138304
The ratio of acceleration due to gravity at a height $h$ above the surface of the earth and at a depth $h$ below the surface of the earth $h \lt R$, radius of earth
1 is constant
2 increases linearly with $h$
3 decreases linearly with $h$
4 decreases parabolically with $\mathrm{h}$
Explanation:
B Gravity at above earth's surface $\mathrm{g}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$ Gravity at below earth's surface $\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$ $\therefore$ Ratio of acceleration due to both gravities $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)}{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \times \frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{GM}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \times \mathrm{R}^{2}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}{\mathrm{R}^{2}}$ $So, $\quad \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$ $=\left(1-\frac{h}{R}\right)\left[1+\frac{2 h}{R}+\left(\frac{h}{R}\right)^{2}\right]$ $=1+\frac{2 h}{R}+\left(\frac{h}{R}\right)^{2}-\frac{h}{R}-2\left(\frac{h}{R}\right)^{2}-\left(\frac{h}{R}\right)^{3}$ $=1+\frac{h}{R}-\left(\frac{h}{R}\right)^{2}-\left(\frac{h}{R}\right)^{3}$ $\because \mathrm{h} \lt \mathrm{R}$, neglecting higher power of $\mathrm{h} / \mathrm{R}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=1+\frac{\mathrm{h}}{\mathrm{R}}$ Thus, it increases linearly with ' $h$ '
UPSEE - 2013
Gravitation
138307
The ratio $\frac{g}{g_{h}}$, where $g$ and $g_{h}$ are the accelerations due to gravity at the surface of the earth and at a height $h$ above the earth's surface respectively, is :
A We know that- $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ The value of acceleration due to gravity $g$ at a height $\mathrm{h}$ above the surface of earth is $\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$ Where, $\mathrm{R}=$ radius of earth $\mathrm{h}=$ height above the earth's surface Dividing by equation (i) by equation (ii) $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\frac{\mathrm{GM}}{\mathrm{R}^{2}}}{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \times \frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{GM}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{R}^{2}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\mathrm{R}^{2}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}{\mathrm{R}^{2}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$
UPSEE - 2005
Gravitation
138309
If the density of a small planet is the same as that of earth, while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration of the surface of that planet is :
1 $0.2 \mathrm{~g}$
2 $0.4 \mathrm{~g}$
3 $2 \mathrm{~g}$
4 $4 \mathrm{~g}$
Explanation:
A Given, Radius of planet $=0.2$ times of the earth's radius Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Mass of planet, $\mathrm{m}=\mathrm{vd}=\mathrm{d} \times \frac{4}{3} \pi \mathrm{R}^{3}$ Where, $d=$ density of the planet for earth $\mathrm{g}=\frac{4}{3} \pi \mathrm{dGR}$ $\because \quad \mathrm{d}^{\prime}=\mathrm{d} \quad \text { and } \quad \mathrm{R}^{\prime}=0.2 \mathrm{R}$ $\therefore \quad \mathrm{g}^{\prime}=\frac{4}{3} \pi \mathrm{d}^{\prime} \mathrm{GR}^{\prime}$ $=\frac{4}{3} \pi \mathrm{dG}(0.2 \mathrm{R})=0.2\left(\frac{4}{3} \pi \mathrm{dGR}\right)$ From equation (i) $\mathrm{g}^{\prime}=0.2 \mathrm{~g}$
UPSEE - 2004
Gravitation
138310
Acceleration due to gravity at a height $h$ is equal to that at a depth $d$ below the surface of the earth, if
1 $\mathrm{d}=\mathrm{h}$
2 $2 \mathrm{~d}=\mathrm{h}$
3 $\mathrm{d}=2 \mathrm{~h}$
4 $3 \mathrm{~d}=\mathrm{h}$
Explanation:
C According to question- Acceleration due to gravity at height $\mathrm{h}$, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ Acceleration due to gravity at depth $\mathrm{d}$, $\mathrm{g}_{\mathrm{d}}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ From the equation (i) and (ii) $\mathrm{g}_{\mathrm{h}}=\mathrm{g}_{\mathrm{d}}$ $\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)=\mathrm{g}_{\mathrm{o}}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\mathrm{d}=2 \mathrm{~h}$
AP EAMCET (21.09.2020) Shift-II
Gravitation
138311
Two bodies each of mass $m$ are hung from a balance whose scale pans differ in a vertical height by $h$. If the mean density of the earth is $\rho$, the error in weighing is
1 $\frac{4 \pi \rho \mathrm{Gmh}}{3}$
2 $\frac{3 \pi \rho \mathrm{Gmh}}{4}$
3 $\frac{8 \pi \rho \mathrm{Gmh}}{3}$
4 $\frac{3 \pi \rho \mathrm{Gmh}}{8}$
Explanation:
C Given, $\text { Mass }=\mathrm{m}$ Density of earth $=\rho$ Let the weight of first body be $\mathrm{mg}_{\mathrm{h}}$ which is $\mathrm{h}$ metre higher and the weight of second body be $\mathrm{mg}$. Then, So, error in weighing $=m g-m_{h}$ $=\mathrm{mg}-\mathrm{mg}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right) \quad\left[\therefore \mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)\right]$ $=\frac{2 \mathrm{mgh}}{\mathrm{R}}$ $=\frac{2 \mathrm{mh}}{\mathrm{R}} \times \frac{\mathrm{GM}}{\mathrm{R}^{2}} \quad\left[\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ $=\frac{2 \mathrm{mh}}{\mathrm{R}^{3}}(\mathrm{G} \rho \mathrm{V})$ $=\frac{2 \mathrm{mh}}{\mathrm{R}^{3}} \times \mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho \quad\left[\therefore \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}\right]$ $=\frac{8}{3} \mathrm{~m} \rho \mathrm{h} \pi \mathrm{G}$
138304
The ratio of acceleration due to gravity at a height $h$ above the surface of the earth and at a depth $h$ below the surface of the earth $h \lt R$, radius of earth
1 is constant
2 increases linearly with $h$
3 decreases linearly with $h$
4 decreases parabolically with $\mathrm{h}$
Explanation:
B Gravity at above earth's surface $\mathrm{g}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$ Gravity at below earth's surface $\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$ $\therefore$ Ratio of acceleration due to both gravities $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)}{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \times \frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{GM}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \times \mathrm{R}^{2}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}{\mathrm{R}^{2}}$ $So, $\quad \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$ $=\left(1-\frac{h}{R}\right)\left[1+\frac{2 h}{R}+\left(\frac{h}{R}\right)^{2}\right]$ $=1+\frac{2 h}{R}+\left(\frac{h}{R}\right)^{2}-\frac{h}{R}-2\left(\frac{h}{R}\right)^{2}-\left(\frac{h}{R}\right)^{3}$ $=1+\frac{h}{R}-\left(\frac{h}{R}\right)^{2}-\left(\frac{h}{R}\right)^{3}$ $\because \mathrm{h} \lt \mathrm{R}$, neglecting higher power of $\mathrm{h} / \mathrm{R}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=1+\frac{\mathrm{h}}{\mathrm{R}}$ Thus, it increases linearly with ' $h$ '
UPSEE - 2013
Gravitation
138307
The ratio $\frac{g}{g_{h}}$, where $g$ and $g_{h}$ are the accelerations due to gravity at the surface of the earth and at a height $h$ above the earth's surface respectively, is :
A We know that- $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ The value of acceleration due to gravity $g$ at a height $\mathrm{h}$ above the surface of earth is $\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$ Where, $\mathrm{R}=$ radius of earth $\mathrm{h}=$ height above the earth's surface Dividing by equation (i) by equation (ii) $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\frac{\mathrm{GM}}{\mathrm{R}^{2}}}{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \times \frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{GM}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{R}^{2}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\mathrm{R}^{2}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}{\mathrm{R}^{2}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$
UPSEE - 2005
Gravitation
138309
If the density of a small planet is the same as that of earth, while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration of the surface of that planet is :
1 $0.2 \mathrm{~g}$
2 $0.4 \mathrm{~g}$
3 $2 \mathrm{~g}$
4 $4 \mathrm{~g}$
Explanation:
A Given, Radius of planet $=0.2$ times of the earth's radius Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Mass of planet, $\mathrm{m}=\mathrm{vd}=\mathrm{d} \times \frac{4}{3} \pi \mathrm{R}^{3}$ Where, $d=$ density of the planet for earth $\mathrm{g}=\frac{4}{3} \pi \mathrm{dGR}$ $\because \quad \mathrm{d}^{\prime}=\mathrm{d} \quad \text { and } \quad \mathrm{R}^{\prime}=0.2 \mathrm{R}$ $\therefore \quad \mathrm{g}^{\prime}=\frac{4}{3} \pi \mathrm{d}^{\prime} \mathrm{GR}^{\prime}$ $=\frac{4}{3} \pi \mathrm{dG}(0.2 \mathrm{R})=0.2\left(\frac{4}{3} \pi \mathrm{dGR}\right)$ From equation (i) $\mathrm{g}^{\prime}=0.2 \mathrm{~g}$
UPSEE - 2004
Gravitation
138310
Acceleration due to gravity at a height $h$ is equal to that at a depth $d$ below the surface of the earth, if
1 $\mathrm{d}=\mathrm{h}$
2 $2 \mathrm{~d}=\mathrm{h}$
3 $\mathrm{d}=2 \mathrm{~h}$
4 $3 \mathrm{~d}=\mathrm{h}$
Explanation:
C According to question- Acceleration due to gravity at height $\mathrm{h}$, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ Acceleration due to gravity at depth $\mathrm{d}$, $\mathrm{g}_{\mathrm{d}}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ From the equation (i) and (ii) $\mathrm{g}_{\mathrm{h}}=\mathrm{g}_{\mathrm{d}}$ $\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)=\mathrm{g}_{\mathrm{o}}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\mathrm{d}=2 \mathrm{~h}$
AP EAMCET (21.09.2020) Shift-II
Gravitation
138311
Two bodies each of mass $m$ are hung from a balance whose scale pans differ in a vertical height by $h$. If the mean density of the earth is $\rho$, the error in weighing is
1 $\frac{4 \pi \rho \mathrm{Gmh}}{3}$
2 $\frac{3 \pi \rho \mathrm{Gmh}}{4}$
3 $\frac{8 \pi \rho \mathrm{Gmh}}{3}$
4 $\frac{3 \pi \rho \mathrm{Gmh}}{8}$
Explanation:
C Given, $\text { Mass }=\mathrm{m}$ Density of earth $=\rho$ Let the weight of first body be $\mathrm{mg}_{\mathrm{h}}$ which is $\mathrm{h}$ metre higher and the weight of second body be $\mathrm{mg}$. Then, So, error in weighing $=m g-m_{h}$ $=\mathrm{mg}-\mathrm{mg}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right) \quad\left[\therefore \mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)\right]$ $=\frac{2 \mathrm{mgh}}{\mathrm{R}}$ $=\frac{2 \mathrm{mh}}{\mathrm{R}} \times \frac{\mathrm{GM}}{\mathrm{R}^{2}} \quad\left[\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ $=\frac{2 \mathrm{mh}}{\mathrm{R}^{3}}(\mathrm{G} \rho \mathrm{V})$ $=\frac{2 \mathrm{mh}}{\mathrm{R}^{3}} \times \mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho \quad\left[\therefore \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}\right]$ $=\frac{8}{3} \mathrm{~m} \rho \mathrm{h} \pi \mathrm{G}$
138304
The ratio of acceleration due to gravity at a height $h$ above the surface of the earth and at a depth $h$ below the surface of the earth $h \lt R$, radius of earth
1 is constant
2 increases linearly with $h$
3 decreases linearly with $h$
4 decreases parabolically with $\mathrm{h}$
Explanation:
B Gravity at above earth's surface $\mathrm{g}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$ Gravity at below earth's surface $\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$ $\therefore$ Ratio of acceleration due to both gravities $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)}{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \times \frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{GM}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \times \mathrm{R}^{2}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}{\mathrm{R}^{2}}$ $So, $\quad \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$ $=\left(1-\frac{h}{R}\right)\left[1+\frac{2 h}{R}+\left(\frac{h}{R}\right)^{2}\right]$ $=1+\frac{2 h}{R}+\left(\frac{h}{R}\right)^{2}-\frac{h}{R}-2\left(\frac{h}{R}\right)^{2}-\left(\frac{h}{R}\right)^{3}$ $=1+\frac{h}{R}-\left(\frac{h}{R}\right)^{2}-\left(\frac{h}{R}\right)^{3}$ $\because \mathrm{h} \lt \mathrm{R}$, neglecting higher power of $\mathrm{h} / \mathrm{R}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=1+\frac{\mathrm{h}}{\mathrm{R}}$ Thus, it increases linearly with ' $h$ '
UPSEE - 2013
Gravitation
138307
The ratio $\frac{g}{g_{h}}$, where $g$ and $g_{h}$ are the accelerations due to gravity at the surface of the earth and at a height $h$ above the earth's surface respectively, is :
A We know that- $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ The value of acceleration due to gravity $g$ at a height $\mathrm{h}$ above the surface of earth is $\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$ Where, $\mathrm{R}=$ radius of earth $\mathrm{h}=$ height above the earth's surface Dividing by equation (i) by equation (ii) $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\frac{\mathrm{GM}}{\mathrm{R}^{2}}}{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \times \frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{GM}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{R}^{2}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\mathrm{R}^{2}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}{\mathrm{R}^{2}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$
UPSEE - 2005
Gravitation
138309
If the density of a small planet is the same as that of earth, while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration of the surface of that planet is :
1 $0.2 \mathrm{~g}$
2 $0.4 \mathrm{~g}$
3 $2 \mathrm{~g}$
4 $4 \mathrm{~g}$
Explanation:
A Given, Radius of planet $=0.2$ times of the earth's radius Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Mass of planet, $\mathrm{m}=\mathrm{vd}=\mathrm{d} \times \frac{4}{3} \pi \mathrm{R}^{3}$ Where, $d=$ density of the planet for earth $\mathrm{g}=\frac{4}{3} \pi \mathrm{dGR}$ $\because \quad \mathrm{d}^{\prime}=\mathrm{d} \quad \text { and } \quad \mathrm{R}^{\prime}=0.2 \mathrm{R}$ $\therefore \quad \mathrm{g}^{\prime}=\frac{4}{3} \pi \mathrm{d}^{\prime} \mathrm{GR}^{\prime}$ $=\frac{4}{3} \pi \mathrm{dG}(0.2 \mathrm{R})=0.2\left(\frac{4}{3} \pi \mathrm{dGR}\right)$ From equation (i) $\mathrm{g}^{\prime}=0.2 \mathrm{~g}$
UPSEE - 2004
Gravitation
138310
Acceleration due to gravity at a height $h$ is equal to that at a depth $d$ below the surface of the earth, if
1 $\mathrm{d}=\mathrm{h}$
2 $2 \mathrm{~d}=\mathrm{h}$
3 $\mathrm{d}=2 \mathrm{~h}$
4 $3 \mathrm{~d}=\mathrm{h}$
Explanation:
C According to question- Acceleration due to gravity at height $\mathrm{h}$, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ Acceleration due to gravity at depth $\mathrm{d}$, $\mathrm{g}_{\mathrm{d}}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ From the equation (i) and (ii) $\mathrm{g}_{\mathrm{h}}=\mathrm{g}_{\mathrm{d}}$ $\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)=\mathrm{g}_{\mathrm{o}}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\mathrm{d}=2 \mathrm{~h}$
AP EAMCET (21.09.2020) Shift-II
Gravitation
138311
Two bodies each of mass $m$ are hung from a balance whose scale pans differ in a vertical height by $h$. If the mean density of the earth is $\rho$, the error in weighing is
1 $\frac{4 \pi \rho \mathrm{Gmh}}{3}$
2 $\frac{3 \pi \rho \mathrm{Gmh}}{4}$
3 $\frac{8 \pi \rho \mathrm{Gmh}}{3}$
4 $\frac{3 \pi \rho \mathrm{Gmh}}{8}$
Explanation:
C Given, $\text { Mass }=\mathrm{m}$ Density of earth $=\rho$ Let the weight of first body be $\mathrm{mg}_{\mathrm{h}}$ which is $\mathrm{h}$ metre higher and the weight of second body be $\mathrm{mg}$. Then, So, error in weighing $=m g-m_{h}$ $=\mathrm{mg}-\mathrm{mg}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right) \quad\left[\therefore \mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)\right]$ $=\frac{2 \mathrm{mgh}}{\mathrm{R}}$ $=\frac{2 \mathrm{mh}}{\mathrm{R}} \times \frac{\mathrm{GM}}{\mathrm{R}^{2}} \quad\left[\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ $=\frac{2 \mathrm{mh}}{\mathrm{R}^{3}}(\mathrm{G} \rho \mathrm{V})$ $=\frac{2 \mathrm{mh}}{\mathrm{R}^{3}} \times \mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho \quad\left[\therefore \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}\right]$ $=\frac{8}{3} \mathrm{~m} \rho \mathrm{h} \pi \mathrm{G}$
138304
The ratio of acceleration due to gravity at a height $h$ above the surface of the earth and at a depth $h$ below the surface of the earth $h \lt R$, radius of earth
1 is constant
2 increases linearly with $h$
3 decreases linearly with $h$
4 decreases parabolically with $\mathrm{h}$
Explanation:
B Gravity at above earth's surface $\mathrm{g}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$ Gravity at below earth's surface $\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$ $\therefore$ Ratio of acceleration due to both gravities $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)}{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \times \frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{GM}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \times \mathrm{R}^{2}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}{\mathrm{R}^{2}}$ $So, $\quad \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$ $=\left(1-\frac{h}{R}\right)\left[1+\frac{2 h}{R}+\left(\frac{h}{R}\right)^{2}\right]$ $=1+\frac{2 h}{R}+\left(\frac{h}{R}\right)^{2}-\frac{h}{R}-2\left(\frac{h}{R}\right)^{2}-\left(\frac{h}{R}\right)^{3}$ $=1+\frac{h}{R}-\left(\frac{h}{R}\right)^{2}-\left(\frac{h}{R}\right)^{3}$ $\because \mathrm{h} \lt \mathrm{R}$, neglecting higher power of $\mathrm{h} / \mathrm{R}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=1+\frac{\mathrm{h}}{\mathrm{R}}$ Thus, it increases linearly with ' $h$ '
UPSEE - 2013
Gravitation
138307
The ratio $\frac{g}{g_{h}}$, where $g$ and $g_{h}$ are the accelerations due to gravity at the surface of the earth and at a height $h$ above the earth's surface respectively, is :
A We know that- $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ The value of acceleration due to gravity $g$ at a height $\mathrm{h}$ above the surface of earth is $\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$ Where, $\mathrm{R}=$ radius of earth $\mathrm{h}=$ height above the earth's surface Dividing by equation (i) by equation (ii) $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\frac{\mathrm{GM}}{\mathrm{R}^{2}}}{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \times \frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{GM}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{R}^{2}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\mathrm{R}^{2}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}{\mathrm{R}^{2}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$
UPSEE - 2005
Gravitation
138309
If the density of a small planet is the same as that of earth, while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration of the surface of that planet is :
1 $0.2 \mathrm{~g}$
2 $0.4 \mathrm{~g}$
3 $2 \mathrm{~g}$
4 $4 \mathrm{~g}$
Explanation:
A Given, Radius of planet $=0.2$ times of the earth's radius Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Mass of planet, $\mathrm{m}=\mathrm{vd}=\mathrm{d} \times \frac{4}{3} \pi \mathrm{R}^{3}$ Where, $d=$ density of the planet for earth $\mathrm{g}=\frac{4}{3} \pi \mathrm{dGR}$ $\because \quad \mathrm{d}^{\prime}=\mathrm{d} \quad \text { and } \quad \mathrm{R}^{\prime}=0.2 \mathrm{R}$ $\therefore \quad \mathrm{g}^{\prime}=\frac{4}{3} \pi \mathrm{d}^{\prime} \mathrm{GR}^{\prime}$ $=\frac{4}{3} \pi \mathrm{dG}(0.2 \mathrm{R})=0.2\left(\frac{4}{3} \pi \mathrm{dGR}\right)$ From equation (i) $\mathrm{g}^{\prime}=0.2 \mathrm{~g}$
UPSEE - 2004
Gravitation
138310
Acceleration due to gravity at a height $h$ is equal to that at a depth $d$ below the surface of the earth, if
1 $\mathrm{d}=\mathrm{h}$
2 $2 \mathrm{~d}=\mathrm{h}$
3 $\mathrm{d}=2 \mathrm{~h}$
4 $3 \mathrm{~d}=\mathrm{h}$
Explanation:
C According to question- Acceleration due to gravity at height $\mathrm{h}$, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ Acceleration due to gravity at depth $\mathrm{d}$, $\mathrm{g}_{\mathrm{d}}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ From the equation (i) and (ii) $\mathrm{g}_{\mathrm{h}}=\mathrm{g}_{\mathrm{d}}$ $\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)=\mathrm{g}_{\mathrm{o}}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\mathrm{d}=2 \mathrm{~h}$
AP EAMCET (21.09.2020) Shift-II
Gravitation
138311
Two bodies each of mass $m$ are hung from a balance whose scale pans differ in a vertical height by $h$. If the mean density of the earth is $\rho$, the error in weighing is
1 $\frac{4 \pi \rho \mathrm{Gmh}}{3}$
2 $\frac{3 \pi \rho \mathrm{Gmh}}{4}$
3 $\frac{8 \pi \rho \mathrm{Gmh}}{3}$
4 $\frac{3 \pi \rho \mathrm{Gmh}}{8}$
Explanation:
C Given, $\text { Mass }=\mathrm{m}$ Density of earth $=\rho$ Let the weight of first body be $\mathrm{mg}_{\mathrm{h}}$ which is $\mathrm{h}$ metre higher and the weight of second body be $\mathrm{mg}$. Then, So, error in weighing $=m g-m_{h}$ $=\mathrm{mg}-\mathrm{mg}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right) \quad\left[\therefore \mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)\right]$ $=\frac{2 \mathrm{mgh}}{\mathrm{R}}$ $=\frac{2 \mathrm{mh}}{\mathrm{R}} \times \frac{\mathrm{GM}}{\mathrm{R}^{2}} \quad\left[\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ $=\frac{2 \mathrm{mh}}{\mathrm{R}^{3}}(\mathrm{G} \rho \mathrm{V})$ $=\frac{2 \mathrm{mh}}{\mathrm{R}^{3}} \times \mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho \quad\left[\therefore \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}\right]$ $=\frac{8}{3} \mathrm{~m} \rho \mathrm{h} \pi \mathrm{G}$
138304
The ratio of acceleration due to gravity at a height $h$ above the surface of the earth and at a depth $h$ below the surface of the earth $h \lt R$, radius of earth
1 is constant
2 increases linearly with $h$
3 decreases linearly with $h$
4 decreases parabolically with $\mathrm{h}$
Explanation:
B Gravity at above earth's surface $\mathrm{g}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$ Gravity at below earth's surface $\mathrm{g}^{\prime}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$ $\therefore$ Ratio of acceleration due to both gravities $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)}{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \times \frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{GM}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \times \mathrm{R}^{2}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}{\mathrm{R}^{2}}$ $So, $\quad \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$ $=\left(1-\frac{h}{R}\right)\left[1+\frac{2 h}{R}+\left(\frac{h}{R}\right)^{2}\right]$ $=1+\frac{2 h}{R}+\left(\frac{h}{R}\right)^{2}-\frac{h}{R}-2\left(\frac{h}{R}\right)^{2}-\left(\frac{h}{R}\right)^{3}$ $=1+\frac{h}{R}-\left(\frac{h}{R}\right)^{2}-\left(\frac{h}{R}\right)^{3}$ $\because \mathrm{h} \lt \mathrm{R}$, neglecting higher power of $\mathrm{h} / \mathrm{R}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=1+\frac{\mathrm{h}}{\mathrm{R}}$ Thus, it increases linearly with ' $h$ '
UPSEE - 2013
Gravitation
138307
The ratio $\frac{g}{g_{h}}$, where $g$ and $g_{h}$ are the accelerations due to gravity at the surface of the earth and at a height $h$ above the earth's surface respectively, is :
A We know that- $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ The value of acceleration due to gravity $g$ at a height $\mathrm{h}$ above the surface of earth is $\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}$ Where, $\mathrm{R}=$ radius of earth $\mathrm{h}=$ height above the earth's surface Dividing by equation (i) by equation (ii) $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\frac{\mathrm{GM}}{\mathrm{R}^{2}}}{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{2}}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \times \frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{GM}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{(\mathrm{R}+\mathrm{h})^{2}}{\mathrm{R}^{2}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\frac{\mathrm{R}^{2}\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}{\mathrm{R}^{2}}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}=\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$
UPSEE - 2005
Gravitation
138309
If the density of a small planet is the same as that of earth, while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration of the surface of that planet is :
1 $0.2 \mathrm{~g}$
2 $0.4 \mathrm{~g}$
3 $2 \mathrm{~g}$
4 $4 \mathrm{~g}$
Explanation:
A Given, Radius of planet $=0.2$ times of the earth's radius Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Mass of planet, $\mathrm{m}=\mathrm{vd}=\mathrm{d} \times \frac{4}{3} \pi \mathrm{R}^{3}$ Where, $d=$ density of the planet for earth $\mathrm{g}=\frac{4}{3} \pi \mathrm{dGR}$ $\because \quad \mathrm{d}^{\prime}=\mathrm{d} \quad \text { and } \quad \mathrm{R}^{\prime}=0.2 \mathrm{R}$ $\therefore \quad \mathrm{g}^{\prime}=\frac{4}{3} \pi \mathrm{d}^{\prime} \mathrm{GR}^{\prime}$ $=\frac{4}{3} \pi \mathrm{dG}(0.2 \mathrm{R})=0.2\left(\frac{4}{3} \pi \mathrm{dGR}\right)$ From equation (i) $\mathrm{g}^{\prime}=0.2 \mathrm{~g}$
UPSEE - 2004
Gravitation
138310
Acceleration due to gravity at a height $h$ is equal to that at a depth $d$ below the surface of the earth, if
1 $\mathrm{d}=\mathrm{h}$
2 $2 \mathrm{~d}=\mathrm{h}$
3 $\mathrm{d}=2 \mathrm{~h}$
4 $3 \mathrm{~d}=\mathrm{h}$
Explanation:
C According to question- Acceleration due to gravity at height $\mathrm{h}$, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ Acceleration due to gravity at depth $\mathrm{d}$, $\mathrm{g}_{\mathrm{d}}=\mathrm{g}_{\mathrm{o}}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ From the equation (i) and (ii) $\mathrm{g}_{\mathrm{h}}=\mathrm{g}_{\mathrm{d}}$ $\mathrm{g}_{\mathrm{o}}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)=\mathrm{g}_{\mathrm{o}}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\mathrm{d}=2 \mathrm{~h}$
AP EAMCET (21.09.2020) Shift-II
Gravitation
138311
Two bodies each of mass $m$ are hung from a balance whose scale pans differ in a vertical height by $h$. If the mean density of the earth is $\rho$, the error in weighing is
1 $\frac{4 \pi \rho \mathrm{Gmh}}{3}$
2 $\frac{3 \pi \rho \mathrm{Gmh}}{4}$
3 $\frac{8 \pi \rho \mathrm{Gmh}}{3}$
4 $\frac{3 \pi \rho \mathrm{Gmh}}{8}$
Explanation:
C Given, $\text { Mass }=\mathrm{m}$ Density of earth $=\rho$ Let the weight of first body be $\mathrm{mg}_{\mathrm{h}}$ which is $\mathrm{h}$ metre higher and the weight of second body be $\mathrm{mg}$. Then, So, error in weighing $=m g-m_{h}$ $=\mathrm{mg}-\mathrm{mg}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right) \quad\left[\therefore \mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)\right]$ $=\frac{2 \mathrm{mgh}}{\mathrm{R}}$ $=\frac{2 \mathrm{mh}}{\mathrm{R}} \times \frac{\mathrm{GM}}{\mathrm{R}^{2}} \quad\left[\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ $=\frac{2 \mathrm{mh}}{\mathrm{R}^{3}}(\mathrm{G} \rho \mathrm{V})$ $=\frac{2 \mathrm{mh}}{\mathrm{R}^{3}} \times \mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^{3} \rho \quad\left[\therefore \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^{3}\right]$ $=\frac{8}{3} \mathrm{~m} \rho \mathrm{h} \pi \mathrm{G}$
