138392
The distance through which one has to dig the earth from its surface so as to reach the point where the acceleration due to gravity is reduced by $40 \%$ of that at the surface of the earth is - (radius of earth is $6400 \mathrm{~km}$ )
1 $2560 \mathrm{~km}$
2 $3000 \mathrm{~km}$
3 $3260 \mathrm{~km}$
4 $1560 \mathrm{~km}$
Explanation:
A Let we have to dig $\mathrm{d} \mathrm{km}$ from the earth's surface Given, $\mathrm{g}_{\mathrm{d}}=(100 \%-40 \%) \text { of } \mathrm{g}$ $\mathrm{g}_{\mathrm{d}}=60 \% \text { of } \mathrm{g}$ $\mathrm{g}_{\mathrm{d}}=0.6 \mathrm{~g}$ $\because \quad \text { At depth ' } \mathrm{d} \quad \mathrm{km} \text { ' value of gravitional }$ $\text { acceleration }$ $\quad \mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $0.6 \mathrm{~g}=\mathrm{g}\left(1-\frac{\mathrm{d}}{6400}\right) \quad(\mathrm{R}=6400 \mathrm{~km})$ $0.4 \mathrm{~g}=\frac{\mathrm{d}}{6400} \Rightarrow \mathrm{d}=2560 \mathrm{~km}$
AP EAMCET-19.08.2021
Gravitation
138393
A body weighs $900 \mathrm{gm}$-wt on the surface of the earth. Its weight on the surface of a planet whose mass is $(1 / 8)^{\text {th }}$ of that of the earth and radius is half that of the earth is
1 400 gm-wt
2 450 gm-wt
3 $200 \mathrm{gm}-\mathrm{wt}$
4 $50 \mathrm{gm}-\mathrm{wt}$
Explanation:
B We know that, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ On the given planet, $M_{P}=\frac{M}{8}$ $r=\frac{R}{2}$ $g_{P}=\frac{G M_{P}}{r^{2}}$ $\mathrm{g}_{\mathrm{P}}=\frac{\frac{\mathrm{GM}}{\frac{8}{\mathrm{R}^{2}}}}{\frac{4}{4}}=\frac{4 \mathrm{~g}}{8}=\frac{\mathrm{g}}{2}$ Hence, weight of the body on the planet $(w)=900 \times \frac{1}{2}$ $=450 \text { gm-wt }$
AP EAMCET-06.09.2021
Gravitation
138394
The radius of earth is about $6400 \mathrm{~km}$ and that of mars is $3200 \mathrm{~km}$. and mass of the earth is about 10 times mass of mars. An object weights $200 \mathrm{~N}$ on the surface of earth. Then its weight on the surface of mars will be
1 $80 \mathrm{~N}$
2 $40 \mathrm{~N}$
3 $20 \mathrm{~N}$
4 $8 \mathrm{~N}$
Explanation:
A Given, Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ Radius of mars $\left(\mathrm{R}_{\mathrm{m}}\right)=3200 \mathrm{~km}$ Then, $\quad \mathrm{R}_{\mathrm{m}}=\frac{\mathrm{R}}{2}$ Mass of earth $=M_{e}$ Mass of mars $=\mathrm{M}_{\mathrm{m}}$ Then, $\mathrm{M}_{\mathrm{e}}=10 \mathrm{M}_{\mathrm{m}}$ Value of $g$ on Mars $\left(g_{m}\right)=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{(\mathrm{R} / 2)^{2}}$ $\mathrm{~g}_{\mathrm{m}}=\frac{4 \mathrm{GM}_{\mathrm{m}}}{\mathrm{R}^{2}}$ And Value of $g$ on earth $\left(g_{e}\right)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}}$ $=\frac{\mathrm{G} \times\left(10 \mathrm{M}_{\mathrm{m}}\right)}{\mathrm{R}^{2}}$ $\mathrm{~g}_{\mathrm{e}} =\frac{10 \mathrm{G} \times \mathrm{M}_{\mathrm{m}}}{\mathrm{R}^{2}}$ By dividing equation (i) with equation (ii), $\frac{g_{m}}{g_{e}}=\frac{\frac{4 \mathrm{GM}_{m}}{\mathrm{R}^{2}}}{\frac{10 \mathrm{GM}_{m}}{\mathrm{R}^{2}}}$ $\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{1}{2.5}$ Given, Weight on earth $=200 \mathrm{~N}$ So, $\operatorname{mass}=200 / \mathrm{g}_{\mathrm{e}}$ Now, $\text { Weight on mars } =\text { mass } \times \mathrm{g}_{\mathrm{m}}$ $=\frac{200}{\mathrm{~g}_{\mathrm{e}}} \times \mathrm{g}_{\mathrm{m}}=\frac{200}{2.5}$ $=80 \mathrm{~N}$ So, weight of object on surface of mars will be $80 \mathrm{~N}$.
AP EAMCET-23.08.2021
Gravitation
138395
A body of mass $100 \mathrm{~kg}$ is lifted from earth's surface to a height $h=5 R$. The change in gravitational potential energy of the body is (Given the radius of earth $R=6400 \mathrm{~km}$. $\overline{\text { Also, }}$ on earth $g=10 \mathrm{~ms}^{-2}$ )
138392
The distance through which one has to dig the earth from its surface so as to reach the point where the acceleration due to gravity is reduced by $40 \%$ of that at the surface of the earth is - (radius of earth is $6400 \mathrm{~km}$ )
1 $2560 \mathrm{~km}$
2 $3000 \mathrm{~km}$
3 $3260 \mathrm{~km}$
4 $1560 \mathrm{~km}$
Explanation:
A Let we have to dig $\mathrm{d} \mathrm{km}$ from the earth's surface Given, $\mathrm{g}_{\mathrm{d}}=(100 \%-40 \%) \text { of } \mathrm{g}$ $\mathrm{g}_{\mathrm{d}}=60 \% \text { of } \mathrm{g}$ $\mathrm{g}_{\mathrm{d}}=0.6 \mathrm{~g}$ $\because \quad \text { At depth ' } \mathrm{d} \quad \mathrm{km} \text { ' value of gravitional }$ $\text { acceleration }$ $\quad \mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $0.6 \mathrm{~g}=\mathrm{g}\left(1-\frac{\mathrm{d}}{6400}\right) \quad(\mathrm{R}=6400 \mathrm{~km})$ $0.4 \mathrm{~g}=\frac{\mathrm{d}}{6400} \Rightarrow \mathrm{d}=2560 \mathrm{~km}$
AP EAMCET-19.08.2021
Gravitation
138393
A body weighs $900 \mathrm{gm}$-wt on the surface of the earth. Its weight on the surface of a planet whose mass is $(1 / 8)^{\text {th }}$ of that of the earth and radius is half that of the earth is
1 400 gm-wt
2 450 gm-wt
3 $200 \mathrm{gm}-\mathrm{wt}$
4 $50 \mathrm{gm}-\mathrm{wt}$
Explanation:
B We know that, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ On the given planet, $M_{P}=\frac{M}{8}$ $r=\frac{R}{2}$ $g_{P}=\frac{G M_{P}}{r^{2}}$ $\mathrm{g}_{\mathrm{P}}=\frac{\frac{\mathrm{GM}}{\frac{8}{\mathrm{R}^{2}}}}{\frac{4}{4}}=\frac{4 \mathrm{~g}}{8}=\frac{\mathrm{g}}{2}$ Hence, weight of the body on the planet $(w)=900 \times \frac{1}{2}$ $=450 \text { gm-wt }$
AP EAMCET-06.09.2021
Gravitation
138394
The radius of earth is about $6400 \mathrm{~km}$ and that of mars is $3200 \mathrm{~km}$. and mass of the earth is about 10 times mass of mars. An object weights $200 \mathrm{~N}$ on the surface of earth. Then its weight on the surface of mars will be
1 $80 \mathrm{~N}$
2 $40 \mathrm{~N}$
3 $20 \mathrm{~N}$
4 $8 \mathrm{~N}$
Explanation:
A Given, Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ Radius of mars $\left(\mathrm{R}_{\mathrm{m}}\right)=3200 \mathrm{~km}$ Then, $\quad \mathrm{R}_{\mathrm{m}}=\frac{\mathrm{R}}{2}$ Mass of earth $=M_{e}$ Mass of mars $=\mathrm{M}_{\mathrm{m}}$ Then, $\mathrm{M}_{\mathrm{e}}=10 \mathrm{M}_{\mathrm{m}}$ Value of $g$ on Mars $\left(g_{m}\right)=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{(\mathrm{R} / 2)^{2}}$ $\mathrm{~g}_{\mathrm{m}}=\frac{4 \mathrm{GM}_{\mathrm{m}}}{\mathrm{R}^{2}}$ And Value of $g$ on earth $\left(g_{e}\right)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}}$ $=\frac{\mathrm{G} \times\left(10 \mathrm{M}_{\mathrm{m}}\right)}{\mathrm{R}^{2}}$ $\mathrm{~g}_{\mathrm{e}} =\frac{10 \mathrm{G} \times \mathrm{M}_{\mathrm{m}}}{\mathrm{R}^{2}}$ By dividing equation (i) with equation (ii), $\frac{g_{m}}{g_{e}}=\frac{\frac{4 \mathrm{GM}_{m}}{\mathrm{R}^{2}}}{\frac{10 \mathrm{GM}_{m}}{\mathrm{R}^{2}}}$ $\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{1}{2.5}$ Given, Weight on earth $=200 \mathrm{~N}$ So, $\operatorname{mass}=200 / \mathrm{g}_{\mathrm{e}}$ Now, $\text { Weight on mars } =\text { mass } \times \mathrm{g}_{\mathrm{m}}$ $=\frac{200}{\mathrm{~g}_{\mathrm{e}}} \times \mathrm{g}_{\mathrm{m}}=\frac{200}{2.5}$ $=80 \mathrm{~N}$ So, weight of object on surface of mars will be $80 \mathrm{~N}$.
AP EAMCET-23.08.2021
Gravitation
138395
A body of mass $100 \mathrm{~kg}$ is lifted from earth's surface to a height $h=5 R$. The change in gravitational potential energy of the body is (Given the radius of earth $R=6400 \mathrm{~km}$. $\overline{\text { Also, }}$ on earth $g=10 \mathrm{~ms}^{-2}$ )
138392
The distance through which one has to dig the earth from its surface so as to reach the point where the acceleration due to gravity is reduced by $40 \%$ of that at the surface of the earth is - (radius of earth is $6400 \mathrm{~km}$ )
1 $2560 \mathrm{~km}$
2 $3000 \mathrm{~km}$
3 $3260 \mathrm{~km}$
4 $1560 \mathrm{~km}$
Explanation:
A Let we have to dig $\mathrm{d} \mathrm{km}$ from the earth's surface Given, $\mathrm{g}_{\mathrm{d}}=(100 \%-40 \%) \text { of } \mathrm{g}$ $\mathrm{g}_{\mathrm{d}}=60 \% \text { of } \mathrm{g}$ $\mathrm{g}_{\mathrm{d}}=0.6 \mathrm{~g}$ $\because \quad \text { At depth ' } \mathrm{d} \quad \mathrm{km} \text { ' value of gravitional }$ $\text { acceleration }$ $\quad \mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $0.6 \mathrm{~g}=\mathrm{g}\left(1-\frac{\mathrm{d}}{6400}\right) \quad(\mathrm{R}=6400 \mathrm{~km})$ $0.4 \mathrm{~g}=\frac{\mathrm{d}}{6400} \Rightarrow \mathrm{d}=2560 \mathrm{~km}$
AP EAMCET-19.08.2021
Gravitation
138393
A body weighs $900 \mathrm{gm}$-wt on the surface of the earth. Its weight on the surface of a planet whose mass is $(1 / 8)^{\text {th }}$ of that of the earth and radius is half that of the earth is
1 400 gm-wt
2 450 gm-wt
3 $200 \mathrm{gm}-\mathrm{wt}$
4 $50 \mathrm{gm}-\mathrm{wt}$
Explanation:
B We know that, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ On the given planet, $M_{P}=\frac{M}{8}$ $r=\frac{R}{2}$ $g_{P}=\frac{G M_{P}}{r^{2}}$ $\mathrm{g}_{\mathrm{P}}=\frac{\frac{\mathrm{GM}}{\frac{8}{\mathrm{R}^{2}}}}{\frac{4}{4}}=\frac{4 \mathrm{~g}}{8}=\frac{\mathrm{g}}{2}$ Hence, weight of the body on the planet $(w)=900 \times \frac{1}{2}$ $=450 \text { gm-wt }$
AP EAMCET-06.09.2021
Gravitation
138394
The radius of earth is about $6400 \mathrm{~km}$ and that of mars is $3200 \mathrm{~km}$. and mass of the earth is about 10 times mass of mars. An object weights $200 \mathrm{~N}$ on the surface of earth. Then its weight on the surface of mars will be
1 $80 \mathrm{~N}$
2 $40 \mathrm{~N}$
3 $20 \mathrm{~N}$
4 $8 \mathrm{~N}$
Explanation:
A Given, Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ Radius of mars $\left(\mathrm{R}_{\mathrm{m}}\right)=3200 \mathrm{~km}$ Then, $\quad \mathrm{R}_{\mathrm{m}}=\frac{\mathrm{R}}{2}$ Mass of earth $=M_{e}$ Mass of mars $=\mathrm{M}_{\mathrm{m}}$ Then, $\mathrm{M}_{\mathrm{e}}=10 \mathrm{M}_{\mathrm{m}}$ Value of $g$ on Mars $\left(g_{m}\right)=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{(\mathrm{R} / 2)^{2}}$ $\mathrm{~g}_{\mathrm{m}}=\frac{4 \mathrm{GM}_{\mathrm{m}}}{\mathrm{R}^{2}}$ And Value of $g$ on earth $\left(g_{e}\right)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}}$ $=\frac{\mathrm{G} \times\left(10 \mathrm{M}_{\mathrm{m}}\right)}{\mathrm{R}^{2}}$ $\mathrm{~g}_{\mathrm{e}} =\frac{10 \mathrm{G} \times \mathrm{M}_{\mathrm{m}}}{\mathrm{R}^{2}}$ By dividing equation (i) with equation (ii), $\frac{g_{m}}{g_{e}}=\frac{\frac{4 \mathrm{GM}_{m}}{\mathrm{R}^{2}}}{\frac{10 \mathrm{GM}_{m}}{\mathrm{R}^{2}}}$ $\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{1}{2.5}$ Given, Weight on earth $=200 \mathrm{~N}$ So, $\operatorname{mass}=200 / \mathrm{g}_{\mathrm{e}}$ Now, $\text { Weight on mars } =\text { mass } \times \mathrm{g}_{\mathrm{m}}$ $=\frac{200}{\mathrm{~g}_{\mathrm{e}}} \times \mathrm{g}_{\mathrm{m}}=\frac{200}{2.5}$ $=80 \mathrm{~N}$ So, weight of object on surface of mars will be $80 \mathrm{~N}$.
AP EAMCET-23.08.2021
Gravitation
138395
A body of mass $100 \mathrm{~kg}$ is lifted from earth's surface to a height $h=5 R$. The change in gravitational potential energy of the body is (Given the radius of earth $R=6400 \mathrm{~km}$. $\overline{\text { Also, }}$ on earth $g=10 \mathrm{~ms}^{-2}$ )
138392
The distance through which one has to dig the earth from its surface so as to reach the point where the acceleration due to gravity is reduced by $40 \%$ of that at the surface of the earth is - (radius of earth is $6400 \mathrm{~km}$ )
1 $2560 \mathrm{~km}$
2 $3000 \mathrm{~km}$
3 $3260 \mathrm{~km}$
4 $1560 \mathrm{~km}$
Explanation:
A Let we have to dig $\mathrm{d} \mathrm{km}$ from the earth's surface Given, $\mathrm{g}_{\mathrm{d}}=(100 \%-40 \%) \text { of } \mathrm{g}$ $\mathrm{g}_{\mathrm{d}}=60 \% \text { of } \mathrm{g}$ $\mathrm{g}_{\mathrm{d}}=0.6 \mathrm{~g}$ $\because \quad \text { At depth ' } \mathrm{d} \quad \mathrm{km} \text { ' value of gravitional }$ $\text { acceleration }$ $\quad \mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $0.6 \mathrm{~g}=\mathrm{g}\left(1-\frac{\mathrm{d}}{6400}\right) \quad(\mathrm{R}=6400 \mathrm{~km})$ $0.4 \mathrm{~g}=\frac{\mathrm{d}}{6400} \Rightarrow \mathrm{d}=2560 \mathrm{~km}$
AP EAMCET-19.08.2021
Gravitation
138393
A body weighs $900 \mathrm{gm}$-wt on the surface of the earth. Its weight on the surface of a planet whose mass is $(1 / 8)^{\text {th }}$ of that of the earth and radius is half that of the earth is
1 400 gm-wt
2 450 gm-wt
3 $200 \mathrm{gm}-\mathrm{wt}$
4 $50 \mathrm{gm}-\mathrm{wt}$
Explanation:
B We know that, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ On the given planet, $M_{P}=\frac{M}{8}$ $r=\frac{R}{2}$ $g_{P}=\frac{G M_{P}}{r^{2}}$ $\mathrm{g}_{\mathrm{P}}=\frac{\frac{\mathrm{GM}}{\frac{8}{\mathrm{R}^{2}}}}{\frac{4}{4}}=\frac{4 \mathrm{~g}}{8}=\frac{\mathrm{g}}{2}$ Hence, weight of the body on the planet $(w)=900 \times \frac{1}{2}$ $=450 \text { gm-wt }$
AP EAMCET-06.09.2021
Gravitation
138394
The radius of earth is about $6400 \mathrm{~km}$ and that of mars is $3200 \mathrm{~km}$. and mass of the earth is about 10 times mass of mars. An object weights $200 \mathrm{~N}$ on the surface of earth. Then its weight on the surface of mars will be
1 $80 \mathrm{~N}$
2 $40 \mathrm{~N}$
3 $20 \mathrm{~N}$
4 $8 \mathrm{~N}$
Explanation:
A Given, Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ Radius of mars $\left(\mathrm{R}_{\mathrm{m}}\right)=3200 \mathrm{~km}$ Then, $\quad \mathrm{R}_{\mathrm{m}}=\frac{\mathrm{R}}{2}$ Mass of earth $=M_{e}$ Mass of mars $=\mathrm{M}_{\mathrm{m}}$ Then, $\mathrm{M}_{\mathrm{e}}=10 \mathrm{M}_{\mathrm{m}}$ Value of $g$ on Mars $\left(g_{m}\right)=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{(\mathrm{R} / 2)^{2}}$ $\mathrm{~g}_{\mathrm{m}}=\frac{4 \mathrm{GM}_{\mathrm{m}}}{\mathrm{R}^{2}}$ And Value of $g$ on earth $\left(g_{e}\right)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}}$ $=\frac{\mathrm{G} \times\left(10 \mathrm{M}_{\mathrm{m}}\right)}{\mathrm{R}^{2}}$ $\mathrm{~g}_{\mathrm{e}} =\frac{10 \mathrm{G} \times \mathrm{M}_{\mathrm{m}}}{\mathrm{R}^{2}}$ By dividing equation (i) with equation (ii), $\frac{g_{m}}{g_{e}}=\frac{\frac{4 \mathrm{GM}_{m}}{\mathrm{R}^{2}}}{\frac{10 \mathrm{GM}_{m}}{\mathrm{R}^{2}}}$ $\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{1}{2.5}$ Given, Weight on earth $=200 \mathrm{~N}$ So, $\operatorname{mass}=200 / \mathrm{g}_{\mathrm{e}}$ Now, $\text { Weight on mars } =\text { mass } \times \mathrm{g}_{\mathrm{m}}$ $=\frac{200}{\mathrm{~g}_{\mathrm{e}}} \times \mathrm{g}_{\mathrm{m}}=\frac{200}{2.5}$ $=80 \mathrm{~N}$ So, weight of object on surface of mars will be $80 \mathrm{~N}$.
AP EAMCET-23.08.2021
Gravitation
138395
A body of mass $100 \mathrm{~kg}$ is lifted from earth's surface to a height $h=5 R$. The change in gravitational potential energy of the body is (Given the radius of earth $R=6400 \mathrm{~km}$. $\overline{\text { Also, }}$ on earth $g=10 \mathrm{~ms}^{-2}$ )
