138396
The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$, the radius of the planet would be
1 $4 \mathrm{R}$
2 $2 \mathrm{R}$
3 $R / 2$
4 $\mathrm{R} / 4$
Explanation:
C Given, $\rho_{P}=2 \rho_{e}$ $g_{e}=g_{P}$ Mass $=$ volume $\times$ density $\mathrm{M}_{\mathrm{e}}=\frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^{3} \rho_{\mathrm{e}}$ From equation of acceleration due to gravity $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}=\frac{\mathrm{G}(4 / 3) \pi \mathrm{R}_{\mathrm{e}}^{3}}{\mathrm{R}_{\mathrm{e}}^{2}} \rho_{\mathrm{e}}$ Here, $\mathrm{g}_{\mathrm{e}} \propto \mathrm{R}_{\mathrm{e}} \rho_{\mathrm{e}}$ Now, acceleration due to gravity of given planet, $\mathrm{g}_{\mathrm{P}} \propto \mathrm{R}_{\mathrm{P}} \rho_{\mathrm{P}}$ given, $g_{e}=g_{P}$ $R_{e} \rho_{e}=R_{P} \rho_{P}$ $R_{e} \rho_{e}=R_{P} 2 \rho_{e} \quad\left(\rho_{P}=2 \rho_{e}\right)$ Hence, $\quad \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{e}}}{2}=\frac{\mathrm{R}}{2} \quad\left\{\because \mathrm{R}_{\mathrm{e}}=\mathrm{R}\right\}$
AMU-2016
Gravitation
138397
Calculate the acceleration due to gravity on the surface of a pulsar of mass $M=1.98 \times 10^{30} \mathrm{~kg}$ and radius $R=12 \mathrm{~km}$ rotating with time period $T=0.041$ seconds. $\left(G=6.67 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\right)$
138399
The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be sphere of radius $R$ of uniform density is shown in figure below. The correct figure is
1
2
3
4
Explanation:
D At depth, when $\mathrm{r} \lt \mathrm{R}$ $g^{\prime}=g\left(1-\frac{d}{R}\right)$ $g^{\prime}=g\left(\frac{R-d}{R}\right)$ $g^{\prime}=g\left(\frac{r}{R}\right)$ $\text { Where } r=R-d$ $\Rightarrow \quad \mathrm{g}^{\prime} \propto \mathrm{r}$ Case-1 when $r=0$ $\mathrm{g}^{\prime}=0$ Case-2 when $\mathrm{r}=\mathrm{R}$ $\mathrm{g}^{\prime}=\mathrm{g}$ At height $\mathrm{h}$, when $\mathrm{r}>\mathrm{R}$ $\mathrm{g}^{\prime}=\frac{\mathrm{gR}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$ $\mathrm{~g}^{\prime}=\frac{\mathrm{gR}^{2}}{\mathrm{r}^{2}} \quad \text { Where } \mathrm{r}=\mathrm{R}+\mathrm{h}$ $\mathrm{g}^{\prime} \propto \frac{1}{\mathrm{r}^{2}}$ Hence, the required graph is-
JEE-MAIN 26.06.2022 Shift-I
Gravitation
138400
If the earth stops rotating, the value of $g$ at the equator
1 Increases
2 decreases
3 no effect
4 none of these
Explanation:
A When earth is rotating, $g_{1}^{\prime}=g-\omega^{2} R \cos ^{2} \alpha$ When earth stop rotating, $g_{2}^{\prime}=g$ $\text { So, } g_{2}^{\prime}>g_{1}^{\prime}$ If the earth stops rotating, the value of $g$ at equator will increase by amount $R \omega^{2} \cos ^{2} \alpha$. with $\alpha=0^{\circ}$ at equator. The effect of rotation of the earth on acceleration due to gravity is to decrease its value. Therefore if the earth stops rotating, the value of $g$ will increase.
DCE-2007
Gravitation
138401
The acceleration due to gravity at a height (take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ on earth surface) above earth's surface is $9 \mathrm{~ms}^{-2}$. It value at a point, at an equal distance below the surface of the earth is
1 $9.7 \mathrm{~ms}^{-2}$
2 $8.5 \mathrm{~ms}^{-2}$
3 $7 . \mathrm{ms}^{-2}$
4 $5.5 \mathrm{~ms}^{-2}$
Explanation:
A Given, $\mathrm{g}_{\mathrm{h}}=9 \mathrm{~m} / \mathrm{s}^{2}$ $\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ We know, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]$ $9=10\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]$ ${\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]=\frac{9}{10}}$ $\frac{1}{10}=\frac{2 \mathrm{~h}}{\mathrm{R}}$ $\mathrm{h}=\frac{\mathrm{R}}{20}$ Now, The value of $g$ below the earth's surface $\mathrm{g}_{\mathrm{d}} =\mathrm{g}\left[1-\frac{\mathrm{d}}{\mathrm{R}}\right]=10\left[1-\frac{\mathrm{R}}{\mathrm{R}}\right]$ $=10\left[1-\frac{1}{20}\right]$ $\mathrm{g}_{\mathrm{d}} =\frac{10 \times 19}{20}=9.5 \mathrm{~ms}^{-2} \approx 9.7 \mathrm{~ms}^{-2}$
138396
The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$, the radius of the planet would be
1 $4 \mathrm{R}$
2 $2 \mathrm{R}$
3 $R / 2$
4 $\mathrm{R} / 4$
Explanation:
C Given, $\rho_{P}=2 \rho_{e}$ $g_{e}=g_{P}$ Mass $=$ volume $\times$ density $\mathrm{M}_{\mathrm{e}}=\frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^{3} \rho_{\mathrm{e}}$ From equation of acceleration due to gravity $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}=\frac{\mathrm{G}(4 / 3) \pi \mathrm{R}_{\mathrm{e}}^{3}}{\mathrm{R}_{\mathrm{e}}^{2}} \rho_{\mathrm{e}}$ Here, $\mathrm{g}_{\mathrm{e}} \propto \mathrm{R}_{\mathrm{e}} \rho_{\mathrm{e}}$ Now, acceleration due to gravity of given planet, $\mathrm{g}_{\mathrm{P}} \propto \mathrm{R}_{\mathrm{P}} \rho_{\mathrm{P}}$ given, $g_{e}=g_{P}$ $R_{e} \rho_{e}=R_{P} \rho_{P}$ $R_{e} \rho_{e}=R_{P} 2 \rho_{e} \quad\left(\rho_{P}=2 \rho_{e}\right)$ Hence, $\quad \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{e}}}{2}=\frac{\mathrm{R}}{2} \quad\left\{\because \mathrm{R}_{\mathrm{e}}=\mathrm{R}\right\}$
AMU-2016
Gravitation
138397
Calculate the acceleration due to gravity on the surface of a pulsar of mass $M=1.98 \times 10^{30} \mathrm{~kg}$ and radius $R=12 \mathrm{~km}$ rotating with time period $T=0.041$ seconds. $\left(G=6.67 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\right)$
138399
The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be sphere of radius $R$ of uniform density is shown in figure below. The correct figure is
1
2
3
4
Explanation:
D At depth, when $\mathrm{r} \lt \mathrm{R}$ $g^{\prime}=g\left(1-\frac{d}{R}\right)$ $g^{\prime}=g\left(\frac{R-d}{R}\right)$ $g^{\prime}=g\left(\frac{r}{R}\right)$ $\text { Where } r=R-d$ $\Rightarrow \quad \mathrm{g}^{\prime} \propto \mathrm{r}$ Case-1 when $r=0$ $\mathrm{g}^{\prime}=0$ Case-2 when $\mathrm{r}=\mathrm{R}$ $\mathrm{g}^{\prime}=\mathrm{g}$ At height $\mathrm{h}$, when $\mathrm{r}>\mathrm{R}$ $\mathrm{g}^{\prime}=\frac{\mathrm{gR}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$ $\mathrm{~g}^{\prime}=\frac{\mathrm{gR}^{2}}{\mathrm{r}^{2}} \quad \text { Where } \mathrm{r}=\mathrm{R}+\mathrm{h}$ $\mathrm{g}^{\prime} \propto \frac{1}{\mathrm{r}^{2}}$ Hence, the required graph is-
JEE-MAIN 26.06.2022 Shift-I
Gravitation
138400
If the earth stops rotating, the value of $g$ at the equator
1 Increases
2 decreases
3 no effect
4 none of these
Explanation:
A When earth is rotating, $g_{1}^{\prime}=g-\omega^{2} R \cos ^{2} \alpha$ When earth stop rotating, $g_{2}^{\prime}=g$ $\text { So, } g_{2}^{\prime}>g_{1}^{\prime}$ If the earth stops rotating, the value of $g$ at equator will increase by amount $R \omega^{2} \cos ^{2} \alpha$. with $\alpha=0^{\circ}$ at equator. The effect of rotation of the earth on acceleration due to gravity is to decrease its value. Therefore if the earth stops rotating, the value of $g$ will increase.
DCE-2007
Gravitation
138401
The acceleration due to gravity at a height (take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ on earth surface) above earth's surface is $9 \mathrm{~ms}^{-2}$. It value at a point, at an equal distance below the surface of the earth is
1 $9.7 \mathrm{~ms}^{-2}$
2 $8.5 \mathrm{~ms}^{-2}$
3 $7 . \mathrm{ms}^{-2}$
4 $5.5 \mathrm{~ms}^{-2}$
Explanation:
A Given, $\mathrm{g}_{\mathrm{h}}=9 \mathrm{~m} / \mathrm{s}^{2}$ $\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ We know, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]$ $9=10\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]$ ${\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]=\frac{9}{10}}$ $\frac{1}{10}=\frac{2 \mathrm{~h}}{\mathrm{R}}$ $\mathrm{h}=\frac{\mathrm{R}}{20}$ Now, The value of $g$ below the earth's surface $\mathrm{g}_{\mathrm{d}} =\mathrm{g}\left[1-\frac{\mathrm{d}}{\mathrm{R}}\right]=10\left[1-\frac{\mathrm{R}}{\mathrm{R}}\right]$ $=10\left[1-\frac{1}{20}\right]$ $\mathrm{g}_{\mathrm{d}} =\frac{10 \times 19}{20}=9.5 \mathrm{~ms}^{-2} \approx 9.7 \mathrm{~ms}^{-2}$
138396
The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$, the radius of the planet would be
1 $4 \mathrm{R}$
2 $2 \mathrm{R}$
3 $R / 2$
4 $\mathrm{R} / 4$
Explanation:
C Given, $\rho_{P}=2 \rho_{e}$ $g_{e}=g_{P}$ Mass $=$ volume $\times$ density $\mathrm{M}_{\mathrm{e}}=\frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^{3} \rho_{\mathrm{e}}$ From equation of acceleration due to gravity $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}=\frac{\mathrm{G}(4 / 3) \pi \mathrm{R}_{\mathrm{e}}^{3}}{\mathrm{R}_{\mathrm{e}}^{2}} \rho_{\mathrm{e}}$ Here, $\mathrm{g}_{\mathrm{e}} \propto \mathrm{R}_{\mathrm{e}} \rho_{\mathrm{e}}$ Now, acceleration due to gravity of given planet, $\mathrm{g}_{\mathrm{P}} \propto \mathrm{R}_{\mathrm{P}} \rho_{\mathrm{P}}$ given, $g_{e}=g_{P}$ $R_{e} \rho_{e}=R_{P} \rho_{P}$ $R_{e} \rho_{e}=R_{P} 2 \rho_{e} \quad\left(\rho_{P}=2 \rho_{e}\right)$ Hence, $\quad \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{e}}}{2}=\frac{\mathrm{R}}{2} \quad\left\{\because \mathrm{R}_{\mathrm{e}}=\mathrm{R}\right\}$
AMU-2016
Gravitation
138397
Calculate the acceleration due to gravity on the surface of a pulsar of mass $M=1.98 \times 10^{30} \mathrm{~kg}$ and radius $R=12 \mathrm{~km}$ rotating with time period $T=0.041$ seconds. $\left(G=6.67 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\right)$
138399
The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be sphere of radius $R$ of uniform density is shown in figure below. The correct figure is
1
2
3
4
Explanation:
D At depth, when $\mathrm{r} \lt \mathrm{R}$ $g^{\prime}=g\left(1-\frac{d}{R}\right)$ $g^{\prime}=g\left(\frac{R-d}{R}\right)$ $g^{\prime}=g\left(\frac{r}{R}\right)$ $\text { Where } r=R-d$ $\Rightarrow \quad \mathrm{g}^{\prime} \propto \mathrm{r}$ Case-1 when $r=0$ $\mathrm{g}^{\prime}=0$ Case-2 when $\mathrm{r}=\mathrm{R}$ $\mathrm{g}^{\prime}=\mathrm{g}$ At height $\mathrm{h}$, when $\mathrm{r}>\mathrm{R}$ $\mathrm{g}^{\prime}=\frac{\mathrm{gR}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$ $\mathrm{~g}^{\prime}=\frac{\mathrm{gR}^{2}}{\mathrm{r}^{2}} \quad \text { Where } \mathrm{r}=\mathrm{R}+\mathrm{h}$ $\mathrm{g}^{\prime} \propto \frac{1}{\mathrm{r}^{2}}$ Hence, the required graph is-
JEE-MAIN 26.06.2022 Shift-I
Gravitation
138400
If the earth stops rotating, the value of $g$ at the equator
1 Increases
2 decreases
3 no effect
4 none of these
Explanation:
A When earth is rotating, $g_{1}^{\prime}=g-\omega^{2} R \cos ^{2} \alpha$ When earth stop rotating, $g_{2}^{\prime}=g$ $\text { So, } g_{2}^{\prime}>g_{1}^{\prime}$ If the earth stops rotating, the value of $g$ at equator will increase by amount $R \omega^{2} \cos ^{2} \alpha$. with $\alpha=0^{\circ}$ at equator. The effect of rotation of the earth on acceleration due to gravity is to decrease its value. Therefore if the earth stops rotating, the value of $g$ will increase.
DCE-2007
Gravitation
138401
The acceleration due to gravity at a height (take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ on earth surface) above earth's surface is $9 \mathrm{~ms}^{-2}$. It value at a point, at an equal distance below the surface of the earth is
1 $9.7 \mathrm{~ms}^{-2}$
2 $8.5 \mathrm{~ms}^{-2}$
3 $7 . \mathrm{ms}^{-2}$
4 $5.5 \mathrm{~ms}^{-2}$
Explanation:
A Given, $\mathrm{g}_{\mathrm{h}}=9 \mathrm{~m} / \mathrm{s}^{2}$ $\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ We know, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]$ $9=10\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]$ ${\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]=\frac{9}{10}}$ $\frac{1}{10}=\frac{2 \mathrm{~h}}{\mathrm{R}}$ $\mathrm{h}=\frac{\mathrm{R}}{20}$ Now, The value of $g$ below the earth's surface $\mathrm{g}_{\mathrm{d}} =\mathrm{g}\left[1-\frac{\mathrm{d}}{\mathrm{R}}\right]=10\left[1-\frac{\mathrm{R}}{\mathrm{R}}\right]$ $=10\left[1-\frac{1}{20}\right]$ $\mathrm{g}_{\mathrm{d}} =\frac{10 \times 19}{20}=9.5 \mathrm{~ms}^{-2} \approx 9.7 \mathrm{~ms}^{-2}$
138396
The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$, the radius of the planet would be
1 $4 \mathrm{R}$
2 $2 \mathrm{R}$
3 $R / 2$
4 $\mathrm{R} / 4$
Explanation:
C Given, $\rho_{P}=2 \rho_{e}$ $g_{e}=g_{P}$ Mass $=$ volume $\times$ density $\mathrm{M}_{\mathrm{e}}=\frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^{3} \rho_{\mathrm{e}}$ From equation of acceleration due to gravity $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}=\frac{\mathrm{G}(4 / 3) \pi \mathrm{R}_{\mathrm{e}}^{3}}{\mathrm{R}_{\mathrm{e}}^{2}} \rho_{\mathrm{e}}$ Here, $\mathrm{g}_{\mathrm{e}} \propto \mathrm{R}_{\mathrm{e}} \rho_{\mathrm{e}}$ Now, acceleration due to gravity of given planet, $\mathrm{g}_{\mathrm{P}} \propto \mathrm{R}_{\mathrm{P}} \rho_{\mathrm{P}}$ given, $g_{e}=g_{P}$ $R_{e} \rho_{e}=R_{P} \rho_{P}$ $R_{e} \rho_{e}=R_{P} 2 \rho_{e} \quad\left(\rho_{P}=2 \rho_{e}\right)$ Hence, $\quad \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{e}}}{2}=\frac{\mathrm{R}}{2} \quad\left\{\because \mathrm{R}_{\mathrm{e}}=\mathrm{R}\right\}$
AMU-2016
Gravitation
138397
Calculate the acceleration due to gravity on the surface of a pulsar of mass $M=1.98 \times 10^{30} \mathrm{~kg}$ and radius $R=12 \mathrm{~km}$ rotating with time period $T=0.041$ seconds. $\left(G=6.67 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\right)$
138399
The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be sphere of radius $R$ of uniform density is shown in figure below. The correct figure is
1
2
3
4
Explanation:
D At depth, when $\mathrm{r} \lt \mathrm{R}$ $g^{\prime}=g\left(1-\frac{d}{R}\right)$ $g^{\prime}=g\left(\frac{R-d}{R}\right)$ $g^{\prime}=g\left(\frac{r}{R}\right)$ $\text { Where } r=R-d$ $\Rightarrow \quad \mathrm{g}^{\prime} \propto \mathrm{r}$ Case-1 when $r=0$ $\mathrm{g}^{\prime}=0$ Case-2 when $\mathrm{r}=\mathrm{R}$ $\mathrm{g}^{\prime}=\mathrm{g}$ At height $\mathrm{h}$, when $\mathrm{r}>\mathrm{R}$ $\mathrm{g}^{\prime}=\frac{\mathrm{gR}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$ $\mathrm{~g}^{\prime}=\frac{\mathrm{gR}^{2}}{\mathrm{r}^{2}} \quad \text { Where } \mathrm{r}=\mathrm{R}+\mathrm{h}$ $\mathrm{g}^{\prime} \propto \frac{1}{\mathrm{r}^{2}}$ Hence, the required graph is-
JEE-MAIN 26.06.2022 Shift-I
Gravitation
138400
If the earth stops rotating, the value of $g$ at the equator
1 Increases
2 decreases
3 no effect
4 none of these
Explanation:
A When earth is rotating, $g_{1}^{\prime}=g-\omega^{2} R \cos ^{2} \alpha$ When earth stop rotating, $g_{2}^{\prime}=g$ $\text { So, } g_{2}^{\prime}>g_{1}^{\prime}$ If the earth stops rotating, the value of $g$ at equator will increase by amount $R \omega^{2} \cos ^{2} \alpha$. with $\alpha=0^{\circ}$ at equator. The effect of rotation of the earth on acceleration due to gravity is to decrease its value. Therefore if the earth stops rotating, the value of $g$ will increase.
DCE-2007
Gravitation
138401
The acceleration due to gravity at a height (take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ on earth surface) above earth's surface is $9 \mathrm{~ms}^{-2}$. It value at a point, at an equal distance below the surface of the earth is
1 $9.7 \mathrm{~ms}^{-2}$
2 $8.5 \mathrm{~ms}^{-2}$
3 $7 . \mathrm{ms}^{-2}$
4 $5.5 \mathrm{~ms}^{-2}$
Explanation:
A Given, $\mathrm{g}_{\mathrm{h}}=9 \mathrm{~m} / \mathrm{s}^{2}$ $\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ We know, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]$ $9=10\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]$ ${\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]=\frac{9}{10}}$ $\frac{1}{10}=\frac{2 \mathrm{~h}}{\mathrm{R}}$ $\mathrm{h}=\frac{\mathrm{R}}{20}$ Now, The value of $g$ below the earth's surface $\mathrm{g}_{\mathrm{d}} =\mathrm{g}\left[1-\frac{\mathrm{d}}{\mathrm{R}}\right]=10\left[1-\frac{\mathrm{R}}{\mathrm{R}}\right]$ $=10\left[1-\frac{1}{20}\right]$ $\mathrm{g}_{\mathrm{d}} =\frac{10 \times 19}{20}=9.5 \mathrm{~ms}^{-2} \approx 9.7 \mathrm{~ms}^{-2}$
138396
The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$, the radius of the planet would be
1 $4 \mathrm{R}$
2 $2 \mathrm{R}$
3 $R / 2$
4 $\mathrm{R} / 4$
Explanation:
C Given, $\rho_{P}=2 \rho_{e}$ $g_{e}=g_{P}$ Mass $=$ volume $\times$ density $\mathrm{M}_{\mathrm{e}}=\frac{4}{3} \pi \mathrm{R}_{\mathrm{e}}^{3} \rho_{\mathrm{e}}$ From equation of acceleration due to gravity $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}=\frac{\mathrm{G}(4 / 3) \pi \mathrm{R}_{\mathrm{e}}^{3}}{\mathrm{R}_{\mathrm{e}}^{2}} \rho_{\mathrm{e}}$ Here, $\mathrm{g}_{\mathrm{e}} \propto \mathrm{R}_{\mathrm{e}} \rho_{\mathrm{e}}$ Now, acceleration due to gravity of given planet, $\mathrm{g}_{\mathrm{P}} \propto \mathrm{R}_{\mathrm{P}} \rho_{\mathrm{P}}$ given, $g_{e}=g_{P}$ $R_{e} \rho_{e}=R_{P} \rho_{P}$ $R_{e} \rho_{e}=R_{P} 2 \rho_{e} \quad\left(\rho_{P}=2 \rho_{e}\right)$ Hence, $\quad \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{e}}}{2}=\frac{\mathrm{R}}{2} \quad\left\{\because \mathrm{R}_{\mathrm{e}}=\mathrm{R}\right\}$
AMU-2016
Gravitation
138397
Calculate the acceleration due to gravity on the surface of a pulsar of mass $M=1.98 \times 10^{30} \mathrm{~kg}$ and radius $R=12 \mathrm{~km}$ rotating with time period $T=0.041$ seconds. $\left(G=6.67 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\right)$
138399
The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth assumed to be sphere of radius $R$ of uniform density is shown in figure below. The correct figure is
1
2
3
4
Explanation:
D At depth, when $\mathrm{r} \lt \mathrm{R}$ $g^{\prime}=g\left(1-\frac{d}{R}\right)$ $g^{\prime}=g\left(\frac{R-d}{R}\right)$ $g^{\prime}=g\left(\frac{r}{R}\right)$ $\text { Where } r=R-d$ $\Rightarrow \quad \mathrm{g}^{\prime} \propto \mathrm{r}$ Case-1 when $r=0$ $\mathrm{g}^{\prime}=0$ Case-2 when $\mathrm{r}=\mathrm{R}$ $\mathrm{g}^{\prime}=\mathrm{g}$ At height $\mathrm{h}$, when $\mathrm{r}>\mathrm{R}$ $\mathrm{g}^{\prime}=\frac{\mathrm{gR}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$ $\mathrm{~g}^{\prime}=\frac{\mathrm{gR}^{2}}{\mathrm{r}^{2}} \quad \text { Where } \mathrm{r}=\mathrm{R}+\mathrm{h}$ $\mathrm{g}^{\prime} \propto \frac{1}{\mathrm{r}^{2}}$ Hence, the required graph is-
JEE-MAIN 26.06.2022 Shift-I
Gravitation
138400
If the earth stops rotating, the value of $g$ at the equator
1 Increases
2 decreases
3 no effect
4 none of these
Explanation:
A When earth is rotating, $g_{1}^{\prime}=g-\omega^{2} R \cos ^{2} \alpha$ When earth stop rotating, $g_{2}^{\prime}=g$ $\text { So, } g_{2}^{\prime}>g_{1}^{\prime}$ If the earth stops rotating, the value of $g$ at equator will increase by amount $R \omega^{2} \cos ^{2} \alpha$. with $\alpha=0^{\circ}$ at equator. The effect of rotation of the earth on acceleration due to gravity is to decrease its value. Therefore if the earth stops rotating, the value of $g$ will increase.
DCE-2007
Gravitation
138401
The acceleration due to gravity at a height (take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ on earth surface) above earth's surface is $9 \mathrm{~ms}^{-2}$. It value at a point, at an equal distance below the surface of the earth is
1 $9.7 \mathrm{~ms}^{-2}$
2 $8.5 \mathrm{~ms}^{-2}$
3 $7 . \mathrm{ms}^{-2}$
4 $5.5 \mathrm{~ms}^{-2}$
Explanation:
A Given, $\mathrm{g}_{\mathrm{h}}=9 \mathrm{~m} / \mathrm{s}^{2}$ $\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ We know, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]$ $9=10\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]$ ${\left[1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right]=\frac{9}{10}}$ $\frac{1}{10}=\frac{2 \mathrm{~h}}{\mathrm{R}}$ $\mathrm{h}=\frac{\mathrm{R}}{20}$ Now, The value of $g$ below the earth's surface $\mathrm{g}_{\mathrm{d}} =\mathrm{g}\left[1-\frac{\mathrm{d}}{\mathrm{R}}\right]=10\left[1-\frac{\mathrm{R}}{\mathrm{R}}\right]$ $=10\left[1-\frac{1}{20}\right]$ $\mathrm{g}_{\mathrm{d}} =\frac{10 \times 19}{20}=9.5 \mathrm{~ms}^{-2} \approx 9.7 \mathrm{~ms}^{-2}$
