138356
Taking the radius of the earth to be $6400 \mathrm{~km}$, by what percentage will the acceleration due to gravity at a height of $100 \mathrm{~km}$ from the surface of the earth differ from that on the surface of the earth?
1 about $1.5 \%$
2 about 5\%
3 about $8 \%$
4 about $3 \%$
Explanation:
D Given that, Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$, Height $(\mathrm{h})=100 \mathrm{~km}$ Acceleration due to gravity at height, $\mathrm{h}$ $g^{\prime}=\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}$ $g^{\prime}=\frac{g R_{e}^{2}}{\left(R_{e}+h\right)^{2}}$ $g^{\prime}=\frac{g(6400)^{2}}{(6400+100)^{2}}$ $g^{\prime}=g\left(\frac{6400}{6500}\right)^{2}$ Percentage change in acceleration due to gravity $=\frac{g-g^{\prime}}{g} \times 100$ $=\left(1-\frac{g^{\prime}}{g}\right) \times 100$ Put the value of $g^{\prime}$ in $\mathrm{eq}^{\mathrm{n}}$. (i) $=\left(1-\frac{\mathrm{g}\left(\frac{6400}{6500}\right)^{2}}{\mathrm{~g}}\right) \times 100 \%$ $=3.05 \%$
J and K-CET-2012
Gravitation
138357
If $R$ is the radius of the Earth then the height above the Earth's surface at which the acceleration due to gravity decreases by $20 \%$ is
1 $\left(\frac{\sqrt{5}}{2}-1\right) \mathrm{R}$
2 $\left(\frac{\sqrt{5}}{2}+1\right) \mathrm{R}$
3 $(5 \sqrt{2}-1) \mathrm{R}$
4 $(5 \sqrt{2}+1) \mathrm{R}$
Explanation:
A We know that $g_{h}=\left[\frac{R^{2}}{(R+h)^{2}}\right] g$ acceleration due to gravity decrease by $20 \%$ $g_{h} =g-\frac{20}{100} \times g$ $g_{h} =\frac{80}{100} g$ $g_{h} =\frac{4}{5} \mathrm{~g}$ $\frac{g_{h}}{g} =\frac{4}{5}$ Putting value of equation (ii) in equation (i), we get $\therefore \quad \frac{4}{5}=\frac{\mathrm{R}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$ $\sqrt{\frac{4}{5}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}$ $\frac{2}{\sqrt{5}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}$ $2 \mathrm{R}+2 \mathrm{~h}=\sqrt{5} \mathrm{R}$ $2 \mathrm{~h}=(\sqrt{5}-2) \mathrm{R}$ $\mathrm{h}=\left(\frac{\sqrt{5}-2}{2}\right) \mathrm{R}$ $\mathrm{h}=\left(\frac{\sqrt{5}}{2}-1\right) \mathrm{R}$
J and K-CET-2016
Gravitation
138358
The earth's mass is 80 times that of moon and their diameters are $1600 \mathrm{~km}$ and $800 \mathrm{~km}$, respectively. If $g$ is the value of acceleration due to gravity on earth, what is its value on moon?
1 $\mathrm{g}$
2 $g / 2$
3 $g / 10$
4 $g / 20$
Explanation:
D Given, Mass of earth $\left(\mathrm{M}_{\mathrm{e}}\right)=80$, Mass of moon $\left(\mathrm{M}_{\mathrm{m}}\right)$ Diameters of earth $\left(\mathrm{d}_{\mathrm{e}}\right)=1600 \mathrm{~km}$ Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=\frac{1600}{2}=800 \mathrm{~km}$ Diameters of moon $\left(\mathrm{d}_{\mathrm{m}}\right)=800 \mathrm{~km}$ Radius of moon $\left(\mathrm{R}_{\mathrm{m}}\right)=\frac{800}{2}=400 \mathrm{~km}$ So, $\quad \mathrm{R}_{\mathrm{e}}=2 \mathrm{R}_{\mathrm{m}}$ Acceleration due to gravity at earth $\left(\mathrm{g}_{\mathrm{e}}\right)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}} \ldots \ldots$ (i) Acceleration due to gravity at $\operatorname{moon}\left(\mathrm{g}_{\mathrm{m}}\right)$ $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ Dividing equation (ii) by (i), we get, $\frac{g_{m}}{g_{e}}=\frac{M_{m}}{R_{m}^{2}} \times \frac{R_{e}{ }^{2}}{M_{e}}$ $\frac{g_{m}}{g_{e}}=\frac{M_{m} \times\left(2 R_{m}\right)^{2}}{\left(R_{m}\right)^{2} \times 80 M_{m}}$ $\frac{g_{m}}{g_{e}}=\frac{4}{80}$ $\frac{g_{m}}{g_{e}}=\frac{1}{20}$ $g_{m}=\frac{g_{e}}{20}$ Then, $g_{m}=\frac{g}{20}$
J and K-CET-2015
Gravitation
138360
Assuming the earth to be a sphere of uniform density, the ratio of acceleration due to gravity on the earth's surface to its value at halfway towards the centre of the earth, will be
1 $1: 1$
2 $1: 2$
3 $2: 3$
4 $2: 1$
Explanation:
D The acceleration due to gravity at a depth $\mathrm{h}$ below the surface of the earth is given by $g^{\prime}=g\left(1-\frac{h}{R_{e}}\right)$ $g^{\prime}=g\left(1-\frac{R_{e}}{2 R_{e}}\right) \quad\left(h=\frac{R_{e}}{2}\right)$ Then $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{1}{2}\right)=\frac{\mathrm{g}}{2}$ $\therefore \quad \frac{\mathrm{g}}{\mathrm{g}^{\prime}}=2$ $\mathrm{~g}: \mathrm{g}^{\prime}=2: 1$
138356
Taking the radius of the earth to be $6400 \mathrm{~km}$, by what percentage will the acceleration due to gravity at a height of $100 \mathrm{~km}$ from the surface of the earth differ from that on the surface of the earth?
1 about $1.5 \%$
2 about 5\%
3 about $8 \%$
4 about $3 \%$
Explanation:
D Given that, Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$, Height $(\mathrm{h})=100 \mathrm{~km}$ Acceleration due to gravity at height, $\mathrm{h}$ $g^{\prime}=\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}$ $g^{\prime}=\frac{g R_{e}^{2}}{\left(R_{e}+h\right)^{2}}$ $g^{\prime}=\frac{g(6400)^{2}}{(6400+100)^{2}}$ $g^{\prime}=g\left(\frac{6400}{6500}\right)^{2}$ Percentage change in acceleration due to gravity $=\frac{g-g^{\prime}}{g} \times 100$ $=\left(1-\frac{g^{\prime}}{g}\right) \times 100$ Put the value of $g^{\prime}$ in $\mathrm{eq}^{\mathrm{n}}$. (i) $=\left(1-\frac{\mathrm{g}\left(\frac{6400}{6500}\right)^{2}}{\mathrm{~g}}\right) \times 100 \%$ $=3.05 \%$
J and K-CET-2012
Gravitation
138357
If $R$ is the radius of the Earth then the height above the Earth's surface at which the acceleration due to gravity decreases by $20 \%$ is
1 $\left(\frac{\sqrt{5}}{2}-1\right) \mathrm{R}$
2 $\left(\frac{\sqrt{5}}{2}+1\right) \mathrm{R}$
3 $(5 \sqrt{2}-1) \mathrm{R}$
4 $(5 \sqrt{2}+1) \mathrm{R}$
Explanation:
A We know that $g_{h}=\left[\frac{R^{2}}{(R+h)^{2}}\right] g$ acceleration due to gravity decrease by $20 \%$ $g_{h} =g-\frac{20}{100} \times g$ $g_{h} =\frac{80}{100} g$ $g_{h} =\frac{4}{5} \mathrm{~g}$ $\frac{g_{h}}{g} =\frac{4}{5}$ Putting value of equation (ii) in equation (i), we get $\therefore \quad \frac{4}{5}=\frac{\mathrm{R}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$ $\sqrt{\frac{4}{5}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}$ $\frac{2}{\sqrt{5}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}$ $2 \mathrm{R}+2 \mathrm{~h}=\sqrt{5} \mathrm{R}$ $2 \mathrm{~h}=(\sqrt{5}-2) \mathrm{R}$ $\mathrm{h}=\left(\frac{\sqrt{5}-2}{2}\right) \mathrm{R}$ $\mathrm{h}=\left(\frac{\sqrt{5}}{2}-1\right) \mathrm{R}$
J and K-CET-2016
Gravitation
138358
The earth's mass is 80 times that of moon and their diameters are $1600 \mathrm{~km}$ and $800 \mathrm{~km}$, respectively. If $g$ is the value of acceleration due to gravity on earth, what is its value on moon?
1 $\mathrm{g}$
2 $g / 2$
3 $g / 10$
4 $g / 20$
Explanation:
D Given, Mass of earth $\left(\mathrm{M}_{\mathrm{e}}\right)=80$, Mass of moon $\left(\mathrm{M}_{\mathrm{m}}\right)$ Diameters of earth $\left(\mathrm{d}_{\mathrm{e}}\right)=1600 \mathrm{~km}$ Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=\frac{1600}{2}=800 \mathrm{~km}$ Diameters of moon $\left(\mathrm{d}_{\mathrm{m}}\right)=800 \mathrm{~km}$ Radius of moon $\left(\mathrm{R}_{\mathrm{m}}\right)=\frac{800}{2}=400 \mathrm{~km}$ So, $\quad \mathrm{R}_{\mathrm{e}}=2 \mathrm{R}_{\mathrm{m}}$ Acceleration due to gravity at earth $\left(\mathrm{g}_{\mathrm{e}}\right)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}} \ldots \ldots$ (i) Acceleration due to gravity at $\operatorname{moon}\left(\mathrm{g}_{\mathrm{m}}\right)$ $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ Dividing equation (ii) by (i), we get, $\frac{g_{m}}{g_{e}}=\frac{M_{m}}{R_{m}^{2}} \times \frac{R_{e}{ }^{2}}{M_{e}}$ $\frac{g_{m}}{g_{e}}=\frac{M_{m} \times\left(2 R_{m}\right)^{2}}{\left(R_{m}\right)^{2} \times 80 M_{m}}$ $\frac{g_{m}}{g_{e}}=\frac{4}{80}$ $\frac{g_{m}}{g_{e}}=\frac{1}{20}$ $g_{m}=\frac{g_{e}}{20}$ Then, $g_{m}=\frac{g}{20}$
J and K-CET-2015
Gravitation
138360
Assuming the earth to be a sphere of uniform density, the ratio of acceleration due to gravity on the earth's surface to its value at halfway towards the centre of the earth, will be
1 $1: 1$
2 $1: 2$
3 $2: 3$
4 $2: 1$
Explanation:
D The acceleration due to gravity at a depth $\mathrm{h}$ below the surface of the earth is given by $g^{\prime}=g\left(1-\frac{h}{R_{e}}\right)$ $g^{\prime}=g\left(1-\frac{R_{e}}{2 R_{e}}\right) \quad\left(h=\frac{R_{e}}{2}\right)$ Then $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{1}{2}\right)=\frac{\mathrm{g}}{2}$ $\therefore \quad \frac{\mathrm{g}}{\mathrm{g}^{\prime}}=2$ $\mathrm{~g}: \mathrm{g}^{\prime}=2: 1$
138356
Taking the radius of the earth to be $6400 \mathrm{~km}$, by what percentage will the acceleration due to gravity at a height of $100 \mathrm{~km}$ from the surface of the earth differ from that on the surface of the earth?
1 about $1.5 \%$
2 about 5\%
3 about $8 \%$
4 about $3 \%$
Explanation:
D Given that, Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$, Height $(\mathrm{h})=100 \mathrm{~km}$ Acceleration due to gravity at height, $\mathrm{h}$ $g^{\prime}=\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}$ $g^{\prime}=\frac{g R_{e}^{2}}{\left(R_{e}+h\right)^{2}}$ $g^{\prime}=\frac{g(6400)^{2}}{(6400+100)^{2}}$ $g^{\prime}=g\left(\frac{6400}{6500}\right)^{2}$ Percentage change in acceleration due to gravity $=\frac{g-g^{\prime}}{g} \times 100$ $=\left(1-\frac{g^{\prime}}{g}\right) \times 100$ Put the value of $g^{\prime}$ in $\mathrm{eq}^{\mathrm{n}}$. (i) $=\left(1-\frac{\mathrm{g}\left(\frac{6400}{6500}\right)^{2}}{\mathrm{~g}}\right) \times 100 \%$ $=3.05 \%$
J and K-CET-2012
Gravitation
138357
If $R$ is the radius of the Earth then the height above the Earth's surface at which the acceleration due to gravity decreases by $20 \%$ is
1 $\left(\frac{\sqrt{5}}{2}-1\right) \mathrm{R}$
2 $\left(\frac{\sqrt{5}}{2}+1\right) \mathrm{R}$
3 $(5 \sqrt{2}-1) \mathrm{R}$
4 $(5 \sqrt{2}+1) \mathrm{R}$
Explanation:
A We know that $g_{h}=\left[\frac{R^{2}}{(R+h)^{2}}\right] g$ acceleration due to gravity decrease by $20 \%$ $g_{h} =g-\frac{20}{100} \times g$ $g_{h} =\frac{80}{100} g$ $g_{h} =\frac{4}{5} \mathrm{~g}$ $\frac{g_{h}}{g} =\frac{4}{5}$ Putting value of equation (ii) in equation (i), we get $\therefore \quad \frac{4}{5}=\frac{\mathrm{R}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$ $\sqrt{\frac{4}{5}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}$ $\frac{2}{\sqrt{5}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}$ $2 \mathrm{R}+2 \mathrm{~h}=\sqrt{5} \mathrm{R}$ $2 \mathrm{~h}=(\sqrt{5}-2) \mathrm{R}$ $\mathrm{h}=\left(\frac{\sqrt{5}-2}{2}\right) \mathrm{R}$ $\mathrm{h}=\left(\frac{\sqrt{5}}{2}-1\right) \mathrm{R}$
J and K-CET-2016
Gravitation
138358
The earth's mass is 80 times that of moon and their diameters are $1600 \mathrm{~km}$ and $800 \mathrm{~km}$, respectively. If $g$ is the value of acceleration due to gravity on earth, what is its value on moon?
1 $\mathrm{g}$
2 $g / 2$
3 $g / 10$
4 $g / 20$
Explanation:
D Given, Mass of earth $\left(\mathrm{M}_{\mathrm{e}}\right)=80$, Mass of moon $\left(\mathrm{M}_{\mathrm{m}}\right)$ Diameters of earth $\left(\mathrm{d}_{\mathrm{e}}\right)=1600 \mathrm{~km}$ Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=\frac{1600}{2}=800 \mathrm{~km}$ Diameters of moon $\left(\mathrm{d}_{\mathrm{m}}\right)=800 \mathrm{~km}$ Radius of moon $\left(\mathrm{R}_{\mathrm{m}}\right)=\frac{800}{2}=400 \mathrm{~km}$ So, $\quad \mathrm{R}_{\mathrm{e}}=2 \mathrm{R}_{\mathrm{m}}$ Acceleration due to gravity at earth $\left(\mathrm{g}_{\mathrm{e}}\right)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}} \ldots \ldots$ (i) Acceleration due to gravity at $\operatorname{moon}\left(\mathrm{g}_{\mathrm{m}}\right)$ $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ Dividing equation (ii) by (i), we get, $\frac{g_{m}}{g_{e}}=\frac{M_{m}}{R_{m}^{2}} \times \frac{R_{e}{ }^{2}}{M_{e}}$ $\frac{g_{m}}{g_{e}}=\frac{M_{m} \times\left(2 R_{m}\right)^{2}}{\left(R_{m}\right)^{2} \times 80 M_{m}}$ $\frac{g_{m}}{g_{e}}=\frac{4}{80}$ $\frac{g_{m}}{g_{e}}=\frac{1}{20}$ $g_{m}=\frac{g_{e}}{20}$ Then, $g_{m}=\frac{g}{20}$
J and K-CET-2015
Gravitation
138360
Assuming the earth to be a sphere of uniform density, the ratio of acceleration due to gravity on the earth's surface to its value at halfway towards the centre of the earth, will be
1 $1: 1$
2 $1: 2$
3 $2: 3$
4 $2: 1$
Explanation:
D The acceleration due to gravity at a depth $\mathrm{h}$ below the surface of the earth is given by $g^{\prime}=g\left(1-\frac{h}{R_{e}}\right)$ $g^{\prime}=g\left(1-\frac{R_{e}}{2 R_{e}}\right) \quad\left(h=\frac{R_{e}}{2}\right)$ Then $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{1}{2}\right)=\frac{\mathrm{g}}{2}$ $\therefore \quad \frac{\mathrm{g}}{\mathrm{g}^{\prime}}=2$ $\mathrm{~g}: \mathrm{g}^{\prime}=2: 1$
138356
Taking the radius of the earth to be $6400 \mathrm{~km}$, by what percentage will the acceleration due to gravity at a height of $100 \mathrm{~km}$ from the surface of the earth differ from that on the surface of the earth?
1 about $1.5 \%$
2 about 5\%
3 about $8 \%$
4 about $3 \%$
Explanation:
D Given that, Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$, Height $(\mathrm{h})=100 \mathrm{~km}$ Acceleration due to gravity at height, $\mathrm{h}$ $g^{\prime}=\frac{g}{\left(1+\frac{h}{R_{e}}\right)^{2}}$ $g^{\prime}=\frac{g R_{e}^{2}}{\left(R_{e}+h\right)^{2}}$ $g^{\prime}=\frac{g(6400)^{2}}{(6400+100)^{2}}$ $g^{\prime}=g\left(\frac{6400}{6500}\right)^{2}$ Percentage change in acceleration due to gravity $=\frac{g-g^{\prime}}{g} \times 100$ $=\left(1-\frac{g^{\prime}}{g}\right) \times 100$ Put the value of $g^{\prime}$ in $\mathrm{eq}^{\mathrm{n}}$. (i) $=\left(1-\frac{\mathrm{g}\left(\frac{6400}{6500}\right)^{2}}{\mathrm{~g}}\right) \times 100 \%$ $=3.05 \%$
J and K-CET-2012
Gravitation
138357
If $R$ is the radius of the Earth then the height above the Earth's surface at which the acceleration due to gravity decreases by $20 \%$ is
1 $\left(\frac{\sqrt{5}}{2}-1\right) \mathrm{R}$
2 $\left(\frac{\sqrt{5}}{2}+1\right) \mathrm{R}$
3 $(5 \sqrt{2}-1) \mathrm{R}$
4 $(5 \sqrt{2}+1) \mathrm{R}$
Explanation:
A We know that $g_{h}=\left[\frac{R^{2}}{(R+h)^{2}}\right] g$ acceleration due to gravity decrease by $20 \%$ $g_{h} =g-\frac{20}{100} \times g$ $g_{h} =\frac{80}{100} g$ $g_{h} =\frac{4}{5} \mathrm{~g}$ $\frac{g_{h}}{g} =\frac{4}{5}$ Putting value of equation (ii) in equation (i), we get $\therefore \quad \frac{4}{5}=\frac{\mathrm{R}^{2}}{(\mathrm{R}+\mathrm{h})^{2}}$ $\sqrt{\frac{4}{5}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}$ $\frac{2}{\sqrt{5}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}$ $2 \mathrm{R}+2 \mathrm{~h}=\sqrt{5} \mathrm{R}$ $2 \mathrm{~h}=(\sqrt{5}-2) \mathrm{R}$ $\mathrm{h}=\left(\frac{\sqrt{5}-2}{2}\right) \mathrm{R}$ $\mathrm{h}=\left(\frac{\sqrt{5}}{2}-1\right) \mathrm{R}$
J and K-CET-2016
Gravitation
138358
The earth's mass is 80 times that of moon and their diameters are $1600 \mathrm{~km}$ and $800 \mathrm{~km}$, respectively. If $g$ is the value of acceleration due to gravity on earth, what is its value on moon?
1 $\mathrm{g}$
2 $g / 2$
3 $g / 10$
4 $g / 20$
Explanation:
D Given, Mass of earth $\left(\mathrm{M}_{\mathrm{e}}\right)=80$, Mass of moon $\left(\mathrm{M}_{\mathrm{m}}\right)$ Diameters of earth $\left(\mathrm{d}_{\mathrm{e}}\right)=1600 \mathrm{~km}$ Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=\frac{1600}{2}=800 \mathrm{~km}$ Diameters of moon $\left(\mathrm{d}_{\mathrm{m}}\right)=800 \mathrm{~km}$ Radius of moon $\left(\mathrm{R}_{\mathrm{m}}\right)=\frac{800}{2}=400 \mathrm{~km}$ So, $\quad \mathrm{R}_{\mathrm{e}}=2 \mathrm{R}_{\mathrm{m}}$ Acceleration due to gravity at earth $\left(\mathrm{g}_{\mathrm{e}}\right)=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}} \ldots \ldots$ (i) Acceleration due to gravity at $\operatorname{moon}\left(\mathrm{g}_{\mathrm{m}}\right)$ $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ Dividing equation (ii) by (i), we get, $\frac{g_{m}}{g_{e}}=\frac{M_{m}}{R_{m}^{2}} \times \frac{R_{e}{ }^{2}}{M_{e}}$ $\frac{g_{m}}{g_{e}}=\frac{M_{m} \times\left(2 R_{m}\right)^{2}}{\left(R_{m}\right)^{2} \times 80 M_{m}}$ $\frac{g_{m}}{g_{e}}=\frac{4}{80}$ $\frac{g_{m}}{g_{e}}=\frac{1}{20}$ $g_{m}=\frac{g_{e}}{20}$ Then, $g_{m}=\frac{g}{20}$
J and K-CET-2015
Gravitation
138360
Assuming the earth to be a sphere of uniform density, the ratio of acceleration due to gravity on the earth's surface to its value at halfway towards the centre of the earth, will be
1 $1: 1$
2 $1: 2$
3 $2: 3$
4 $2: 1$
Explanation:
D The acceleration due to gravity at a depth $\mathrm{h}$ below the surface of the earth is given by $g^{\prime}=g\left(1-\frac{h}{R_{e}}\right)$ $g^{\prime}=g\left(1-\frac{R_{e}}{2 R_{e}}\right) \quad\left(h=\frac{R_{e}}{2}\right)$ Then $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{1}{2}\right)=\frac{\mathrm{g}}{2}$ $\therefore \quad \frac{\mathrm{g}}{\mathrm{g}^{\prime}}=2$ $\mathrm{~g}: \mathrm{g}^{\prime}=2: 1$
