NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
138300
The ratio of radii of earth to another planet is $\frac{2}{3}$ and the ratio of their mean densities is $\frac{4}{5}$, If an astronaut can jump to a maximum height of $1.5 \mathrm{~m}$ on the earth, with the same effort, the maximum height he can jump on the planet is
1 $1 \mathrm{~m}$
2 $0.8 \mathrm{~m}$
3 $0.5 \mathrm{~m}$
4 $1.25 \mathrm{~m}$
5 $2 \mathrm{~m}$
Explanation:
B Given, Ratio of radii of earth to another planet, $\frac{R_{e}}{R_{p}}=\frac{2}{3}$ Ratio of their mean densities, $\frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{p}}}=\frac{4}{5}$ On the earth $\left(\mathrm{h}_{\max }\right)=1.5 \mathrm{~m}$ On the planet $\left(\mathrm{h}_{\text {max }}^{\prime}\right)=$ ? Mass of the earth $\left(M_{e}\right)=\rho_{e} \times V$ $M_{e}=\rho_{e} \times \frac{4}{3} \pi R_{e}^{3}$ Gravitational acceleration at earth's surface $g_{e} =\frac{G M_{e}}{R_{e}^{2}}$ $g_{e} =\frac{G \times \rho_{e} \times \frac{4}{3} \pi R_{e}^{3}}{R_{e}^{2}}$ $g_{e} =G \times \rho_{e} \times \frac{4}{3} \pi R_{e}$ Gravitational acceleration at the planet's surface $g_{p}=G \times \rho_{p} \times \frac{4}{3} \pi R_{p}$ Astronaut mass on both planets $=\mathrm{m}$ At the maximum height, this energy will be converted to potential energy. $\mathrm{mg}_{\mathrm{e}} \mathrm{h}_{\max }=\mathrm{mg}_{\mathrm{p}} \mathrm{h}_{\text {max }}^{\prime}$ $\mathrm{h}_{\text {max }}^{\prime}=\mathrm{h}_{\max } \frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{p}}}$ Putting value of $\mathrm{eq}^{\mathrm{n}}$ (i) and (ii) in equation, we get $\mathrm{h}_{\text {max }}^{\prime}=\mathrm{h}_{\text {max }} \frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{p}}} \times \frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{p}}}$ $\mathrm{h}_{\text {max }}^{\prime}=1.5 \times \frac{4}{5} \times \frac{2}{3}$ $\mathrm{~h}_{\text {max }}^{\prime}=0.8 \mathrm{~m}$ Thus, at the maximum height $0.8 \mathrm{~m}$ astronaut can jump on the planet.
Kerala CEE - 2009
Gravitation
138301
Two planets have radii $r_{1}$ and $r_{2}$ and densities $d_{1}$ and $d_{2}$ respectively. Then the ratio of acceleration due to gravity on them will be :
A Given, radii $=\mathrm{r}_{1}$ and $\mathrm{r}_{2}$ Densities $=\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ Planet -1 Acceleration due to gravity on planet-1 $\mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{r}_{1}^{2}}$ $\because \mathrm{M}_{1}=\mathrm{d}_{1} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}$ $\therefore \mathrm{g}_{1}=\frac{\mathrm{G}\left(\mathrm{d}_{1} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}\right)}{\mathrm{r}_{1}^{2}}$ Planet -2 Acceleration due to gravity on planet-2 $\mathrm{g}_{2}=\frac{\mathrm{GM}_{2}}{\mathrm{r}_{2}^{2}}$ $\because \mathrm{M}_{2}=\mathrm{d}_{2} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}$ $\therefore \quad \mathrm{g}_{2}=\frac{\mathrm{G}\left(\mathrm{d}_{2} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}\right)}{\mathrm{r}_{2}^{2}}$ Then, On dividing equation (i) by equation (ii) The ratio of acceleration due to gravity on the both planets- \(\frac{\mathrm{g}_1}{\mathrm{~g}_2}=\frac{\frac{\mathrm{G}\left(\mathrm{d}_1 \times \frac{4}{3} \pi \mathrm{r}_1^3\right)}{\mathrm{r}_1^2}}{\frac{\mathrm{G}\left(\mathrm{d}_2 \times \frac{4}{3} \pi \mathrm{r}_2^3\right)}{\mathrm{r}_2^2}}\) $\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\mathrm{d}_{1} \mathrm{r}_{1}}{\mathrm{~d}_{2} \mathrm{r}_{2}}$ $\mathrm{g}_{1}: \mathrm{g}_{2}=\mathrm{d}_{1} \mathrm{r}_{1}: \mathrm{d}_{2} \mathrm{r}_{2}$
Kerala CEE 2006
Gravitation
138302
Acceleration due to gravity is $g$ on the surface of the earth. Then the value of the acceleration due to gravity at a height of $32 \mathrm{~km}$ above earth's surface is: (Assume radius of earth to be $6400 \mathrm{~km}$ )
1 $0.99 \mathrm{~g}$
2 $0.8 \mathrm{~g}$
3 $1.01 \mathrm{~g}$
4 $0.9 \mathrm{~g}$
5 $9 \mathrm{~g}$
Explanation:
A Given, $\mathrm{h}=32 \mathrm{~km}$ The radius of the earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ We know that $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}_{\mathrm{e}}}\right)$ Where, $\mathrm{g}^{\prime}=$ acceleration due to gravity at height $\mathrm{h}$. $\mathrm{g}=$ acceleration due to gravity on the earth's surface $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \times 32}{6400}\right)$ $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{64}{6400}\right)$ $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{1}{100}\right)$ $\mathrm{g}^{\prime}=(1-0.01)$ $\mathrm{g}^{\prime}=0.99 \mathrm{~g}$
Kerala CEE 2004
Gravitation
138303
The effect of rotation of the Earth on the value of acceleration due to gravity is
1 $\mathrm{g}$ is maximum at both poles
2 $g$ is minimum at both poles
3 $\mathrm{g}$ is maximum at equator and minimum at the poles
4 $\mathrm{g}$ is minimum at the equator and maximum at the poles
Explanation:
C We know that $g^{\prime}=g-\omega^{2} R \cos ^{2} \lambda$ At pole $\lambda=90^{\circ}$ $g_{\text {Pole }}^{\prime}=\mathrm{g}-\omega^{2} \operatorname{Rcos}^{2} 90^{\circ}=\mathrm{g}$ $\mathrm{g}_{\text {Pole }}^{\prime}=\mathrm{g}$ The value of $g^{\prime}$ not effected at the pole, i.e, There is zero effect at poles. At equator $\lambda=0^{\circ}$ $g_{\text {equator }}=\mathrm{g}-\omega^{2} R \cos ^{2} 0^{\circ}$ $\mathrm{g}^{\prime}=\mathrm{g}-\omega^{2} \mathrm{R}$ So, There is maximum effect at equator. Thus, the effect of rotation of the earth on the value of acceleration due to gravity is that $\mathrm{g}$ is maximum at the equator and minimum at the pole.
138300
The ratio of radii of earth to another planet is $\frac{2}{3}$ and the ratio of their mean densities is $\frac{4}{5}$, If an astronaut can jump to a maximum height of $1.5 \mathrm{~m}$ on the earth, with the same effort, the maximum height he can jump on the planet is
1 $1 \mathrm{~m}$
2 $0.8 \mathrm{~m}$
3 $0.5 \mathrm{~m}$
4 $1.25 \mathrm{~m}$
5 $2 \mathrm{~m}$
Explanation:
B Given, Ratio of radii of earth to another planet, $\frac{R_{e}}{R_{p}}=\frac{2}{3}$ Ratio of their mean densities, $\frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{p}}}=\frac{4}{5}$ On the earth $\left(\mathrm{h}_{\max }\right)=1.5 \mathrm{~m}$ On the planet $\left(\mathrm{h}_{\text {max }}^{\prime}\right)=$ ? Mass of the earth $\left(M_{e}\right)=\rho_{e} \times V$ $M_{e}=\rho_{e} \times \frac{4}{3} \pi R_{e}^{3}$ Gravitational acceleration at earth's surface $g_{e} =\frac{G M_{e}}{R_{e}^{2}}$ $g_{e} =\frac{G \times \rho_{e} \times \frac{4}{3} \pi R_{e}^{3}}{R_{e}^{2}}$ $g_{e} =G \times \rho_{e} \times \frac{4}{3} \pi R_{e}$ Gravitational acceleration at the planet's surface $g_{p}=G \times \rho_{p} \times \frac{4}{3} \pi R_{p}$ Astronaut mass on both planets $=\mathrm{m}$ At the maximum height, this energy will be converted to potential energy. $\mathrm{mg}_{\mathrm{e}} \mathrm{h}_{\max }=\mathrm{mg}_{\mathrm{p}} \mathrm{h}_{\text {max }}^{\prime}$ $\mathrm{h}_{\text {max }}^{\prime}=\mathrm{h}_{\max } \frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{p}}}$ Putting value of $\mathrm{eq}^{\mathrm{n}}$ (i) and (ii) in equation, we get $\mathrm{h}_{\text {max }}^{\prime}=\mathrm{h}_{\text {max }} \frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{p}}} \times \frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{p}}}$ $\mathrm{h}_{\text {max }}^{\prime}=1.5 \times \frac{4}{5} \times \frac{2}{3}$ $\mathrm{~h}_{\text {max }}^{\prime}=0.8 \mathrm{~m}$ Thus, at the maximum height $0.8 \mathrm{~m}$ astronaut can jump on the planet.
Kerala CEE - 2009
Gravitation
138301
Two planets have radii $r_{1}$ and $r_{2}$ and densities $d_{1}$ and $d_{2}$ respectively. Then the ratio of acceleration due to gravity on them will be :
A Given, radii $=\mathrm{r}_{1}$ and $\mathrm{r}_{2}$ Densities $=\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ Planet -1 Acceleration due to gravity on planet-1 $\mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{r}_{1}^{2}}$ $\because \mathrm{M}_{1}=\mathrm{d}_{1} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}$ $\therefore \mathrm{g}_{1}=\frac{\mathrm{G}\left(\mathrm{d}_{1} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}\right)}{\mathrm{r}_{1}^{2}}$ Planet -2 Acceleration due to gravity on planet-2 $\mathrm{g}_{2}=\frac{\mathrm{GM}_{2}}{\mathrm{r}_{2}^{2}}$ $\because \mathrm{M}_{2}=\mathrm{d}_{2} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}$ $\therefore \quad \mathrm{g}_{2}=\frac{\mathrm{G}\left(\mathrm{d}_{2} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}\right)}{\mathrm{r}_{2}^{2}}$ Then, On dividing equation (i) by equation (ii) The ratio of acceleration due to gravity on the both planets- \(\frac{\mathrm{g}_1}{\mathrm{~g}_2}=\frac{\frac{\mathrm{G}\left(\mathrm{d}_1 \times \frac{4}{3} \pi \mathrm{r}_1^3\right)}{\mathrm{r}_1^2}}{\frac{\mathrm{G}\left(\mathrm{d}_2 \times \frac{4}{3} \pi \mathrm{r}_2^3\right)}{\mathrm{r}_2^2}}\) $\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\mathrm{d}_{1} \mathrm{r}_{1}}{\mathrm{~d}_{2} \mathrm{r}_{2}}$ $\mathrm{g}_{1}: \mathrm{g}_{2}=\mathrm{d}_{1} \mathrm{r}_{1}: \mathrm{d}_{2} \mathrm{r}_{2}$
Kerala CEE 2006
Gravitation
138302
Acceleration due to gravity is $g$ on the surface of the earth. Then the value of the acceleration due to gravity at a height of $32 \mathrm{~km}$ above earth's surface is: (Assume radius of earth to be $6400 \mathrm{~km}$ )
1 $0.99 \mathrm{~g}$
2 $0.8 \mathrm{~g}$
3 $1.01 \mathrm{~g}$
4 $0.9 \mathrm{~g}$
5 $9 \mathrm{~g}$
Explanation:
A Given, $\mathrm{h}=32 \mathrm{~km}$ The radius of the earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ We know that $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}_{\mathrm{e}}}\right)$ Where, $\mathrm{g}^{\prime}=$ acceleration due to gravity at height $\mathrm{h}$. $\mathrm{g}=$ acceleration due to gravity on the earth's surface $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \times 32}{6400}\right)$ $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{64}{6400}\right)$ $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{1}{100}\right)$ $\mathrm{g}^{\prime}=(1-0.01)$ $\mathrm{g}^{\prime}=0.99 \mathrm{~g}$
Kerala CEE 2004
Gravitation
138303
The effect of rotation of the Earth on the value of acceleration due to gravity is
1 $\mathrm{g}$ is maximum at both poles
2 $g$ is minimum at both poles
3 $\mathrm{g}$ is maximum at equator and minimum at the poles
4 $\mathrm{g}$ is minimum at the equator and maximum at the poles
Explanation:
C We know that $g^{\prime}=g-\omega^{2} R \cos ^{2} \lambda$ At pole $\lambda=90^{\circ}$ $g_{\text {Pole }}^{\prime}=\mathrm{g}-\omega^{2} \operatorname{Rcos}^{2} 90^{\circ}=\mathrm{g}$ $\mathrm{g}_{\text {Pole }}^{\prime}=\mathrm{g}$ The value of $g^{\prime}$ not effected at the pole, i.e, There is zero effect at poles. At equator $\lambda=0^{\circ}$ $g_{\text {equator }}=\mathrm{g}-\omega^{2} R \cos ^{2} 0^{\circ}$ $\mathrm{g}^{\prime}=\mathrm{g}-\omega^{2} \mathrm{R}$ So, There is maximum effect at equator. Thus, the effect of rotation of the earth on the value of acceleration due to gravity is that $\mathrm{g}$ is maximum at the equator and minimum at the pole.
138300
The ratio of radii of earth to another planet is $\frac{2}{3}$ and the ratio of their mean densities is $\frac{4}{5}$, If an astronaut can jump to a maximum height of $1.5 \mathrm{~m}$ on the earth, with the same effort, the maximum height he can jump on the planet is
1 $1 \mathrm{~m}$
2 $0.8 \mathrm{~m}$
3 $0.5 \mathrm{~m}$
4 $1.25 \mathrm{~m}$
5 $2 \mathrm{~m}$
Explanation:
B Given, Ratio of radii of earth to another planet, $\frac{R_{e}}{R_{p}}=\frac{2}{3}$ Ratio of their mean densities, $\frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{p}}}=\frac{4}{5}$ On the earth $\left(\mathrm{h}_{\max }\right)=1.5 \mathrm{~m}$ On the planet $\left(\mathrm{h}_{\text {max }}^{\prime}\right)=$ ? Mass of the earth $\left(M_{e}\right)=\rho_{e} \times V$ $M_{e}=\rho_{e} \times \frac{4}{3} \pi R_{e}^{3}$ Gravitational acceleration at earth's surface $g_{e} =\frac{G M_{e}}{R_{e}^{2}}$ $g_{e} =\frac{G \times \rho_{e} \times \frac{4}{3} \pi R_{e}^{3}}{R_{e}^{2}}$ $g_{e} =G \times \rho_{e} \times \frac{4}{3} \pi R_{e}$ Gravitational acceleration at the planet's surface $g_{p}=G \times \rho_{p} \times \frac{4}{3} \pi R_{p}$ Astronaut mass on both planets $=\mathrm{m}$ At the maximum height, this energy will be converted to potential energy. $\mathrm{mg}_{\mathrm{e}} \mathrm{h}_{\max }=\mathrm{mg}_{\mathrm{p}} \mathrm{h}_{\text {max }}^{\prime}$ $\mathrm{h}_{\text {max }}^{\prime}=\mathrm{h}_{\max } \frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{p}}}$ Putting value of $\mathrm{eq}^{\mathrm{n}}$ (i) and (ii) in equation, we get $\mathrm{h}_{\text {max }}^{\prime}=\mathrm{h}_{\text {max }} \frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{p}}} \times \frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{p}}}$ $\mathrm{h}_{\text {max }}^{\prime}=1.5 \times \frac{4}{5} \times \frac{2}{3}$ $\mathrm{~h}_{\text {max }}^{\prime}=0.8 \mathrm{~m}$ Thus, at the maximum height $0.8 \mathrm{~m}$ astronaut can jump on the planet.
Kerala CEE - 2009
Gravitation
138301
Two planets have radii $r_{1}$ and $r_{2}$ and densities $d_{1}$ and $d_{2}$ respectively. Then the ratio of acceleration due to gravity on them will be :
A Given, radii $=\mathrm{r}_{1}$ and $\mathrm{r}_{2}$ Densities $=\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ Planet -1 Acceleration due to gravity on planet-1 $\mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{r}_{1}^{2}}$ $\because \mathrm{M}_{1}=\mathrm{d}_{1} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}$ $\therefore \mathrm{g}_{1}=\frac{\mathrm{G}\left(\mathrm{d}_{1} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}\right)}{\mathrm{r}_{1}^{2}}$ Planet -2 Acceleration due to gravity on planet-2 $\mathrm{g}_{2}=\frac{\mathrm{GM}_{2}}{\mathrm{r}_{2}^{2}}$ $\because \mathrm{M}_{2}=\mathrm{d}_{2} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}$ $\therefore \quad \mathrm{g}_{2}=\frac{\mathrm{G}\left(\mathrm{d}_{2} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}\right)}{\mathrm{r}_{2}^{2}}$ Then, On dividing equation (i) by equation (ii) The ratio of acceleration due to gravity on the both planets- \(\frac{\mathrm{g}_1}{\mathrm{~g}_2}=\frac{\frac{\mathrm{G}\left(\mathrm{d}_1 \times \frac{4}{3} \pi \mathrm{r}_1^3\right)}{\mathrm{r}_1^2}}{\frac{\mathrm{G}\left(\mathrm{d}_2 \times \frac{4}{3} \pi \mathrm{r}_2^3\right)}{\mathrm{r}_2^2}}\) $\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\mathrm{d}_{1} \mathrm{r}_{1}}{\mathrm{~d}_{2} \mathrm{r}_{2}}$ $\mathrm{g}_{1}: \mathrm{g}_{2}=\mathrm{d}_{1} \mathrm{r}_{1}: \mathrm{d}_{2} \mathrm{r}_{2}$
Kerala CEE 2006
Gravitation
138302
Acceleration due to gravity is $g$ on the surface of the earth. Then the value of the acceleration due to gravity at a height of $32 \mathrm{~km}$ above earth's surface is: (Assume radius of earth to be $6400 \mathrm{~km}$ )
1 $0.99 \mathrm{~g}$
2 $0.8 \mathrm{~g}$
3 $1.01 \mathrm{~g}$
4 $0.9 \mathrm{~g}$
5 $9 \mathrm{~g}$
Explanation:
A Given, $\mathrm{h}=32 \mathrm{~km}$ The radius of the earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ We know that $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}_{\mathrm{e}}}\right)$ Where, $\mathrm{g}^{\prime}=$ acceleration due to gravity at height $\mathrm{h}$. $\mathrm{g}=$ acceleration due to gravity on the earth's surface $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \times 32}{6400}\right)$ $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{64}{6400}\right)$ $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{1}{100}\right)$ $\mathrm{g}^{\prime}=(1-0.01)$ $\mathrm{g}^{\prime}=0.99 \mathrm{~g}$
Kerala CEE 2004
Gravitation
138303
The effect of rotation of the Earth on the value of acceleration due to gravity is
1 $\mathrm{g}$ is maximum at both poles
2 $g$ is minimum at both poles
3 $\mathrm{g}$ is maximum at equator and minimum at the poles
4 $\mathrm{g}$ is minimum at the equator and maximum at the poles
Explanation:
C We know that $g^{\prime}=g-\omega^{2} R \cos ^{2} \lambda$ At pole $\lambda=90^{\circ}$ $g_{\text {Pole }}^{\prime}=\mathrm{g}-\omega^{2} \operatorname{Rcos}^{2} 90^{\circ}=\mathrm{g}$ $\mathrm{g}_{\text {Pole }}^{\prime}=\mathrm{g}$ The value of $g^{\prime}$ not effected at the pole, i.e, There is zero effect at poles. At equator $\lambda=0^{\circ}$ $g_{\text {equator }}=\mathrm{g}-\omega^{2} R \cos ^{2} 0^{\circ}$ $\mathrm{g}^{\prime}=\mathrm{g}-\omega^{2} \mathrm{R}$ So, There is maximum effect at equator. Thus, the effect of rotation of the earth on the value of acceleration due to gravity is that $\mathrm{g}$ is maximum at the equator and minimum at the pole.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
138300
The ratio of radii of earth to another planet is $\frac{2}{3}$ and the ratio of their mean densities is $\frac{4}{5}$, If an astronaut can jump to a maximum height of $1.5 \mathrm{~m}$ on the earth, with the same effort, the maximum height he can jump on the planet is
1 $1 \mathrm{~m}$
2 $0.8 \mathrm{~m}$
3 $0.5 \mathrm{~m}$
4 $1.25 \mathrm{~m}$
5 $2 \mathrm{~m}$
Explanation:
B Given, Ratio of radii of earth to another planet, $\frac{R_{e}}{R_{p}}=\frac{2}{3}$ Ratio of their mean densities, $\frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{p}}}=\frac{4}{5}$ On the earth $\left(\mathrm{h}_{\max }\right)=1.5 \mathrm{~m}$ On the planet $\left(\mathrm{h}_{\text {max }}^{\prime}\right)=$ ? Mass of the earth $\left(M_{e}\right)=\rho_{e} \times V$ $M_{e}=\rho_{e} \times \frac{4}{3} \pi R_{e}^{3}$ Gravitational acceleration at earth's surface $g_{e} =\frac{G M_{e}}{R_{e}^{2}}$ $g_{e} =\frac{G \times \rho_{e} \times \frac{4}{3} \pi R_{e}^{3}}{R_{e}^{2}}$ $g_{e} =G \times \rho_{e} \times \frac{4}{3} \pi R_{e}$ Gravitational acceleration at the planet's surface $g_{p}=G \times \rho_{p} \times \frac{4}{3} \pi R_{p}$ Astronaut mass on both planets $=\mathrm{m}$ At the maximum height, this energy will be converted to potential energy. $\mathrm{mg}_{\mathrm{e}} \mathrm{h}_{\max }=\mathrm{mg}_{\mathrm{p}} \mathrm{h}_{\text {max }}^{\prime}$ $\mathrm{h}_{\text {max }}^{\prime}=\mathrm{h}_{\max } \frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{p}}}$ Putting value of $\mathrm{eq}^{\mathrm{n}}$ (i) and (ii) in equation, we get $\mathrm{h}_{\text {max }}^{\prime}=\mathrm{h}_{\text {max }} \frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{p}}} \times \frac{\mathrm{R}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{p}}}$ $\mathrm{h}_{\text {max }}^{\prime}=1.5 \times \frac{4}{5} \times \frac{2}{3}$ $\mathrm{~h}_{\text {max }}^{\prime}=0.8 \mathrm{~m}$ Thus, at the maximum height $0.8 \mathrm{~m}$ astronaut can jump on the planet.
Kerala CEE - 2009
Gravitation
138301
Two planets have radii $r_{1}$ and $r_{2}$ and densities $d_{1}$ and $d_{2}$ respectively. Then the ratio of acceleration due to gravity on them will be :
A Given, radii $=\mathrm{r}_{1}$ and $\mathrm{r}_{2}$ Densities $=\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ Planet -1 Acceleration due to gravity on planet-1 $\mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{r}_{1}^{2}}$ $\because \mathrm{M}_{1}=\mathrm{d}_{1} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}$ $\therefore \mathrm{g}_{1}=\frac{\mathrm{G}\left(\mathrm{d}_{1} \times \frac{4}{3} \pi \mathrm{r}_{1}^{3}\right)}{\mathrm{r}_{1}^{2}}$ Planet -2 Acceleration due to gravity on planet-2 $\mathrm{g}_{2}=\frac{\mathrm{GM}_{2}}{\mathrm{r}_{2}^{2}}$ $\because \mathrm{M}_{2}=\mathrm{d}_{2} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}$ $\therefore \quad \mathrm{g}_{2}=\frac{\mathrm{G}\left(\mathrm{d}_{2} \times \frac{4}{3} \pi \mathrm{r}_{2}^{3}\right)}{\mathrm{r}_{2}^{2}}$ Then, On dividing equation (i) by equation (ii) The ratio of acceleration due to gravity on the both planets- \(\frac{\mathrm{g}_1}{\mathrm{~g}_2}=\frac{\frac{\mathrm{G}\left(\mathrm{d}_1 \times \frac{4}{3} \pi \mathrm{r}_1^3\right)}{\mathrm{r}_1^2}}{\frac{\mathrm{G}\left(\mathrm{d}_2 \times \frac{4}{3} \pi \mathrm{r}_2^3\right)}{\mathrm{r}_2^2}}\) $\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\mathrm{d}_{1} \mathrm{r}_{1}}{\mathrm{~d}_{2} \mathrm{r}_{2}}$ $\mathrm{g}_{1}: \mathrm{g}_{2}=\mathrm{d}_{1} \mathrm{r}_{1}: \mathrm{d}_{2} \mathrm{r}_{2}$
Kerala CEE 2006
Gravitation
138302
Acceleration due to gravity is $g$ on the surface of the earth. Then the value of the acceleration due to gravity at a height of $32 \mathrm{~km}$ above earth's surface is: (Assume radius of earth to be $6400 \mathrm{~km}$ )
1 $0.99 \mathrm{~g}$
2 $0.8 \mathrm{~g}$
3 $1.01 \mathrm{~g}$
4 $0.9 \mathrm{~g}$
5 $9 \mathrm{~g}$
Explanation:
A Given, $\mathrm{h}=32 \mathrm{~km}$ The radius of the earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ We know that $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}_{\mathrm{e}}}\right)$ Where, $\mathrm{g}^{\prime}=$ acceleration due to gravity at height $\mathrm{h}$. $\mathrm{g}=$ acceleration due to gravity on the earth's surface $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \times 32}{6400}\right)$ $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{64}{6400}\right)$ $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{1}{100}\right)$ $\mathrm{g}^{\prime}=(1-0.01)$ $\mathrm{g}^{\prime}=0.99 \mathrm{~g}$
Kerala CEE 2004
Gravitation
138303
The effect of rotation of the Earth on the value of acceleration due to gravity is
1 $\mathrm{g}$ is maximum at both poles
2 $g$ is minimum at both poles
3 $\mathrm{g}$ is maximum at equator and minimum at the poles
4 $\mathrm{g}$ is minimum at the equator and maximum at the poles
Explanation:
C We know that $g^{\prime}=g-\omega^{2} R \cos ^{2} \lambda$ At pole $\lambda=90^{\circ}$ $g_{\text {Pole }}^{\prime}=\mathrm{g}-\omega^{2} \operatorname{Rcos}^{2} 90^{\circ}=\mathrm{g}$ $\mathrm{g}_{\text {Pole }}^{\prime}=\mathrm{g}$ The value of $g^{\prime}$ not effected at the pole, i.e, There is zero effect at poles. At equator $\lambda=0^{\circ}$ $g_{\text {equator }}=\mathrm{g}-\omega^{2} R \cos ^{2} 0^{\circ}$ $\mathrm{g}^{\prime}=\mathrm{g}-\omega^{2} \mathrm{R}$ So, There is maximum effect at equator. Thus, the effect of rotation of the earth on the value of acceleration due to gravity is that $\mathrm{g}$ is maximum at the equator and minimum at the pole.
