138403
If the acceleration due to gravity $g$ at the earth's surface is $9.8 \mathrm{~ms}^{-2}$ and mass of earth is 80 times that of moon and radius of earth 4 times that of moon, the value of $g$ at the moon's surface will be
1 $1.96 \mathrm{~ms}^{-2}$
2 $2.96 \mathrm{~ms}^{-2}$
3 $0.96 \mathrm{~ms}^{-2}$
4 $3.96 \mathrm{~ms}^{-2}$
Explanation:
A Given, $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ Let mass of moon be ' $M$ ' and radius of moon is 'R'. $M_{e}=80 M$ $R_{e}=4 \times R$ We know, For moon, $\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ For Earth, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ \(g=\frac{G \times 80 \times M}{(4 R)^2} \quad\left(\begin{array}{c}\because M_e=80 M \\ R_e=4 R\end{array}\right.\) $=\frac{80}{16} \frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{~g}=5 \frac{\mathrm{GM}}{\mathrm{R}^{2}}$ By equation (i) \& (ii) $\mathrm{g}=5 \mathrm{~g}^{\prime}$ $\text { Then, } \quad \mathrm{g}^{\prime}=\frac{\mathrm{g}}{5}=\frac{9.8}{5}=1.96 \mathrm{~ms}^{-2}$
EAMCET-1993
Gravitation
138404
The mass of a planet is $1 / 9$ of the mass of the earth and its radius is half that of the earth. If a body weighs $9 \mathrm{~N}$ on the earth, its weight on the planet would be
1 $6 \mathrm{~N}$
2 $4 \mathrm{~N}$
3 $2 \mathrm{~N}$
4 $1 \mathrm{~N}$
Explanation:
B Given, $M^{\prime}=\frac{M_{e}}{9}$ $R^{\prime}=\frac{R_{e}}{2}$ For earth, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ For planet, $\mathrm{g}^{\prime}=\frac{\mathrm{GM}^{\prime}}{\left(\mathrm{R}^{\prime}\right)^{2}}$ $\mathrm{~g}^{\prime}=\mathrm{G} \times \frac{1}{9} \frac{\mathrm{M}_{\mathrm{e}}}{\left(\mathrm{R}_{\mathrm{e}} / 2\right)^{2}}=\frac{4}{9}\left(\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^{2}}\right)$ $\mathrm{g}^{\prime}=\frac{4}{9} \mathrm{~g}$ Multiplying both side with $\mathrm{m}$, $\mathrm{mg}^{\prime}=\frac{4}{9} \mathrm{mg}$ Now, $\text { Weight on planet }\left(\mathrm{w}^{\prime}\right)=\frac{4}{9} \times 9 \quad[\mathrm{w}=\mathrm{mg}]$ $\mathrm{w}^{\prime}=4 \mathrm{~N}$
EAMCET-1997
Gravitation
138405
Value of $g$ is
1 maximum at poles
2 maximum at equator
3 same everywhere
4 minimum at poles
Explanation:
A As shown in the figure, $r_{e}>r_{p}$ We know that, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{r}^{2}}$ $\Rightarrow \quad \mathrm{g} \propto \frac{1}{\mathrm{r}^{2}}$ $\therefore \quad \mathrm{g}_{\text {at pole }}>\mathrm{g}_{\text {at equator }}$ Value of $g$ is maximum at poles and minimum at equator because the distance from the poles to the centre of the earth is lesser than the distance from the equator to the centre of the earth. Thus, the acceleration due to gravity is greater at the poles than at the equator.
EAMCET-1998
Gravitation
138406
$R$ and $r$ are the radii of the earth and moon respectively, $\rho_{e}$ and $\rho_{m}$ are densities of earth and moon respectively. The ratio of the acceleration due to gravity on the surfaces of the moon and earth, is
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Gravitation
138403
If the acceleration due to gravity $g$ at the earth's surface is $9.8 \mathrm{~ms}^{-2}$ and mass of earth is 80 times that of moon and radius of earth 4 times that of moon, the value of $g$ at the moon's surface will be
1 $1.96 \mathrm{~ms}^{-2}$
2 $2.96 \mathrm{~ms}^{-2}$
3 $0.96 \mathrm{~ms}^{-2}$
4 $3.96 \mathrm{~ms}^{-2}$
Explanation:
A Given, $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ Let mass of moon be ' $M$ ' and radius of moon is 'R'. $M_{e}=80 M$ $R_{e}=4 \times R$ We know, For moon, $\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ For Earth, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ \(g=\frac{G \times 80 \times M}{(4 R)^2} \quad\left(\begin{array}{c}\because M_e=80 M \\ R_e=4 R\end{array}\right.\) $=\frac{80}{16} \frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{~g}=5 \frac{\mathrm{GM}}{\mathrm{R}^{2}}$ By equation (i) \& (ii) $\mathrm{g}=5 \mathrm{~g}^{\prime}$ $\text { Then, } \quad \mathrm{g}^{\prime}=\frac{\mathrm{g}}{5}=\frac{9.8}{5}=1.96 \mathrm{~ms}^{-2}$
EAMCET-1993
Gravitation
138404
The mass of a planet is $1 / 9$ of the mass of the earth and its radius is half that of the earth. If a body weighs $9 \mathrm{~N}$ on the earth, its weight on the planet would be
1 $6 \mathrm{~N}$
2 $4 \mathrm{~N}$
3 $2 \mathrm{~N}$
4 $1 \mathrm{~N}$
Explanation:
B Given, $M^{\prime}=\frac{M_{e}}{9}$ $R^{\prime}=\frac{R_{e}}{2}$ For earth, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ For planet, $\mathrm{g}^{\prime}=\frac{\mathrm{GM}^{\prime}}{\left(\mathrm{R}^{\prime}\right)^{2}}$ $\mathrm{~g}^{\prime}=\mathrm{G} \times \frac{1}{9} \frac{\mathrm{M}_{\mathrm{e}}}{\left(\mathrm{R}_{\mathrm{e}} / 2\right)^{2}}=\frac{4}{9}\left(\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^{2}}\right)$ $\mathrm{g}^{\prime}=\frac{4}{9} \mathrm{~g}$ Multiplying both side with $\mathrm{m}$, $\mathrm{mg}^{\prime}=\frac{4}{9} \mathrm{mg}$ Now, $\text { Weight on planet }\left(\mathrm{w}^{\prime}\right)=\frac{4}{9} \times 9 \quad[\mathrm{w}=\mathrm{mg}]$ $\mathrm{w}^{\prime}=4 \mathrm{~N}$
EAMCET-1997
Gravitation
138405
Value of $g$ is
1 maximum at poles
2 maximum at equator
3 same everywhere
4 minimum at poles
Explanation:
A As shown in the figure, $r_{e}>r_{p}$ We know that, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{r}^{2}}$ $\Rightarrow \quad \mathrm{g} \propto \frac{1}{\mathrm{r}^{2}}$ $\therefore \quad \mathrm{g}_{\text {at pole }}>\mathrm{g}_{\text {at equator }}$ Value of $g$ is maximum at poles and minimum at equator because the distance from the poles to the centre of the earth is lesser than the distance from the equator to the centre of the earth. Thus, the acceleration due to gravity is greater at the poles than at the equator.
EAMCET-1998
Gravitation
138406
$R$ and $r$ are the radii of the earth and moon respectively, $\rho_{e}$ and $\rho_{m}$ are densities of earth and moon respectively. The ratio of the acceleration due to gravity on the surfaces of the moon and earth, is
138403
If the acceleration due to gravity $g$ at the earth's surface is $9.8 \mathrm{~ms}^{-2}$ and mass of earth is 80 times that of moon and radius of earth 4 times that of moon, the value of $g$ at the moon's surface will be
1 $1.96 \mathrm{~ms}^{-2}$
2 $2.96 \mathrm{~ms}^{-2}$
3 $0.96 \mathrm{~ms}^{-2}$
4 $3.96 \mathrm{~ms}^{-2}$
Explanation:
A Given, $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ Let mass of moon be ' $M$ ' and radius of moon is 'R'. $M_{e}=80 M$ $R_{e}=4 \times R$ We know, For moon, $\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ For Earth, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ \(g=\frac{G \times 80 \times M}{(4 R)^2} \quad\left(\begin{array}{c}\because M_e=80 M \\ R_e=4 R\end{array}\right.\) $=\frac{80}{16} \frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{~g}=5 \frac{\mathrm{GM}}{\mathrm{R}^{2}}$ By equation (i) \& (ii) $\mathrm{g}=5 \mathrm{~g}^{\prime}$ $\text { Then, } \quad \mathrm{g}^{\prime}=\frac{\mathrm{g}}{5}=\frac{9.8}{5}=1.96 \mathrm{~ms}^{-2}$
EAMCET-1993
Gravitation
138404
The mass of a planet is $1 / 9$ of the mass of the earth and its radius is half that of the earth. If a body weighs $9 \mathrm{~N}$ on the earth, its weight on the planet would be
1 $6 \mathrm{~N}$
2 $4 \mathrm{~N}$
3 $2 \mathrm{~N}$
4 $1 \mathrm{~N}$
Explanation:
B Given, $M^{\prime}=\frac{M_{e}}{9}$ $R^{\prime}=\frac{R_{e}}{2}$ For earth, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ For planet, $\mathrm{g}^{\prime}=\frac{\mathrm{GM}^{\prime}}{\left(\mathrm{R}^{\prime}\right)^{2}}$ $\mathrm{~g}^{\prime}=\mathrm{G} \times \frac{1}{9} \frac{\mathrm{M}_{\mathrm{e}}}{\left(\mathrm{R}_{\mathrm{e}} / 2\right)^{2}}=\frac{4}{9}\left(\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^{2}}\right)$ $\mathrm{g}^{\prime}=\frac{4}{9} \mathrm{~g}$ Multiplying both side with $\mathrm{m}$, $\mathrm{mg}^{\prime}=\frac{4}{9} \mathrm{mg}$ Now, $\text { Weight on planet }\left(\mathrm{w}^{\prime}\right)=\frac{4}{9} \times 9 \quad[\mathrm{w}=\mathrm{mg}]$ $\mathrm{w}^{\prime}=4 \mathrm{~N}$
EAMCET-1997
Gravitation
138405
Value of $g$ is
1 maximum at poles
2 maximum at equator
3 same everywhere
4 minimum at poles
Explanation:
A As shown in the figure, $r_{e}>r_{p}$ We know that, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{r}^{2}}$ $\Rightarrow \quad \mathrm{g} \propto \frac{1}{\mathrm{r}^{2}}$ $\therefore \quad \mathrm{g}_{\text {at pole }}>\mathrm{g}_{\text {at equator }}$ Value of $g$ is maximum at poles and minimum at equator because the distance from the poles to the centre of the earth is lesser than the distance from the equator to the centre of the earth. Thus, the acceleration due to gravity is greater at the poles than at the equator.
EAMCET-1998
Gravitation
138406
$R$ and $r$ are the radii of the earth and moon respectively, $\rho_{e}$ and $\rho_{m}$ are densities of earth and moon respectively. The ratio of the acceleration due to gravity on the surfaces of the moon and earth, is
138403
If the acceleration due to gravity $g$ at the earth's surface is $9.8 \mathrm{~ms}^{-2}$ and mass of earth is 80 times that of moon and radius of earth 4 times that of moon, the value of $g$ at the moon's surface will be
1 $1.96 \mathrm{~ms}^{-2}$
2 $2.96 \mathrm{~ms}^{-2}$
3 $0.96 \mathrm{~ms}^{-2}$
4 $3.96 \mathrm{~ms}^{-2}$
Explanation:
A Given, $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$ Let mass of moon be ' $M$ ' and radius of moon is 'R'. $M_{e}=80 M$ $R_{e}=4 \times R$ We know, For moon, $\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ For Earth, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ \(g=\frac{G \times 80 \times M}{(4 R)^2} \quad\left(\begin{array}{c}\because M_e=80 M \\ R_e=4 R\end{array}\right.\) $=\frac{80}{16} \frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{~g}=5 \frac{\mathrm{GM}}{\mathrm{R}^{2}}$ By equation (i) \& (ii) $\mathrm{g}=5 \mathrm{~g}^{\prime}$ $\text { Then, } \quad \mathrm{g}^{\prime}=\frac{\mathrm{g}}{5}=\frac{9.8}{5}=1.96 \mathrm{~ms}^{-2}$
EAMCET-1993
Gravitation
138404
The mass of a planet is $1 / 9$ of the mass of the earth and its radius is half that of the earth. If a body weighs $9 \mathrm{~N}$ on the earth, its weight on the planet would be
1 $6 \mathrm{~N}$
2 $4 \mathrm{~N}$
3 $2 \mathrm{~N}$
4 $1 \mathrm{~N}$
Explanation:
B Given, $M^{\prime}=\frac{M_{e}}{9}$ $R^{\prime}=\frac{R_{e}}{2}$ For earth, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ For planet, $\mathrm{g}^{\prime}=\frac{\mathrm{GM}^{\prime}}{\left(\mathrm{R}^{\prime}\right)^{2}}$ $\mathrm{~g}^{\prime}=\mathrm{G} \times \frac{1}{9} \frac{\mathrm{M}_{\mathrm{e}}}{\left(\mathrm{R}_{\mathrm{e}} / 2\right)^{2}}=\frac{4}{9}\left(\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^{2}}\right)$ $\mathrm{g}^{\prime}=\frac{4}{9} \mathrm{~g}$ Multiplying both side with $\mathrm{m}$, $\mathrm{mg}^{\prime}=\frac{4}{9} \mathrm{mg}$ Now, $\text { Weight on planet }\left(\mathrm{w}^{\prime}\right)=\frac{4}{9} \times 9 \quad[\mathrm{w}=\mathrm{mg}]$ $\mathrm{w}^{\prime}=4 \mathrm{~N}$
EAMCET-1997
Gravitation
138405
Value of $g$ is
1 maximum at poles
2 maximum at equator
3 same everywhere
4 minimum at poles
Explanation:
A As shown in the figure, $r_{e}>r_{p}$ We know that, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{r}^{2}}$ $\Rightarrow \quad \mathrm{g} \propto \frac{1}{\mathrm{r}^{2}}$ $\therefore \quad \mathrm{g}_{\text {at pole }}>\mathrm{g}_{\text {at equator }}$ Value of $g$ is maximum at poles and minimum at equator because the distance from the poles to the centre of the earth is lesser than the distance from the equator to the centre of the earth. Thus, the acceleration due to gravity is greater at the poles than at the equator.
EAMCET-1998
Gravitation
138406
$R$ and $r$ are the radii of the earth and moon respectively, $\rho_{e}$ and $\rho_{m}$ are densities of earth and moon respectively. The ratio of the acceleration due to gravity on the surfaces of the moon and earth, is
