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 $g=10 \mathrm{~ms}^{-2}$ )
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
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 $g=10 \mathrm{~ms}^{-2}$ )