NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138286
A body weighs $w$ Newton at the surface of the earth. Its weight at a height equals to half the radius of the earth, will be
1 $\frac{\mathrm{W}}{2}$
2 $\frac{2 \mathrm{w}}{3}$
3 $\frac{4 \mathrm{w}}{9}$
4 $\frac{8 \mathrm{w}}{27}$
Explanation:
C Given that, A body weights w Newton at the surface of the earth. Height $(\mathrm{h})=$ Half of the radius of the earth $\mathrm{h}=\frac{\mathrm{R}}{2}$ We know that, Weight on surface of earth (w) $=\mathrm{mg}$ $\left[\because \mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}\right]$ Then, $w=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}$ Weight at distance from earth surface $\left(\mathrm{w}^{\prime}\right)=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{(\mathrm{R}+\mathrm{h})^2}$ $\mathrm{w}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\left(\mathrm{R}+\frac{\mathrm{R}}{2}\right)^2}$ $\mathrm{w}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\left(\frac{3 \mathrm{R}}{2}\right)^2}$ $\mathrm{w}^{\prime}=\frac{4}{9} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}$ On dividing equation (ii) by equation (i) we get $\frac{\mathrm{w}^{\prime}}{\mathrm{w}}=\frac{\frac{4}{9} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}}{\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}}$ $\frac{\mathrm{w}^{\prime}}{\mathrm{w}}=\frac{4}{9}$ $\mathrm{w}^{\prime}=\frac{4 \mathrm{w}}{9}$
JCECE-2010
Gravitation
138287
If the mass of moon is $\frac{M}{81}$, where $M$ is the mass of earth, find the distance of the point from the moon, where gravitational field due to earth and moon cancel each other. Given that distance between earth and moon is $60 \mathrm{R}$, where $R$ is the radius of earth.
1 $6 \mathrm{R}$
2 $8 \mathrm{R}$
3 $2 \mathrm{R}$
4 $4 \mathrm{R}$
Explanation:
A Given that, Mass of the moon is $\frac{M}{81}$, where $M$ is the mass of earth The distance between earth and moon is $60 \mathrm{R}$, where $\mathrm{R}$ is the radius of earth Let the distance of the point $(\mathrm{P})$ from the moon is $\mathrm{y}$. $P$ is at a distance $y$ from the moon and $(60 R-y)$ from the earth We know that, So, $F_{1}=\frac{G\left(\frac{M}{81}\right) m}{y^{2}}$ and $F_{2}=\frac{G M m}{(60 R-y)^{2}}$ Here, it is given that gravitational field due to earth and moon cancel to each other, the gravitational force will be equal. $F_{1}=F_{2}$ $\frac{G\left(\frac{M}{81}\right) m}{y^{2}}=\frac{G M m}{(60 R-y)^{2}}$ $\frac{1}{81 y^{2}}=\frac{1}{(60 R-y)^{2}}$ $\quad(60 R-y)^{2}=81 y^{2}$ $\Rightarrow 60 R-y=9 y$ $\Rightarrow 10 y=60 R$ $\Rightarrow y=\frac{60}{10} R$ $\Rightarrow y=6 R$ Hence, distance of that point from moon is 6R.
JCECE-2007
Gravitation
138288
The escapes velocity from the Earth is $11 \mathrm{kms}^{-1}$. The escape velocity from a planet having twice the radius and same mean density as that of Earth is
1 $11 \mathrm{kms}^{-1}$
2 $5.5 \mathrm{kms}^{-1}$
3 $22 \mathrm{kms}^{-1}$
4 $10 \mathrm{kms}^{-1}$
Explanation:
C Given that, The escapes velocity from the Earth $\left(\mathrm{v}_{\mathrm{e}}\right)=11 \mathrm{~km} / \mathrm{s}$ Let, Radius of Earth $=\mathrm{R}_{\mathrm{e}}$ According to questions, Radius of planet $\left(\mathrm{R}_{\mathrm{P}}\right)=2 \mathrm{R}_{\mathrm{e}}$ Density of earth $=$ Density of Planet $=\rho$ We know that, Escape Velocity $(\mathrm{v})=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ Then, $v=\sqrt{\frac{2 G \times\left(\frac{4}{3} \pi R^{3} \rho\right)}{R}}$ $[\because$ Mass $(\mathrm{m})=$ Volume $\times$ Density $]$ $\mathrm{v}=\sqrt{\frac{8}{3} \pi \mathrm{GR}^{2} \rho}$ $\mathrm{v} \propto \sqrt{\mathrm{R}^{2} \rho}$ $\therefore$ The escapes velocity from the earth- $\mathrm{v}_{\mathrm{e}} \propto \sqrt{\mathrm{R}_{\mathrm{e}}^{2} \rho}$ The escapes velocity from the planet- $\mathrm{v}_{\mathrm{p}} \propto \sqrt{\mathrm{R}_{\mathrm{p}}^{2} \rho}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\sqrt{\frac{\mathrm{R}_{\mathrm{e}}^{2}}{\mathrm{R}_{\mathrm{p}}^{2}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\sqrt{\frac{\mathrm{R}_{\mathrm{e}}^{2}}{\left(2 \mathrm{R}_{\mathrm{e}}\right)^{2}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\frac{1}{2}$ $\mathrm{v}_{\mathrm{p}} =2 \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{p}} =2 \times 11 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{p}} =22 \mathrm{kms}^{-1}$
UPSEE - 2015
Gravitation
138228
Two identical spheres of radius $R$ made of the same material are kept at a distance $d$ apart. Then the gravitational attraction between them is proportional to
1 $\mathrm{d}^{-2}$
2 $d^{2}$
3 $\mathrm{d}^{4}$
4 $d$
5 $\mathrm{d}^{-4}$
Explanation:
A According to Newton's law of gravitation $F=G \frac{m m}{d^{2}}$ $F \propto \frac{1}{d^{2}} \text { or } F \propto d^{-2} \text { and } F \propto m^{2}$ The gravitation attraction between the identical spheres is proportional to $\mathrm{d}^{-2}$.
138286
A body weighs $w$ Newton at the surface of the earth. Its weight at a height equals to half the radius of the earth, will be
1 $\frac{\mathrm{W}}{2}$
2 $\frac{2 \mathrm{w}}{3}$
3 $\frac{4 \mathrm{w}}{9}$
4 $\frac{8 \mathrm{w}}{27}$
Explanation:
C Given that, A body weights w Newton at the surface of the earth. Height $(\mathrm{h})=$ Half of the radius of the earth $\mathrm{h}=\frac{\mathrm{R}}{2}$ We know that, Weight on surface of earth (w) $=\mathrm{mg}$ $\left[\because \mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}\right]$ Then, $w=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}$ Weight at distance from earth surface $\left(\mathrm{w}^{\prime}\right)=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{(\mathrm{R}+\mathrm{h})^2}$ $\mathrm{w}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\left(\mathrm{R}+\frac{\mathrm{R}}{2}\right)^2}$ $\mathrm{w}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\left(\frac{3 \mathrm{R}}{2}\right)^2}$ $\mathrm{w}^{\prime}=\frac{4}{9} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}$ On dividing equation (ii) by equation (i) we get $\frac{\mathrm{w}^{\prime}}{\mathrm{w}}=\frac{\frac{4}{9} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}}{\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}}$ $\frac{\mathrm{w}^{\prime}}{\mathrm{w}}=\frac{4}{9}$ $\mathrm{w}^{\prime}=\frac{4 \mathrm{w}}{9}$
JCECE-2010
Gravitation
138287
If the mass of moon is $\frac{M}{81}$, where $M$ is the mass of earth, find the distance of the point from the moon, where gravitational field due to earth and moon cancel each other. Given that distance between earth and moon is $60 \mathrm{R}$, where $R$ is the radius of earth.
1 $6 \mathrm{R}$
2 $8 \mathrm{R}$
3 $2 \mathrm{R}$
4 $4 \mathrm{R}$
Explanation:
A Given that, Mass of the moon is $\frac{M}{81}$, where $M$ is the mass of earth The distance between earth and moon is $60 \mathrm{R}$, where $\mathrm{R}$ is the radius of earth Let the distance of the point $(\mathrm{P})$ from the moon is $\mathrm{y}$. $P$ is at a distance $y$ from the moon and $(60 R-y)$ from the earth We know that, So, $F_{1}=\frac{G\left(\frac{M}{81}\right) m}{y^{2}}$ and $F_{2}=\frac{G M m}{(60 R-y)^{2}}$ Here, it is given that gravitational field due to earth and moon cancel to each other, the gravitational force will be equal. $F_{1}=F_{2}$ $\frac{G\left(\frac{M}{81}\right) m}{y^{2}}=\frac{G M m}{(60 R-y)^{2}}$ $\frac{1}{81 y^{2}}=\frac{1}{(60 R-y)^{2}}$ $\quad(60 R-y)^{2}=81 y^{2}$ $\Rightarrow 60 R-y=9 y$ $\Rightarrow 10 y=60 R$ $\Rightarrow y=\frac{60}{10} R$ $\Rightarrow y=6 R$ Hence, distance of that point from moon is 6R.
JCECE-2007
Gravitation
138288
The escapes velocity from the Earth is $11 \mathrm{kms}^{-1}$. The escape velocity from a planet having twice the radius and same mean density as that of Earth is
1 $11 \mathrm{kms}^{-1}$
2 $5.5 \mathrm{kms}^{-1}$
3 $22 \mathrm{kms}^{-1}$
4 $10 \mathrm{kms}^{-1}$
Explanation:
C Given that, The escapes velocity from the Earth $\left(\mathrm{v}_{\mathrm{e}}\right)=11 \mathrm{~km} / \mathrm{s}$ Let, Radius of Earth $=\mathrm{R}_{\mathrm{e}}$ According to questions, Radius of planet $\left(\mathrm{R}_{\mathrm{P}}\right)=2 \mathrm{R}_{\mathrm{e}}$ Density of earth $=$ Density of Planet $=\rho$ We know that, Escape Velocity $(\mathrm{v})=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ Then, $v=\sqrt{\frac{2 G \times\left(\frac{4}{3} \pi R^{3} \rho\right)}{R}}$ $[\because$ Mass $(\mathrm{m})=$ Volume $\times$ Density $]$ $\mathrm{v}=\sqrt{\frac{8}{3} \pi \mathrm{GR}^{2} \rho}$ $\mathrm{v} \propto \sqrt{\mathrm{R}^{2} \rho}$ $\therefore$ The escapes velocity from the earth- $\mathrm{v}_{\mathrm{e}} \propto \sqrt{\mathrm{R}_{\mathrm{e}}^{2} \rho}$ The escapes velocity from the planet- $\mathrm{v}_{\mathrm{p}} \propto \sqrt{\mathrm{R}_{\mathrm{p}}^{2} \rho}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\sqrt{\frac{\mathrm{R}_{\mathrm{e}}^{2}}{\mathrm{R}_{\mathrm{p}}^{2}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\sqrt{\frac{\mathrm{R}_{\mathrm{e}}^{2}}{\left(2 \mathrm{R}_{\mathrm{e}}\right)^{2}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\frac{1}{2}$ $\mathrm{v}_{\mathrm{p}} =2 \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{p}} =2 \times 11 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{p}} =22 \mathrm{kms}^{-1}$
UPSEE - 2015
Gravitation
138228
Two identical spheres of radius $R$ made of the same material are kept at a distance $d$ apart. Then the gravitational attraction between them is proportional to
1 $\mathrm{d}^{-2}$
2 $d^{2}$
3 $\mathrm{d}^{4}$
4 $d$
5 $\mathrm{d}^{-4}$
Explanation:
A According to Newton's law of gravitation $F=G \frac{m m}{d^{2}}$ $F \propto \frac{1}{d^{2}} \text { or } F \propto d^{-2} \text { and } F \propto m^{2}$ The gravitation attraction between the identical spheres is proportional to $\mathrm{d}^{-2}$.
138286
A body weighs $w$ Newton at the surface of the earth. Its weight at a height equals to half the radius of the earth, will be
1 $\frac{\mathrm{W}}{2}$
2 $\frac{2 \mathrm{w}}{3}$
3 $\frac{4 \mathrm{w}}{9}$
4 $\frac{8 \mathrm{w}}{27}$
Explanation:
C Given that, A body weights w Newton at the surface of the earth. Height $(\mathrm{h})=$ Half of the radius of the earth $\mathrm{h}=\frac{\mathrm{R}}{2}$ We know that, Weight on surface of earth (w) $=\mathrm{mg}$ $\left[\because \mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}\right]$ Then, $w=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}$ Weight at distance from earth surface $\left(\mathrm{w}^{\prime}\right)=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{(\mathrm{R}+\mathrm{h})^2}$ $\mathrm{w}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\left(\mathrm{R}+\frac{\mathrm{R}}{2}\right)^2}$ $\mathrm{w}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\left(\frac{3 \mathrm{R}}{2}\right)^2}$ $\mathrm{w}^{\prime}=\frac{4}{9} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}$ On dividing equation (ii) by equation (i) we get $\frac{\mathrm{w}^{\prime}}{\mathrm{w}}=\frac{\frac{4}{9} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}}{\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}}$ $\frac{\mathrm{w}^{\prime}}{\mathrm{w}}=\frac{4}{9}$ $\mathrm{w}^{\prime}=\frac{4 \mathrm{w}}{9}$
JCECE-2010
Gravitation
138287
If the mass of moon is $\frac{M}{81}$, where $M$ is the mass of earth, find the distance of the point from the moon, where gravitational field due to earth and moon cancel each other. Given that distance between earth and moon is $60 \mathrm{R}$, where $R$ is the radius of earth.
1 $6 \mathrm{R}$
2 $8 \mathrm{R}$
3 $2 \mathrm{R}$
4 $4 \mathrm{R}$
Explanation:
A Given that, Mass of the moon is $\frac{M}{81}$, where $M$ is the mass of earth The distance between earth and moon is $60 \mathrm{R}$, where $\mathrm{R}$ is the radius of earth Let the distance of the point $(\mathrm{P})$ from the moon is $\mathrm{y}$. $P$ is at a distance $y$ from the moon and $(60 R-y)$ from the earth We know that, So, $F_{1}=\frac{G\left(\frac{M}{81}\right) m}{y^{2}}$ and $F_{2}=\frac{G M m}{(60 R-y)^{2}}$ Here, it is given that gravitational field due to earth and moon cancel to each other, the gravitational force will be equal. $F_{1}=F_{2}$ $\frac{G\left(\frac{M}{81}\right) m}{y^{2}}=\frac{G M m}{(60 R-y)^{2}}$ $\frac{1}{81 y^{2}}=\frac{1}{(60 R-y)^{2}}$ $\quad(60 R-y)^{2}=81 y^{2}$ $\Rightarrow 60 R-y=9 y$ $\Rightarrow 10 y=60 R$ $\Rightarrow y=\frac{60}{10} R$ $\Rightarrow y=6 R$ Hence, distance of that point from moon is 6R.
JCECE-2007
Gravitation
138288
The escapes velocity from the Earth is $11 \mathrm{kms}^{-1}$. The escape velocity from a planet having twice the radius and same mean density as that of Earth is
1 $11 \mathrm{kms}^{-1}$
2 $5.5 \mathrm{kms}^{-1}$
3 $22 \mathrm{kms}^{-1}$
4 $10 \mathrm{kms}^{-1}$
Explanation:
C Given that, The escapes velocity from the Earth $\left(\mathrm{v}_{\mathrm{e}}\right)=11 \mathrm{~km} / \mathrm{s}$ Let, Radius of Earth $=\mathrm{R}_{\mathrm{e}}$ According to questions, Radius of planet $\left(\mathrm{R}_{\mathrm{P}}\right)=2 \mathrm{R}_{\mathrm{e}}$ Density of earth $=$ Density of Planet $=\rho$ We know that, Escape Velocity $(\mathrm{v})=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ Then, $v=\sqrt{\frac{2 G \times\left(\frac{4}{3} \pi R^{3} \rho\right)}{R}}$ $[\because$ Mass $(\mathrm{m})=$ Volume $\times$ Density $]$ $\mathrm{v}=\sqrt{\frac{8}{3} \pi \mathrm{GR}^{2} \rho}$ $\mathrm{v} \propto \sqrt{\mathrm{R}^{2} \rho}$ $\therefore$ The escapes velocity from the earth- $\mathrm{v}_{\mathrm{e}} \propto \sqrt{\mathrm{R}_{\mathrm{e}}^{2} \rho}$ The escapes velocity from the planet- $\mathrm{v}_{\mathrm{p}} \propto \sqrt{\mathrm{R}_{\mathrm{p}}^{2} \rho}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\sqrt{\frac{\mathrm{R}_{\mathrm{e}}^{2}}{\mathrm{R}_{\mathrm{p}}^{2}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\sqrt{\frac{\mathrm{R}_{\mathrm{e}}^{2}}{\left(2 \mathrm{R}_{\mathrm{e}}\right)^{2}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\frac{1}{2}$ $\mathrm{v}_{\mathrm{p}} =2 \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{p}} =2 \times 11 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{p}} =22 \mathrm{kms}^{-1}$
UPSEE - 2015
Gravitation
138228
Two identical spheres of radius $R$ made of the same material are kept at a distance $d$ apart. Then the gravitational attraction between them is proportional to
1 $\mathrm{d}^{-2}$
2 $d^{2}$
3 $\mathrm{d}^{4}$
4 $d$
5 $\mathrm{d}^{-4}$
Explanation:
A According to Newton's law of gravitation $F=G \frac{m m}{d^{2}}$ $F \propto \frac{1}{d^{2}} \text { or } F \propto d^{-2} \text { and } F \propto m^{2}$ The gravitation attraction between the identical spheres is proportional to $\mathrm{d}^{-2}$.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
138286
A body weighs $w$ Newton at the surface of the earth. Its weight at a height equals to half the radius of the earth, will be
1 $\frac{\mathrm{W}}{2}$
2 $\frac{2 \mathrm{w}}{3}$
3 $\frac{4 \mathrm{w}}{9}$
4 $\frac{8 \mathrm{w}}{27}$
Explanation:
C Given that, A body weights w Newton at the surface of the earth. Height $(\mathrm{h})=$ Half of the radius of the earth $\mathrm{h}=\frac{\mathrm{R}}{2}$ We know that, Weight on surface of earth (w) $=\mathrm{mg}$ $\left[\because \mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}\right]$ Then, $w=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}$ Weight at distance from earth surface $\left(\mathrm{w}^{\prime}\right)=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{(\mathrm{R}+\mathrm{h})^2}$ $\mathrm{w}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\left(\mathrm{R}+\frac{\mathrm{R}}{2}\right)^2}$ $\mathrm{w}^{\prime}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\left(\frac{3 \mathrm{R}}{2}\right)^2}$ $\mathrm{w}^{\prime}=\frac{4}{9} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}$ On dividing equation (ii) by equation (i) we get $\frac{\mathrm{w}^{\prime}}{\mathrm{w}}=\frac{\frac{4}{9} \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}}{\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}^2}}$ $\frac{\mathrm{w}^{\prime}}{\mathrm{w}}=\frac{4}{9}$ $\mathrm{w}^{\prime}=\frac{4 \mathrm{w}}{9}$
JCECE-2010
Gravitation
138287
If the mass of moon is $\frac{M}{81}$, where $M$ is the mass of earth, find the distance of the point from the moon, where gravitational field due to earth and moon cancel each other. Given that distance between earth and moon is $60 \mathrm{R}$, where $R$ is the radius of earth.
1 $6 \mathrm{R}$
2 $8 \mathrm{R}$
3 $2 \mathrm{R}$
4 $4 \mathrm{R}$
Explanation:
A Given that, Mass of the moon is $\frac{M}{81}$, where $M$ is the mass of earth The distance between earth and moon is $60 \mathrm{R}$, where $\mathrm{R}$ is the radius of earth Let the distance of the point $(\mathrm{P})$ from the moon is $\mathrm{y}$. $P$ is at a distance $y$ from the moon and $(60 R-y)$ from the earth We know that, So, $F_{1}=\frac{G\left(\frac{M}{81}\right) m}{y^{2}}$ and $F_{2}=\frac{G M m}{(60 R-y)^{2}}$ Here, it is given that gravitational field due to earth and moon cancel to each other, the gravitational force will be equal. $F_{1}=F_{2}$ $\frac{G\left(\frac{M}{81}\right) m}{y^{2}}=\frac{G M m}{(60 R-y)^{2}}$ $\frac{1}{81 y^{2}}=\frac{1}{(60 R-y)^{2}}$ $\quad(60 R-y)^{2}=81 y^{2}$ $\Rightarrow 60 R-y=9 y$ $\Rightarrow 10 y=60 R$ $\Rightarrow y=\frac{60}{10} R$ $\Rightarrow y=6 R$ Hence, distance of that point from moon is 6R.
JCECE-2007
Gravitation
138288
The escapes velocity from the Earth is $11 \mathrm{kms}^{-1}$. The escape velocity from a planet having twice the radius and same mean density as that of Earth is
1 $11 \mathrm{kms}^{-1}$
2 $5.5 \mathrm{kms}^{-1}$
3 $22 \mathrm{kms}^{-1}$
4 $10 \mathrm{kms}^{-1}$
Explanation:
C Given that, The escapes velocity from the Earth $\left(\mathrm{v}_{\mathrm{e}}\right)=11 \mathrm{~km} / \mathrm{s}$ Let, Radius of Earth $=\mathrm{R}_{\mathrm{e}}$ According to questions, Radius of planet $\left(\mathrm{R}_{\mathrm{P}}\right)=2 \mathrm{R}_{\mathrm{e}}$ Density of earth $=$ Density of Planet $=\rho$ We know that, Escape Velocity $(\mathrm{v})=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ Then, $v=\sqrt{\frac{2 G \times\left(\frac{4}{3} \pi R^{3} \rho\right)}{R}}$ $[\because$ Mass $(\mathrm{m})=$ Volume $\times$ Density $]$ $\mathrm{v}=\sqrt{\frac{8}{3} \pi \mathrm{GR}^{2} \rho}$ $\mathrm{v} \propto \sqrt{\mathrm{R}^{2} \rho}$ $\therefore$ The escapes velocity from the earth- $\mathrm{v}_{\mathrm{e}} \propto \sqrt{\mathrm{R}_{\mathrm{e}}^{2} \rho}$ The escapes velocity from the planet- $\mathrm{v}_{\mathrm{p}} \propto \sqrt{\mathrm{R}_{\mathrm{p}}^{2} \rho}$ Dividing equation (1) by equation (2) we get- $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\sqrt{\frac{\mathrm{R}_{\mathrm{e}}^{2}}{\mathrm{R}_{\mathrm{p}}^{2}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\sqrt{\frac{\mathrm{R}_{\mathrm{e}}^{2}}{\left(2 \mathrm{R}_{\mathrm{e}}\right)^{2}}}$ $\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}} =\frac{1}{2}$ $\mathrm{v}_{\mathrm{p}} =2 \mathrm{v}_{\mathrm{e}}$ $\mathrm{v}_{\mathrm{p}} =2 \times 11 \mathrm{~km} / \mathrm{s}$ $\mathrm{v}_{\mathrm{p}} =22 \mathrm{kms}^{-1}$
UPSEE - 2015
Gravitation
138228
Two identical spheres of radius $R$ made of the same material are kept at a distance $d$ apart. Then the gravitational attraction between them is proportional to
1 $\mathrm{d}^{-2}$
2 $d^{2}$
3 $\mathrm{d}^{4}$
4 $d$
5 $\mathrm{d}^{-4}$
Explanation:
A According to Newton's law of gravitation $F=G \frac{m m}{d^{2}}$ $F \propto \frac{1}{d^{2}} \text { or } F \propto d^{-2} \text { and } F \propto m^{2}$ The gravitation attraction between the identical spheres is proportional to $\mathrm{d}^{-2}$.
