NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138345
$\mathrm{g}_{\mathrm{e}}$ and $\mathrm{g}_{\mathrm{p}}$ denote the acceleration due to gravity on the surface of the earth and another planet whose mass and radius are twice to that of the earth, then:
A Given, Mass of earth $=M_{e}$ Radius of planet $=\mathrm{R}_{\mathrm{e}}$ Mass of planet $\left(\mathrm{M}_{\mathrm{P}}\right)=2 \mathrm{M}_{\mathrm{e}}$ Radius of planet $\left(\mathrm{R}_{\mathrm{P}}\right)=2 \mathrm{R}_{\mathrm{e}}$ We know, Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ For earth, $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ For planet, $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ Therefore, $\frac{g_{e}}{g_{P}}=\frac{G M_{e} / R_{e}^{2}}{G M_{P} / R_{P}^{2}}$ $\frac{g_{e}}{g_{P}}=\frac{M_{e}}{M_{P}} \times \frac{R_{P}^{2}}{R_{e}^{2}}$ Putting the value of $M_{P}$ and $R_{P}$ in the equation (i), we get $\frac{g_{e}}{g_{p}}=\frac{M_{e}}{2 M_{e}} \times \frac{\left(2 R_{e}\right)^{2}}{R_{e}^{2}}$ $\frac{g_{e}}{g_{P}}=\frac{1}{2} \times 4$ $\frac{g_{e}}{g_{P}}=2$ $g_{P}=\frac{g_{e}}{2}$
BCECE-2003
Gravitation
138346
A body situated on earth's surface at its equator becomes weightless when the rotational kinetic energy of the earth reaches a critical value which is given by ( $M$ and $R$ be the mass and radius of earth respectively)
1 $\frac{\mathrm{MgR}}{2}$
2 $\frac{\mathrm{MgR}}{3}$
3 $\frac{\mathrm{MgR}}{4}$
4 $\frac{\mathrm{MgR}}{5}$
Explanation:
D For the body on the equator to become weight less, \(\mathrm{mg}^{\prime}=0 \quad\left[\begin{array}{l}\text { As, } \mathrm{m} \neq 0 \\ \text { So, } \mathrm{g}^{\prime}=0\end{array}\right]\) $\mathrm{So}, \mathrm{g}^{\prime}=0$ We know, $g^{\prime}=g-R \omega^{2}$ $0=g-R \omega^{2}$ $\omega^{2}=\frac{g}{R}$ Rotational K.E. of the earth when reach a critical value, $\text { K.E. }=\frac{1}{2} \mathrm{I}^{2}=\frac{1}{2} \times \frac{2}{5} \mathrm{MR}^{2} \times \frac{\mathrm{g}}{\mathrm{R}} \quad\left[\therefore \mathrm{I}=\frac{2}{5} \mathrm{MR}^{2}\right]$ $\text { K.E. }=\frac{\mathrm{MgR}}{5}$
MHT-CET 2020
Gravitation
138347
At what height over the Earth's pole does the free fall acceleration decrease by $1 \%$ ? (Radius of the Earth $=6400 \mathrm{~km}$ )
1 $32 \mathrm{~km}$
2 $64 \mathrm{~km}$
3 $320 \mathrm{~km}$
4 $640 \mathrm{~km}$
Explanation:
A Given, Decrease in acceleration due to gravity $\left(\frac{\mathrm{g}-\mathrm{g}^{\prime}}{\mathrm{g}}\right)=1 \%$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}$ If acceleration due to gravity at a height (h) is $\mathrm{g}^{\prime}$. Then, $g^{\prime}=g\left(1-\frac{2 h}{R}\right)$ $g^{\prime}=g-\frac{2 g h}{R}$ $\frac{2 g h}{R}=g-g^{\prime}$ $h=\frac{R}{2}\left(\frac{g-g^{\prime}}{g}\right)$ $h=\frac{6400}{2} \times \frac{1}{100}$ $h=32 \mathrm{~km}$
SCRA-2010
Gravitation
138348
The value of acceleration due to gravity at a depth of $1600 \mathrm{~km}$ is equal to:
1 $4.8 \mathrm{~ms}^{-2}$
2 $9.8 \mathrm{~ms}^{-2}$
3 $7.35 \mathrm{~ms}^{-2}$
4 $19.6 \mathrm{~ms}^{-2}$
Explanation:
C Given, Depth $(\mathrm{d})=1600 \mathrm{~km}$ Acceleration due to gravity $(\mathrm{g})=9.8 \mathrm{~m} / \mathrm{s}^{2}$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}$ We know, Acceleration due to gravity at depth 'd' $\left(g_{d}\right)=g\left(1-\frac{d}{R}\right)$ $g_{d} =9.8\left(1-\frac{1600}{6400}\right)$ $=9.8\left(1-\frac{1}{4}\right)$ $=\frac{9.8 \times 3}{4}$ $g_{d} =7.35 \mathrm{~ms}^{-2}$
138345
$\mathrm{g}_{\mathrm{e}}$ and $\mathrm{g}_{\mathrm{p}}$ denote the acceleration due to gravity on the surface of the earth and another planet whose mass and radius are twice to that of the earth, then:
A Given, Mass of earth $=M_{e}$ Radius of planet $=\mathrm{R}_{\mathrm{e}}$ Mass of planet $\left(\mathrm{M}_{\mathrm{P}}\right)=2 \mathrm{M}_{\mathrm{e}}$ Radius of planet $\left(\mathrm{R}_{\mathrm{P}}\right)=2 \mathrm{R}_{\mathrm{e}}$ We know, Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ For earth, $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ For planet, $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ Therefore, $\frac{g_{e}}{g_{P}}=\frac{G M_{e} / R_{e}^{2}}{G M_{P} / R_{P}^{2}}$ $\frac{g_{e}}{g_{P}}=\frac{M_{e}}{M_{P}} \times \frac{R_{P}^{2}}{R_{e}^{2}}$ Putting the value of $M_{P}$ and $R_{P}$ in the equation (i), we get $\frac{g_{e}}{g_{p}}=\frac{M_{e}}{2 M_{e}} \times \frac{\left(2 R_{e}\right)^{2}}{R_{e}^{2}}$ $\frac{g_{e}}{g_{P}}=\frac{1}{2} \times 4$ $\frac{g_{e}}{g_{P}}=2$ $g_{P}=\frac{g_{e}}{2}$
BCECE-2003
Gravitation
138346
A body situated on earth's surface at its equator becomes weightless when the rotational kinetic energy of the earth reaches a critical value which is given by ( $M$ and $R$ be the mass and radius of earth respectively)
1 $\frac{\mathrm{MgR}}{2}$
2 $\frac{\mathrm{MgR}}{3}$
3 $\frac{\mathrm{MgR}}{4}$
4 $\frac{\mathrm{MgR}}{5}$
Explanation:
D For the body on the equator to become weight less, \(\mathrm{mg}^{\prime}=0 \quad\left[\begin{array}{l}\text { As, } \mathrm{m} \neq 0 \\ \text { So, } \mathrm{g}^{\prime}=0\end{array}\right]\) $\mathrm{So}, \mathrm{g}^{\prime}=0$ We know, $g^{\prime}=g-R \omega^{2}$ $0=g-R \omega^{2}$ $\omega^{2}=\frac{g}{R}$ Rotational K.E. of the earth when reach a critical value, $\text { K.E. }=\frac{1}{2} \mathrm{I}^{2}=\frac{1}{2} \times \frac{2}{5} \mathrm{MR}^{2} \times \frac{\mathrm{g}}{\mathrm{R}} \quad\left[\therefore \mathrm{I}=\frac{2}{5} \mathrm{MR}^{2}\right]$ $\text { K.E. }=\frac{\mathrm{MgR}}{5}$
MHT-CET 2020
Gravitation
138347
At what height over the Earth's pole does the free fall acceleration decrease by $1 \%$ ? (Radius of the Earth $=6400 \mathrm{~km}$ )
1 $32 \mathrm{~km}$
2 $64 \mathrm{~km}$
3 $320 \mathrm{~km}$
4 $640 \mathrm{~km}$
Explanation:
A Given, Decrease in acceleration due to gravity $\left(\frac{\mathrm{g}-\mathrm{g}^{\prime}}{\mathrm{g}}\right)=1 \%$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}$ If acceleration due to gravity at a height (h) is $\mathrm{g}^{\prime}$. Then, $g^{\prime}=g\left(1-\frac{2 h}{R}\right)$ $g^{\prime}=g-\frac{2 g h}{R}$ $\frac{2 g h}{R}=g-g^{\prime}$ $h=\frac{R}{2}\left(\frac{g-g^{\prime}}{g}\right)$ $h=\frac{6400}{2} \times \frac{1}{100}$ $h=32 \mathrm{~km}$
SCRA-2010
Gravitation
138348
The value of acceleration due to gravity at a depth of $1600 \mathrm{~km}$ is equal to:
1 $4.8 \mathrm{~ms}^{-2}$
2 $9.8 \mathrm{~ms}^{-2}$
3 $7.35 \mathrm{~ms}^{-2}$
4 $19.6 \mathrm{~ms}^{-2}$
Explanation:
C Given, Depth $(\mathrm{d})=1600 \mathrm{~km}$ Acceleration due to gravity $(\mathrm{g})=9.8 \mathrm{~m} / \mathrm{s}^{2}$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}$ We know, Acceleration due to gravity at depth 'd' $\left(g_{d}\right)=g\left(1-\frac{d}{R}\right)$ $g_{d} =9.8\left(1-\frac{1600}{6400}\right)$ $=9.8\left(1-\frac{1}{4}\right)$ $=\frac{9.8 \times 3}{4}$ $g_{d} =7.35 \mathrm{~ms}^{-2}$
138345
$\mathrm{g}_{\mathrm{e}}$ and $\mathrm{g}_{\mathrm{p}}$ denote the acceleration due to gravity on the surface of the earth and another planet whose mass and radius are twice to that of the earth, then:
A Given, Mass of earth $=M_{e}$ Radius of planet $=\mathrm{R}_{\mathrm{e}}$ Mass of planet $\left(\mathrm{M}_{\mathrm{P}}\right)=2 \mathrm{M}_{\mathrm{e}}$ Radius of planet $\left(\mathrm{R}_{\mathrm{P}}\right)=2 \mathrm{R}_{\mathrm{e}}$ We know, Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ For earth, $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ For planet, $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ Therefore, $\frac{g_{e}}{g_{P}}=\frac{G M_{e} / R_{e}^{2}}{G M_{P} / R_{P}^{2}}$ $\frac{g_{e}}{g_{P}}=\frac{M_{e}}{M_{P}} \times \frac{R_{P}^{2}}{R_{e}^{2}}$ Putting the value of $M_{P}$ and $R_{P}$ in the equation (i), we get $\frac{g_{e}}{g_{p}}=\frac{M_{e}}{2 M_{e}} \times \frac{\left(2 R_{e}\right)^{2}}{R_{e}^{2}}$ $\frac{g_{e}}{g_{P}}=\frac{1}{2} \times 4$ $\frac{g_{e}}{g_{P}}=2$ $g_{P}=\frac{g_{e}}{2}$
BCECE-2003
Gravitation
138346
A body situated on earth's surface at its equator becomes weightless when the rotational kinetic energy of the earth reaches a critical value which is given by ( $M$ and $R$ be the mass and radius of earth respectively)
1 $\frac{\mathrm{MgR}}{2}$
2 $\frac{\mathrm{MgR}}{3}$
3 $\frac{\mathrm{MgR}}{4}$
4 $\frac{\mathrm{MgR}}{5}$
Explanation:
D For the body on the equator to become weight less, \(\mathrm{mg}^{\prime}=0 \quad\left[\begin{array}{l}\text { As, } \mathrm{m} \neq 0 \\ \text { So, } \mathrm{g}^{\prime}=0\end{array}\right]\) $\mathrm{So}, \mathrm{g}^{\prime}=0$ We know, $g^{\prime}=g-R \omega^{2}$ $0=g-R \omega^{2}$ $\omega^{2}=\frac{g}{R}$ Rotational K.E. of the earth when reach a critical value, $\text { K.E. }=\frac{1}{2} \mathrm{I}^{2}=\frac{1}{2} \times \frac{2}{5} \mathrm{MR}^{2} \times \frac{\mathrm{g}}{\mathrm{R}} \quad\left[\therefore \mathrm{I}=\frac{2}{5} \mathrm{MR}^{2}\right]$ $\text { K.E. }=\frac{\mathrm{MgR}}{5}$
MHT-CET 2020
Gravitation
138347
At what height over the Earth's pole does the free fall acceleration decrease by $1 \%$ ? (Radius of the Earth $=6400 \mathrm{~km}$ )
1 $32 \mathrm{~km}$
2 $64 \mathrm{~km}$
3 $320 \mathrm{~km}$
4 $640 \mathrm{~km}$
Explanation:
A Given, Decrease in acceleration due to gravity $\left(\frac{\mathrm{g}-\mathrm{g}^{\prime}}{\mathrm{g}}\right)=1 \%$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}$ If acceleration due to gravity at a height (h) is $\mathrm{g}^{\prime}$. Then, $g^{\prime}=g\left(1-\frac{2 h}{R}\right)$ $g^{\prime}=g-\frac{2 g h}{R}$ $\frac{2 g h}{R}=g-g^{\prime}$ $h=\frac{R}{2}\left(\frac{g-g^{\prime}}{g}\right)$ $h=\frac{6400}{2} \times \frac{1}{100}$ $h=32 \mathrm{~km}$
SCRA-2010
Gravitation
138348
The value of acceleration due to gravity at a depth of $1600 \mathrm{~km}$ is equal to:
1 $4.8 \mathrm{~ms}^{-2}$
2 $9.8 \mathrm{~ms}^{-2}$
3 $7.35 \mathrm{~ms}^{-2}$
4 $19.6 \mathrm{~ms}^{-2}$
Explanation:
C Given, Depth $(\mathrm{d})=1600 \mathrm{~km}$ Acceleration due to gravity $(\mathrm{g})=9.8 \mathrm{~m} / \mathrm{s}^{2}$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}$ We know, Acceleration due to gravity at depth 'd' $\left(g_{d}\right)=g\left(1-\frac{d}{R}\right)$ $g_{d} =9.8\left(1-\frac{1600}{6400}\right)$ $=9.8\left(1-\frac{1}{4}\right)$ $=\frac{9.8 \times 3}{4}$ $g_{d} =7.35 \mathrm{~ms}^{-2}$
138345
$\mathrm{g}_{\mathrm{e}}$ and $\mathrm{g}_{\mathrm{p}}$ denote the acceleration due to gravity on the surface of the earth and another planet whose mass and radius are twice to that of the earth, then:
A Given, Mass of earth $=M_{e}$ Radius of planet $=\mathrm{R}_{\mathrm{e}}$ Mass of planet $\left(\mathrm{M}_{\mathrm{P}}\right)=2 \mathrm{M}_{\mathrm{e}}$ Radius of planet $\left(\mathrm{R}_{\mathrm{P}}\right)=2 \mathrm{R}_{\mathrm{e}}$ We know, Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ For earth, $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ For planet, $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ Therefore, $\frac{g_{e}}{g_{P}}=\frac{G M_{e} / R_{e}^{2}}{G M_{P} / R_{P}^{2}}$ $\frac{g_{e}}{g_{P}}=\frac{M_{e}}{M_{P}} \times \frac{R_{P}^{2}}{R_{e}^{2}}$ Putting the value of $M_{P}$ and $R_{P}$ in the equation (i), we get $\frac{g_{e}}{g_{p}}=\frac{M_{e}}{2 M_{e}} \times \frac{\left(2 R_{e}\right)^{2}}{R_{e}^{2}}$ $\frac{g_{e}}{g_{P}}=\frac{1}{2} \times 4$ $\frac{g_{e}}{g_{P}}=2$ $g_{P}=\frac{g_{e}}{2}$
BCECE-2003
Gravitation
138346
A body situated on earth's surface at its equator becomes weightless when the rotational kinetic energy of the earth reaches a critical value which is given by ( $M$ and $R$ be the mass and radius of earth respectively)
1 $\frac{\mathrm{MgR}}{2}$
2 $\frac{\mathrm{MgR}}{3}$
3 $\frac{\mathrm{MgR}}{4}$
4 $\frac{\mathrm{MgR}}{5}$
Explanation:
D For the body on the equator to become weight less, \(\mathrm{mg}^{\prime}=0 \quad\left[\begin{array}{l}\text { As, } \mathrm{m} \neq 0 \\ \text { So, } \mathrm{g}^{\prime}=0\end{array}\right]\) $\mathrm{So}, \mathrm{g}^{\prime}=0$ We know, $g^{\prime}=g-R \omega^{2}$ $0=g-R \omega^{2}$ $\omega^{2}=\frac{g}{R}$ Rotational K.E. of the earth when reach a critical value, $\text { K.E. }=\frac{1}{2} \mathrm{I}^{2}=\frac{1}{2} \times \frac{2}{5} \mathrm{MR}^{2} \times \frac{\mathrm{g}}{\mathrm{R}} \quad\left[\therefore \mathrm{I}=\frac{2}{5} \mathrm{MR}^{2}\right]$ $\text { K.E. }=\frac{\mathrm{MgR}}{5}$
MHT-CET 2020
Gravitation
138347
At what height over the Earth's pole does the free fall acceleration decrease by $1 \%$ ? (Radius of the Earth $=6400 \mathrm{~km}$ )
1 $32 \mathrm{~km}$
2 $64 \mathrm{~km}$
3 $320 \mathrm{~km}$
4 $640 \mathrm{~km}$
Explanation:
A Given, Decrease in acceleration due to gravity $\left(\frac{\mathrm{g}-\mathrm{g}^{\prime}}{\mathrm{g}}\right)=1 \%$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}$ If acceleration due to gravity at a height (h) is $\mathrm{g}^{\prime}$. Then, $g^{\prime}=g\left(1-\frac{2 h}{R}\right)$ $g^{\prime}=g-\frac{2 g h}{R}$ $\frac{2 g h}{R}=g-g^{\prime}$ $h=\frac{R}{2}\left(\frac{g-g^{\prime}}{g}\right)$ $h=\frac{6400}{2} \times \frac{1}{100}$ $h=32 \mathrm{~km}$
SCRA-2010
Gravitation
138348
The value of acceleration due to gravity at a depth of $1600 \mathrm{~km}$ is equal to:
1 $4.8 \mathrm{~ms}^{-2}$
2 $9.8 \mathrm{~ms}^{-2}$
3 $7.35 \mathrm{~ms}^{-2}$
4 $19.6 \mathrm{~ms}^{-2}$
Explanation:
C Given, Depth $(\mathrm{d})=1600 \mathrm{~km}$ Acceleration due to gravity $(\mathrm{g})=9.8 \mathrm{~m} / \mathrm{s}^{2}$ Radius of earth $(\mathrm{R})=6400 \mathrm{~km}$ We know, Acceleration due to gravity at depth 'd' $\left(g_{d}\right)=g\left(1-\frac{d}{R}\right)$ $g_{d} =9.8\left(1-\frac{1600}{6400}\right)$ $=9.8\left(1-\frac{1}{4}\right)$ $=\frac{9.8 \times 3}{4}$ $g_{d} =7.35 \mathrm{~ms}^{-2}$
