138376
What will be the acceleration due to gravity at a depth $\mathrm{d}$, where $\mathrm{g}$ is acceleration due to gravity on the surface of earth ?
D Variation of $g$ with depth In this figure $\mathrm{m}=\text { mass }$ $\mathrm{d}=\text { distance below the earth's surface }$ We know that on the surface of the earth $\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{RG} \quad\left[\therefore \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ At a distance (d) below the earth's surface $\mathrm{g}_{\mathrm{d}}=\frac{4}{3} \pi \rho(\mathrm{R}-\mathrm{d}) \mathrm{G}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{d}}}=\frac{\mathrm{R}}{(\mathrm{R}-\mathrm{d})}$ $\mathrm{g}_{\mathrm{d}}=\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \mathrm{g}$ When, $d=0$ then value of $g$ on the surface And $\mathrm{g}_{\mathrm{d}}=\mathrm{g}$ When, $d=R$ then value of $g$ at the centre of the earth. $\mathrm{g}_{\mathrm{d}}=0$
UP CPMT-2010
Gravitation
138377
The period of oscillation of a simple pendulum of constant length at surface of the earth is $T$ its time period inside a mine will be
1 cannot be compared
2 equal to $\mathrm{T}$
3 less than $\mathrm{T}$
4 more than $\mathrm{T}$
Explanation:
D Time Period $(\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}}$ Let, mine be at a depth ' $h$ ' below the surface of the earth having radius $\mathrm{R}$, Then, $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$ Hence, $g$ decreases, Therefore ' $T$ ' increases So, time period of simple pendulum is greater than ' $\mathrm{T}$ ' as value of ' $\mathrm{g}$ ' is less in mine.
UP CPMT-2005
Gravitation
138378
The value of acceleration due to gravity $g$ at distance $r$ from earth's centre such that $r \lt R$ depend on $r$ according to relation: $(R=$ radius of earth)
1 $g \propto \frac{1}{\mathrm{r}^{2}}$
2 $g \propto \frac{1}{\mathrm{r}}$
3 $g \propto r$
4 $g \propto r^{2}$ [UP CMPT-2002]
Explanation:
C Gravitational force $(\mathrm{F})=\mathrm{mg}$ $\mathrm{g}=\frac{\mathrm{F}}{\mathrm{m}}$ $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{3}} \mathrm{r}$ $\mathrm{g}=\frac{\mathrm{GMmr}}{\mathrm{R}^{3} \mathrm{~m}}=\frac{\mathrm{Gm}}{\mathrm{R}^{3}} \mathrm{r}$ $\mathrm{g} \propto \mathrm{r}$
Gravitation
138379
At what depth below the surface of earth, the acceleration due to gravity $g$ will be half of its value $1600 \mathrm{~km}$ above the surface of earth? (Radius of earth $=6400 \mathrm{~km}$ )
1 $1600 \mathrm{~km}$
2 $2400 \mathrm{~km}$
3 $3200 \mathrm{~km}$
4 $4352 \mathrm{~km}$
Explanation:
D Given, $\mathrm{h}=1600 \mathrm{~km}, \mathrm{R}_{\mathrm{e}}=6400 \mathrm{~km}$ At a height $h$ above the earth's surface $g^{\prime}=\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}$ And below the earth's surface at a depth $\mathrm{x}$, $\mathrm{g}^{\prime \prime}=\mathrm{g}\left(1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}\right)$ According to question $\mathrm{g}^{\prime \prime}=\frac{\mathrm{g}^{\prime}}{2}$ Then, $\mathrm{g}^{\prime \prime}=\frac{1}{2} \frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ From equation (i) and (ii) $\mathrm{g}\left(1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}\right)=\frac{1}{2} \frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ $1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}=\frac{1}{2\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ $1-\frac{\mathrm{x}}{6400}=\frac{1}{2\left(1+\frac{1600}{6400}\right)^{2}}$ $1-\frac{\mathrm{x}}{6400}=\frac{8}{25}$ $\frac{\mathrm{x}}{6400}=1-\frac{8}{25}$ $\mathrm{x}=\frac{17}{25} \times 6400$ $\mathrm{x}=4352 \mathrm{~km}$
UP CPMT-2001
Gravitation
138381
The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is
138376
What will be the acceleration due to gravity at a depth $\mathrm{d}$, where $\mathrm{g}$ is acceleration due to gravity on the surface of earth ?
D Variation of $g$ with depth In this figure $\mathrm{m}=\text { mass }$ $\mathrm{d}=\text { distance below the earth's surface }$ We know that on the surface of the earth $\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{RG} \quad\left[\therefore \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ At a distance (d) below the earth's surface $\mathrm{g}_{\mathrm{d}}=\frac{4}{3} \pi \rho(\mathrm{R}-\mathrm{d}) \mathrm{G}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{d}}}=\frac{\mathrm{R}}{(\mathrm{R}-\mathrm{d})}$ $\mathrm{g}_{\mathrm{d}}=\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \mathrm{g}$ When, $d=0$ then value of $g$ on the surface And $\mathrm{g}_{\mathrm{d}}=\mathrm{g}$ When, $d=R$ then value of $g$ at the centre of the earth. $\mathrm{g}_{\mathrm{d}}=0$
UP CPMT-2010
Gravitation
138377
The period of oscillation of a simple pendulum of constant length at surface of the earth is $T$ its time period inside a mine will be
1 cannot be compared
2 equal to $\mathrm{T}$
3 less than $\mathrm{T}$
4 more than $\mathrm{T}$
Explanation:
D Time Period $(\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}}$ Let, mine be at a depth ' $h$ ' below the surface of the earth having radius $\mathrm{R}$, Then, $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$ Hence, $g$ decreases, Therefore ' $T$ ' increases So, time period of simple pendulum is greater than ' $\mathrm{T}$ ' as value of ' $\mathrm{g}$ ' is less in mine.
UP CPMT-2005
Gravitation
138378
The value of acceleration due to gravity $g$ at distance $r$ from earth's centre such that $r \lt R$ depend on $r$ according to relation: $(R=$ radius of earth)
1 $g \propto \frac{1}{\mathrm{r}^{2}}$
2 $g \propto \frac{1}{\mathrm{r}}$
3 $g \propto r$
4 $g \propto r^{2}$ [UP CMPT-2002]
Explanation:
C Gravitational force $(\mathrm{F})=\mathrm{mg}$ $\mathrm{g}=\frac{\mathrm{F}}{\mathrm{m}}$ $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{3}} \mathrm{r}$ $\mathrm{g}=\frac{\mathrm{GMmr}}{\mathrm{R}^{3} \mathrm{~m}}=\frac{\mathrm{Gm}}{\mathrm{R}^{3}} \mathrm{r}$ $\mathrm{g} \propto \mathrm{r}$
Gravitation
138379
At what depth below the surface of earth, the acceleration due to gravity $g$ will be half of its value $1600 \mathrm{~km}$ above the surface of earth? (Radius of earth $=6400 \mathrm{~km}$ )
1 $1600 \mathrm{~km}$
2 $2400 \mathrm{~km}$
3 $3200 \mathrm{~km}$
4 $4352 \mathrm{~km}$
Explanation:
D Given, $\mathrm{h}=1600 \mathrm{~km}, \mathrm{R}_{\mathrm{e}}=6400 \mathrm{~km}$ At a height $h$ above the earth's surface $g^{\prime}=\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}$ And below the earth's surface at a depth $\mathrm{x}$, $\mathrm{g}^{\prime \prime}=\mathrm{g}\left(1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}\right)$ According to question $\mathrm{g}^{\prime \prime}=\frac{\mathrm{g}^{\prime}}{2}$ Then, $\mathrm{g}^{\prime \prime}=\frac{1}{2} \frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ From equation (i) and (ii) $\mathrm{g}\left(1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}\right)=\frac{1}{2} \frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ $1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}=\frac{1}{2\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ $1-\frac{\mathrm{x}}{6400}=\frac{1}{2\left(1+\frac{1600}{6400}\right)^{2}}$ $1-\frac{\mathrm{x}}{6400}=\frac{8}{25}$ $\frac{\mathrm{x}}{6400}=1-\frac{8}{25}$ $\mathrm{x}=\frac{17}{25} \times 6400$ $\mathrm{x}=4352 \mathrm{~km}$
UP CPMT-2001
Gravitation
138381
The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is
138376
What will be the acceleration due to gravity at a depth $\mathrm{d}$, where $\mathrm{g}$ is acceleration due to gravity on the surface of earth ?
D Variation of $g$ with depth In this figure $\mathrm{m}=\text { mass }$ $\mathrm{d}=\text { distance below the earth's surface }$ We know that on the surface of the earth $\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{RG} \quad\left[\therefore \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ At a distance (d) below the earth's surface $\mathrm{g}_{\mathrm{d}}=\frac{4}{3} \pi \rho(\mathrm{R}-\mathrm{d}) \mathrm{G}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{d}}}=\frac{\mathrm{R}}{(\mathrm{R}-\mathrm{d})}$ $\mathrm{g}_{\mathrm{d}}=\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \mathrm{g}$ When, $d=0$ then value of $g$ on the surface And $\mathrm{g}_{\mathrm{d}}=\mathrm{g}$ When, $d=R$ then value of $g$ at the centre of the earth. $\mathrm{g}_{\mathrm{d}}=0$
UP CPMT-2010
Gravitation
138377
The period of oscillation of a simple pendulum of constant length at surface of the earth is $T$ its time period inside a mine will be
1 cannot be compared
2 equal to $\mathrm{T}$
3 less than $\mathrm{T}$
4 more than $\mathrm{T}$
Explanation:
D Time Period $(\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}}$ Let, mine be at a depth ' $h$ ' below the surface of the earth having radius $\mathrm{R}$, Then, $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$ Hence, $g$ decreases, Therefore ' $T$ ' increases So, time period of simple pendulum is greater than ' $\mathrm{T}$ ' as value of ' $\mathrm{g}$ ' is less in mine.
UP CPMT-2005
Gravitation
138378
The value of acceleration due to gravity $g$ at distance $r$ from earth's centre such that $r \lt R$ depend on $r$ according to relation: $(R=$ radius of earth)
1 $g \propto \frac{1}{\mathrm{r}^{2}}$
2 $g \propto \frac{1}{\mathrm{r}}$
3 $g \propto r$
4 $g \propto r^{2}$ [UP CMPT-2002]
Explanation:
C Gravitational force $(\mathrm{F})=\mathrm{mg}$ $\mathrm{g}=\frac{\mathrm{F}}{\mathrm{m}}$ $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{3}} \mathrm{r}$ $\mathrm{g}=\frac{\mathrm{GMmr}}{\mathrm{R}^{3} \mathrm{~m}}=\frac{\mathrm{Gm}}{\mathrm{R}^{3}} \mathrm{r}$ $\mathrm{g} \propto \mathrm{r}$
Gravitation
138379
At what depth below the surface of earth, the acceleration due to gravity $g$ will be half of its value $1600 \mathrm{~km}$ above the surface of earth? (Radius of earth $=6400 \mathrm{~km}$ )
1 $1600 \mathrm{~km}$
2 $2400 \mathrm{~km}$
3 $3200 \mathrm{~km}$
4 $4352 \mathrm{~km}$
Explanation:
D Given, $\mathrm{h}=1600 \mathrm{~km}, \mathrm{R}_{\mathrm{e}}=6400 \mathrm{~km}$ At a height $h$ above the earth's surface $g^{\prime}=\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}$ And below the earth's surface at a depth $\mathrm{x}$, $\mathrm{g}^{\prime \prime}=\mathrm{g}\left(1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}\right)$ According to question $\mathrm{g}^{\prime \prime}=\frac{\mathrm{g}^{\prime}}{2}$ Then, $\mathrm{g}^{\prime \prime}=\frac{1}{2} \frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ From equation (i) and (ii) $\mathrm{g}\left(1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}\right)=\frac{1}{2} \frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ $1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}=\frac{1}{2\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ $1-\frac{\mathrm{x}}{6400}=\frac{1}{2\left(1+\frac{1600}{6400}\right)^{2}}$ $1-\frac{\mathrm{x}}{6400}=\frac{8}{25}$ $\frac{\mathrm{x}}{6400}=1-\frac{8}{25}$ $\mathrm{x}=\frac{17}{25} \times 6400$ $\mathrm{x}=4352 \mathrm{~km}$
UP CPMT-2001
Gravitation
138381
The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is
138376
What will be the acceleration due to gravity at a depth $\mathrm{d}$, where $\mathrm{g}$ is acceleration due to gravity on the surface of earth ?
D Variation of $g$ with depth In this figure $\mathrm{m}=\text { mass }$ $\mathrm{d}=\text { distance below the earth's surface }$ We know that on the surface of the earth $\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{RG} \quad\left[\therefore \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ At a distance (d) below the earth's surface $\mathrm{g}_{\mathrm{d}}=\frac{4}{3} \pi \rho(\mathrm{R}-\mathrm{d}) \mathrm{G}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{d}}}=\frac{\mathrm{R}}{(\mathrm{R}-\mathrm{d})}$ $\mathrm{g}_{\mathrm{d}}=\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \mathrm{g}$ When, $d=0$ then value of $g$ on the surface And $\mathrm{g}_{\mathrm{d}}=\mathrm{g}$ When, $d=R$ then value of $g$ at the centre of the earth. $\mathrm{g}_{\mathrm{d}}=0$
UP CPMT-2010
Gravitation
138377
The period of oscillation of a simple pendulum of constant length at surface of the earth is $T$ its time period inside a mine will be
1 cannot be compared
2 equal to $\mathrm{T}$
3 less than $\mathrm{T}$
4 more than $\mathrm{T}$
Explanation:
D Time Period $(\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}}$ Let, mine be at a depth ' $h$ ' below the surface of the earth having radius $\mathrm{R}$, Then, $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$ Hence, $g$ decreases, Therefore ' $T$ ' increases So, time period of simple pendulum is greater than ' $\mathrm{T}$ ' as value of ' $\mathrm{g}$ ' is less in mine.
UP CPMT-2005
Gravitation
138378
The value of acceleration due to gravity $g$ at distance $r$ from earth's centre such that $r \lt R$ depend on $r$ according to relation: $(R=$ radius of earth)
1 $g \propto \frac{1}{\mathrm{r}^{2}}$
2 $g \propto \frac{1}{\mathrm{r}}$
3 $g \propto r$
4 $g \propto r^{2}$ [UP CMPT-2002]
Explanation:
C Gravitational force $(\mathrm{F})=\mathrm{mg}$ $\mathrm{g}=\frac{\mathrm{F}}{\mathrm{m}}$ $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{3}} \mathrm{r}$ $\mathrm{g}=\frac{\mathrm{GMmr}}{\mathrm{R}^{3} \mathrm{~m}}=\frac{\mathrm{Gm}}{\mathrm{R}^{3}} \mathrm{r}$ $\mathrm{g} \propto \mathrm{r}$
Gravitation
138379
At what depth below the surface of earth, the acceleration due to gravity $g$ will be half of its value $1600 \mathrm{~km}$ above the surface of earth? (Radius of earth $=6400 \mathrm{~km}$ )
1 $1600 \mathrm{~km}$
2 $2400 \mathrm{~km}$
3 $3200 \mathrm{~km}$
4 $4352 \mathrm{~km}$
Explanation:
D Given, $\mathrm{h}=1600 \mathrm{~km}, \mathrm{R}_{\mathrm{e}}=6400 \mathrm{~km}$ At a height $h$ above the earth's surface $g^{\prime}=\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}$ And below the earth's surface at a depth $\mathrm{x}$, $\mathrm{g}^{\prime \prime}=\mathrm{g}\left(1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}\right)$ According to question $\mathrm{g}^{\prime \prime}=\frac{\mathrm{g}^{\prime}}{2}$ Then, $\mathrm{g}^{\prime \prime}=\frac{1}{2} \frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ From equation (i) and (ii) $\mathrm{g}\left(1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}\right)=\frac{1}{2} \frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ $1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}=\frac{1}{2\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ $1-\frac{\mathrm{x}}{6400}=\frac{1}{2\left(1+\frac{1600}{6400}\right)^{2}}$ $1-\frac{\mathrm{x}}{6400}=\frac{8}{25}$ $\frac{\mathrm{x}}{6400}=1-\frac{8}{25}$ $\mathrm{x}=\frac{17}{25} \times 6400$ $\mathrm{x}=4352 \mathrm{~km}$
UP CPMT-2001
Gravitation
138381
The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is
138376
What will be the acceleration due to gravity at a depth $\mathrm{d}$, where $\mathrm{g}$ is acceleration due to gravity on the surface of earth ?
D Variation of $g$ with depth In this figure $\mathrm{m}=\text { mass }$ $\mathrm{d}=\text { distance below the earth's surface }$ We know that on the surface of the earth $\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{RG} \quad\left[\therefore \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]$ At a distance (d) below the earth's surface $\mathrm{g}_{\mathrm{d}}=\frac{4}{3} \pi \rho(\mathrm{R}-\mathrm{d}) \mathrm{G}$ $\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{d}}}=\frac{\mathrm{R}}{(\mathrm{R}-\mathrm{d})}$ $\mathrm{g}_{\mathrm{d}}=\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \mathrm{g}$ When, $d=0$ then value of $g$ on the surface And $\mathrm{g}_{\mathrm{d}}=\mathrm{g}$ When, $d=R$ then value of $g$ at the centre of the earth. $\mathrm{g}_{\mathrm{d}}=0$
UP CPMT-2010
Gravitation
138377
The period of oscillation of a simple pendulum of constant length at surface of the earth is $T$ its time period inside a mine will be
1 cannot be compared
2 equal to $\mathrm{T}$
3 less than $\mathrm{T}$
4 more than $\mathrm{T}$
Explanation:
D Time Period $(\mathrm{T})=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}}$ Let, mine be at a depth ' $h$ ' below the surface of the earth having radius $\mathrm{R}$, Then, $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)$ Hence, $g$ decreases, Therefore ' $T$ ' increases So, time period of simple pendulum is greater than ' $\mathrm{T}$ ' as value of ' $\mathrm{g}$ ' is less in mine.
UP CPMT-2005
Gravitation
138378
The value of acceleration due to gravity $g$ at distance $r$ from earth's centre such that $r \lt R$ depend on $r$ according to relation: $(R=$ radius of earth)
1 $g \propto \frac{1}{\mathrm{r}^{2}}$
2 $g \propto \frac{1}{\mathrm{r}}$
3 $g \propto r$
4 $g \propto r^{2}$ [UP CMPT-2002]
Explanation:
C Gravitational force $(\mathrm{F})=\mathrm{mg}$ $\mathrm{g}=\frac{\mathrm{F}}{\mathrm{m}}$ $\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{3}} \mathrm{r}$ $\mathrm{g}=\frac{\mathrm{GMmr}}{\mathrm{R}^{3} \mathrm{~m}}=\frac{\mathrm{Gm}}{\mathrm{R}^{3}} \mathrm{r}$ $\mathrm{g} \propto \mathrm{r}$
Gravitation
138379
At what depth below the surface of earth, the acceleration due to gravity $g$ will be half of its value $1600 \mathrm{~km}$ above the surface of earth? (Radius of earth $=6400 \mathrm{~km}$ )
1 $1600 \mathrm{~km}$
2 $2400 \mathrm{~km}$
3 $3200 \mathrm{~km}$
4 $4352 \mathrm{~km}$
Explanation:
D Given, $\mathrm{h}=1600 \mathrm{~km}, \mathrm{R}_{\mathrm{e}}=6400 \mathrm{~km}$ At a height $h$ above the earth's surface $g^{\prime}=\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}$ And below the earth's surface at a depth $\mathrm{x}$, $\mathrm{g}^{\prime \prime}=\mathrm{g}\left(1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}\right)$ According to question $\mathrm{g}^{\prime \prime}=\frac{\mathrm{g}^{\prime}}{2}$ Then, $\mathrm{g}^{\prime \prime}=\frac{1}{2} \frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ From equation (i) and (ii) $\mathrm{g}\left(1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}\right)=\frac{1}{2} \frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ $1-\frac{\mathrm{x}}{\mathrm{R}_{\mathrm{e}}}=\frac{1}{2\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^{2}}$ $1-\frac{\mathrm{x}}{6400}=\frac{1}{2\left(1+\frac{1600}{6400}\right)^{2}}$ $1-\frac{\mathrm{x}}{6400}=\frac{8}{25}$ $\frac{\mathrm{x}}{6400}=1-\frac{8}{25}$ $\mathrm{x}=\frac{17}{25} \times 6400$ $\mathrm{x}=4352 \mathrm{~km}$
UP CPMT-2001
Gravitation
138381
The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is
