus, when a brass ball is brought to this planet its weight becomes one fourth.
Kerala CEE - 2016
Gravitation
138298
A body hanging from a massless spring stretches is by $3 \mathbf{c m}$ on Earth's surface. At a place $800 \mathrm{~km}$ above the Earth's surface, the same body will stretch the spring by (Radius of Earth $=6400 \mathrm{~km}$ )
1 $\left(\frac{34}{27}\right) \mathrm{cm}$
2 $\left(\frac{64}{27}\right) \mathrm{cm}$
3 $\left(\frac{27}{64}\right) \mathrm{cm}$
4 $\left(\frac{27}{34}\right) \mathrm{cm}$
5 $\left(\frac{35}{81}\right) \mathrm{cm}$
Explanation:
B Given, $\mathrm{x}=3 \mathrm{~cm}$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}=6400 \times 10^{3} \mathrm{~m}$ Height from the earth surface (h) $=800 \mathrm{~km}=800 \times 10^{3} \mathrm{~m}$ When a body hanging from a massless spring on earth surface. So, Acceleration due to gravity, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{~F}=-\mathrm{kx}=\mathrm{mg}$ At a place $800 \mathrm{~km}$ above the earth's surface, the same body will stretch the spring by $x^{\prime}$ and the value of $g^{\prime}$ $\mathrm{F} =-k x^{\prime}=m g^{\prime}$ $g^{\prime} =\frac{G M}{(R+h)^{2}}$ Dividing equation (iii) by equation (ii), we get- $\frac{-\mathrm{kx}^{\prime}}{-\mathrm{kx}}=\frac{\mathrm{mg}}{\mathrm{mg}}$ $\frac{\mathrm{x}^{\prime}}{\mathrm{x}}=\frac{\mathrm{g}^{\prime}}{\mathrm{g}}$ $\frac{x^{\prime}}{x}=\frac{(R+h)^2}{\frac{G M}{R^2}}$ $\frac{x^{\prime}}{x}=\frac{R^2}{(R+h)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{h}{R}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{800 \times 10^3}{6400 \times 10^3}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{1}{\left(1+\frac{1}{8}\right)^2}$ $\frac{x^{\prime}}{x}=\frac{64}{81}$ $x^{\prime}=\frac{64}{81} x$ $x^{\prime}=\frac{64}{81} \times 3$ $x^{\prime}=\left(\frac{64}{27}\right) \mathrm{cm}$
Kerala CEE - 2016
Gravitation
138299
At what depth below the surface of the earth, the value of $g$ is the same as that at a height of 5 km?
1 $1.25 \mathrm{~km}$
2 $2.5 \mathrm{~km}$
3 $5 \mathrm{~km}$
4 $7.5 \mathrm{~km}$
5 $10 \mathrm{~km}$
Explanation:
E Given, Height $(\mathrm{h})=5 \mathrm{~km}$ Variation of $\mathrm{g}$ with depth $g=g_{o}\left(1-\frac{d}{R}\right)$ Variation of $\mathrm{g}$ with height $\mathrm{g}^{\prime}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ On Solving eq ${ }^{\mathrm{n}}$ (i) and (ii) We get $g_{o}\left(1-\frac{2 h}{R}\right)=g_{o}\left(1-\frac{d}{R}\right) \quad\left[\because g=g^{\prime}\right]$ $\frac{2 h}{R}=\frac{d}{R}$ $d=2 h$ $d=2 \times 5$ $d=10 k m .$
