138261
A uniform ring of mass $m$ and radius $a$ is placed directly above a uniform sphere of mass $M$ and equal radius. The center of the ring is at a distance $\sqrt{3}$ a from the centre of the sphere. The gravitational force $[\mathrm{F}]$ exerted by the sphere on the ring is
A Given that, Mass of the ring $\mathrm{m}$ and radius $=\mathrm{a}$ Gravitational field due to the ring at a distance $\mathrm{d}=\sqrt{3} \mathrm{a}$ on its axis - $\mathrm{E}=\frac{\mathrm{Gmd}}{\left(\mathrm{a}^{2}+\mathrm{d}^{2}\right)^{3 / 2}}$ Here, $d=\sqrt{3} \mathrm{a}$ $\mathrm{E}=\frac{\operatorname{Gm}(\sqrt{3} \mathrm{a})}{\left(\mathrm{a}^{2}+(\sqrt{3} \mathrm{a})^{2}\right)^{3 / 2}}$ $\mathrm{E}=\frac{\mathrm{Gm}(\sqrt{3} \mathrm{a})}{\left(\mathrm{a}^{2}+3 \mathrm{a}^{2}\right)^{3 / 2}}$ $\mathrm{E}=\frac{\mathrm{Gm}(\sqrt{3} \mathrm{a})}{8 \mathrm{a}^{3}}$ $\mathrm{E}=\frac{\sqrt{3} \mathrm{Gm}}{8 \mathrm{a}^{2}}$ The force in a particle of mass $\mathrm{M}$ placed here is $\mathrm{F}=\mathrm{ME}$ $\mathrm{F}=\frac{\sqrt{3} \mathrm{GMm}}{8 \mathrm{a}^{2}}$
JIPMER-2015
Gravitation
138264
If $\mathbf{R}$ is the radius of earth, the height, at which the weight of a body becomes $\frac{1}{4}$ of its weight on the surface of earth, is
1 $2 \mathrm{R}$
2 $\mathrm{R}$
3 $\frac{R}{2}$
4 $\frac{\mathrm{R}}{4}$
Explanation:
B Weight of the body at earth's surface- $\mathrm{W}_{\mathrm{s}}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}$ Weight of the body and height $\mathrm{h}$ from earth's surface- $\mathrm{W}_{\mathrm{h}}=\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}}$ Now, according to question - $\mathrm{W}_{\mathrm{h}}=\frac{\mathrm{W}_{\mathrm{s}}}{4}$ $\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}}=\frac{\mathrm{GMm}}{4 \mathrm{R}^{2}}$ $\frac{1}{\mathrm{R}+\mathrm{h}}=\frac{1}{2 \mathrm{R}}$ $\mathrm{R}+\mathrm{h}=2 \mathrm{R}$ $\mathrm{h}=\mathrm{R}$
AP EAMCET-1999
Gravitation
138265
Mass $M$ is divided into two parts $\mathrm{xm}$ and (1$\mathbf{x}) \mathbf{m}$. For a given separation the value of $x$ for which the gravitational attraction between the two pieces becomes maximum is
1 $\frac{1}{2}$
2 $\frac{3}{5}$
3 1
4 2
Explanation:
A We know that Gravitational force - $F=\frac{\mathrm{Gm}_{1} m_{2}}{\mathrm{r}^{2}}$ $\mathrm{~F}=\frac{\mathrm{Gm}(1-\mathrm{x}) \mathrm{mx}}{\mathrm{r}^{2}}$ $\mathrm{~F}=\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \mathrm{x}(1-\mathrm{x})$ For maximum value of force $\frac{\mathrm{dF}}{\mathrm{dx}}=0$ $\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \mathrm{x}(1-\mathrm{x})\right]=0$ ${\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-\mathrm{x}^{2}\right)\right]=0}$ ${\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}(1-2 \mathrm{x})\right]=0}$ $1-2 \mathrm{x}=0$ $\mathrm{x}=\frac{1}{2}$
AP EAMCET-2001
Gravitation
138266
Two identical objects each of mass $50 \mathrm{~kg}$ are kept at a distance of separation of $50 \mathrm{~cm}$ apart on a horizontal table. The net gravitational force at the mid-point of the line joining their centres is
1 zero
2 $6.6733 \times 10^{-9} \mathrm{~N}$
3 $13.346 \times 10^{-9} \mathrm{~N}$
4 $3.336 \times 10^{-9} \mathrm{~N}$
5 $6.673 \times 10^{6} \mathrm{~N}$
Explanation:
A Given data, $\mathrm{m}=50 \mathrm{~kg}, \mathrm{~d}=50 \mathrm{~cm}=0.5 \mathrm{~m}$ Let $\mathrm{m}_{1}$ be the mass at the midpoint. The net gravitational force at mid-point becomes, $\mathrm{F}_{\mathrm{G}}=\mathrm{F}_{1}-\mathrm{F}_{2}=\frac{50 \times \mathrm{G} \times \mathrm{m}_{1}}{(0.5)^{2}}-\frac{50 \times \mathrm{G} \times \mathrm{m}_{1}}{(0.5)^{2}}=0$ The force exerted by the object on either side of the central object is same and it is opposite in direction. Hence, they cancel out each other and the resultant force is zero.
138261
A uniform ring of mass $m$ and radius $a$ is placed directly above a uniform sphere of mass $M$ and equal radius. The center of the ring is at a distance $\sqrt{3}$ a from the centre of the sphere. The gravitational force $[\mathrm{F}]$ exerted by the sphere on the ring is
A Given that, Mass of the ring $\mathrm{m}$ and radius $=\mathrm{a}$ Gravitational field due to the ring at a distance $\mathrm{d}=\sqrt{3} \mathrm{a}$ on its axis - $\mathrm{E}=\frac{\mathrm{Gmd}}{\left(\mathrm{a}^{2}+\mathrm{d}^{2}\right)^{3 / 2}}$ Here, $d=\sqrt{3} \mathrm{a}$ $\mathrm{E}=\frac{\operatorname{Gm}(\sqrt{3} \mathrm{a})}{\left(\mathrm{a}^{2}+(\sqrt{3} \mathrm{a})^{2}\right)^{3 / 2}}$ $\mathrm{E}=\frac{\mathrm{Gm}(\sqrt{3} \mathrm{a})}{\left(\mathrm{a}^{2}+3 \mathrm{a}^{2}\right)^{3 / 2}}$ $\mathrm{E}=\frac{\mathrm{Gm}(\sqrt{3} \mathrm{a})}{8 \mathrm{a}^{3}}$ $\mathrm{E}=\frac{\sqrt{3} \mathrm{Gm}}{8 \mathrm{a}^{2}}$ The force in a particle of mass $\mathrm{M}$ placed here is $\mathrm{F}=\mathrm{ME}$ $\mathrm{F}=\frac{\sqrt{3} \mathrm{GMm}}{8 \mathrm{a}^{2}}$
JIPMER-2015
Gravitation
138264
If $\mathbf{R}$ is the radius of earth, the height, at which the weight of a body becomes $\frac{1}{4}$ of its weight on the surface of earth, is
1 $2 \mathrm{R}$
2 $\mathrm{R}$
3 $\frac{R}{2}$
4 $\frac{\mathrm{R}}{4}$
Explanation:
B Weight of the body at earth's surface- $\mathrm{W}_{\mathrm{s}}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}$ Weight of the body and height $\mathrm{h}$ from earth's surface- $\mathrm{W}_{\mathrm{h}}=\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}}$ Now, according to question - $\mathrm{W}_{\mathrm{h}}=\frac{\mathrm{W}_{\mathrm{s}}}{4}$ $\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}}=\frac{\mathrm{GMm}}{4 \mathrm{R}^{2}}$ $\frac{1}{\mathrm{R}+\mathrm{h}}=\frac{1}{2 \mathrm{R}}$ $\mathrm{R}+\mathrm{h}=2 \mathrm{R}$ $\mathrm{h}=\mathrm{R}$
AP EAMCET-1999
Gravitation
138265
Mass $M$ is divided into two parts $\mathrm{xm}$ and (1$\mathbf{x}) \mathbf{m}$. For a given separation the value of $x$ for which the gravitational attraction between the two pieces becomes maximum is
1 $\frac{1}{2}$
2 $\frac{3}{5}$
3 1
4 2
Explanation:
A We know that Gravitational force - $F=\frac{\mathrm{Gm}_{1} m_{2}}{\mathrm{r}^{2}}$ $\mathrm{~F}=\frac{\mathrm{Gm}(1-\mathrm{x}) \mathrm{mx}}{\mathrm{r}^{2}}$ $\mathrm{~F}=\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \mathrm{x}(1-\mathrm{x})$ For maximum value of force $\frac{\mathrm{dF}}{\mathrm{dx}}=0$ $\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \mathrm{x}(1-\mathrm{x})\right]=0$ ${\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-\mathrm{x}^{2}\right)\right]=0}$ ${\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}(1-2 \mathrm{x})\right]=0}$ $1-2 \mathrm{x}=0$ $\mathrm{x}=\frac{1}{2}$
AP EAMCET-2001
Gravitation
138266
Two identical objects each of mass $50 \mathrm{~kg}$ are kept at a distance of separation of $50 \mathrm{~cm}$ apart on a horizontal table. The net gravitational force at the mid-point of the line joining their centres is
1 zero
2 $6.6733 \times 10^{-9} \mathrm{~N}$
3 $13.346 \times 10^{-9} \mathrm{~N}$
4 $3.336 \times 10^{-9} \mathrm{~N}$
5 $6.673 \times 10^{6} \mathrm{~N}$
Explanation:
A Given data, $\mathrm{m}=50 \mathrm{~kg}, \mathrm{~d}=50 \mathrm{~cm}=0.5 \mathrm{~m}$ Let $\mathrm{m}_{1}$ be the mass at the midpoint. The net gravitational force at mid-point becomes, $\mathrm{F}_{\mathrm{G}}=\mathrm{F}_{1}-\mathrm{F}_{2}=\frac{50 \times \mathrm{G} \times \mathrm{m}_{1}}{(0.5)^{2}}-\frac{50 \times \mathrm{G} \times \mathrm{m}_{1}}{(0.5)^{2}}=0$ The force exerted by the object on either side of the central object is same and it is opposite in direction. Hence, they cancel out each other and the resultant force is zero.
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Gravitation
138261
A uniform ring of mass $m$ and radius $a$ is placed directly above a uniform sphere of mass $M$ and equal radius. The center of the ring is at a distance $\sqrt{3}$ a from the centre of the sphere. The gravitational force $[\mathrm{F}]$ exerted by the sphere on the ring is
A Given that, Mass of the ring $\mathrm{m}$ and radius $=\mathrm{a}$ Gravitational field due to the ring at a distance $\mathrm{d}=\sqrt{3} \mathrm{a}$ on its axis - $\mathrm{E}=\frac{\mathrm{Gmd}}{\left(\mathrm{a}^{2}+\mathrm{d}^{2}\right)^{3 / 2}}$ Here, $d=\sqrt{3} \mathrm{a}$ $\mathrm{E}=\frac{\operatorname{Gm}(\sqrt{3} \mathrm{a})}{\left(\mathrm{a}^{2}+(\sqrt{3} \mathrm{a})^{2}\right)^{3 / 2}}$ $\mathrm{E}=\frac{\mathrm{Gm}(\sqrt{3} \mathrm{a})}{\left(\mathrm{a}^{2}+3 \mathrm{a}^{2}\right)^{3 / 2}}$ $\mathrm{E}=\frac{\mathrm{Gm}(\sqrt{3} \mathrm{a})}{8 \mathrm{a}^{3}}$ $\mathrm{E}=\frac{\sqrt{3} \mathrm{Gm}}{8 \mathrm{a}^{2}}$ The force in a particle of mass $\mathrm{M}$ placed here is $\mathrm{F}=\mathrm{ME}$ $\mathrm{F}=\frac{\sqrt{3} \mathrm{GMm}}{8 \mathrm{a}^{2}}$
JIPMER-2015
Gravitation
138264
If $\mathbf{R}$ is the radius of earth, the height, at which the weight of a body becomes $\frac{1}{4}$ of its weight on the surface of earth, is
1 $2 \mathrm{R}$
2 $\mathrm{R}$
3 $\frac{R}{2}$
4 $\frac{\mathrm{R}}{4}$
Explanation:
B Weight of the body at earth's surface- $\mathrm{W}_{\mathrm{s}}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}$ Weight of the body and height $\mathrm{h}$ from earth's surface- $\mathrm{W}_{\mathrm{h}}=\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}}$ Now, according to question - $\mathrm{W}_{\mathrm{h}}=\frac{\mathrm{W}_{\mathrm{s}}}{4}$ $\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}}=\frac{\mathrm{GMm}}{4 \mathrm{R}^{2}}$ $\frac{1}{\mathrm{R}+\mathrm{h}}=\frac{1}{2 \mathrm{R}}$ $\mathrm{R}+\mathrm{h}=2 \mathrm{R}$ $\mathrm{h}=\mathrm{R}$
AP EAMCET-1999
Gravitation
138265
Mass $M$ is divided into two parts $\mathrm{xm}$ and (1$\mathbf{x}) \mathbf{m}$. For a given separation the value of $x$ for which the gravitational attraction between the two pieces becomes maximum is
1 $\frac{1}{2}$
2 $\frac{3}{5}$
3 1
4 2
Explanation:
A We know that Gravitational force - $F=\frac{\mathrm{Gm}_{1} m_{2}}{\mathrm{r}^{2}}$ $\mathrm{~F}=\frac{\mathrm{Gm}(1-\mathrm{x}) \mathrm{mx}}{\mathrm{r}^{2}}$ $\mathrm{~F}=\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \mathrm{x}(1-\mathrm{x})$ For maximum value of force $\frac{\mathrm{dF}}{\mathrm{dx}}=0$ $\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \mathrm{x}(1-\mathrm{x})\right]=0$ ${\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-\mathrm{x}^{2}\right)\right]=0}$ ${\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}(1-2 \mathrm{x})\right]=0}$ $1-2 \mathrm{x}=0$ $\mathrm{x}=\frac{1}{2}$
AP EAMCET-2001
Gravitation
138266
Two identical objects each of mass $50 \mathrm{~kg}$ are kept at a distance of separation of $50 \mathrm{~cm}$ apart on a horizontal table. The net gravitational force at the mid-point of the line joining their centres is
1 zero
2 $6.6733 \times 10^{-9} \mathrm{~N}$
3 $13.346 \times 10^{-9} \mathrm{~N}$
4 $3.336 \times 10^{-9} \mathrm{~N}$
5 $6.673 \times 10^{6} \mathrm{~N}$
Explanation:
A Given data, $\mathrm{m}=50 \mathrm{~kg}, \mathrm{~d}=50 \mathrm{~cm}=0.5 \mathrm{~m}$ Let $\mathrm{m}_{1}$ be the mass at the midpoint. The net gravitational force at mid-point becomes, $\mathrm{F}_{\mathrm{G}}=\mathrm{F}_{1}-\mathrm{F}_{2}=\frac{50 \times \mathrm{G} \times \mathrm{m}_{1}}{(0.5)^{2}}-\frac{50 \times \mathrm{G} \times \mathrm{m}_{1}}{(0.5)^{2}}=0$ The force exerted by the object on either side of the central object is same and it is opposite in direction. Hence, they cancel out each other and the resultant force is zero.
138261
A uniform ring of mass $m$ and radius $a$ is placed directly above a uniform sphere of mass $M$ and equal radius. The center of the ring is at a distance $\sqrt{3}$ a from the centre of the sphere. The gravitational force $[\mathrm{F}]$ exerted by the sphere on the ring is
A Given that, Mass of the ring $\mathrm{m}$ and radius $=\mathrm{a}$ Gravitational field due to the ring at a distance $\mathrm{d}=\sqrt{3} \mathrm{a}$ on its axis - $\mathrm{E}=\frac{\mathrm{Gmd}}{\left(\mathrm{a}^{2}+\mathrm{d}^{2}\right)^{3 / 2}}$ Here, $d=\sqrt{3} \mathrm{a}$ $\mathrm{E}=\frac{\operatorname{Gm}(\sqrt{3} \mathrm{a})}{\left(\mathrm{a}^{2}+(\sqrt{3} \mathrm{a})^{2}\right)^{3 / 2}}$ $\mathrm{E}=\frac{\mathrm{Gm}(\sqrt{3} \mathrm{a})}{\left(\mathrm{a}^{2}+3 \mathrm{a}^{2}\right)^{3 / 2}}$ $\mathrm{E}=\frac{\mathrm{Gm}(\sqrt{3} \mathrm{a})}{8 \mathrm{a}^{3}}$ $\mathrm{E}=\frac{\sqrt{3} \mathrm{Gm}}{8 \mathrm{a}^{2}}$ The force in a particle of mass $\mathrm{M}$ placed here is $\mathrm{F}=\mathrm{ME}$ $\mathrm{F}=\frac{\sqrt{3} \mathrm{GMm}}{8 \mathrm{a}^{2}}$
JIPMER-2015
Gravitation
138264
If $\mathbf{R}$ is the radius of earth, the height, at which the weight of a body becomes $\frac{1}{4}$ of its weight on the surface of earth, is
1 $2 \mathrm{R}$
2 $\mathrm{R}$
3 $\frac{R}{2}$
4 $\frac{\mathrm{R}}{4}$
Explanation:
B Weight of the body at earth's surface- $\mathrm{W}_{\mathrm{s}}=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}$ Weight of the body and height $\mathrm{h}$ from earth's surface- $\mathrm{W}_{\mathrm{h}}=\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}}$ Now, according to question - $\mathrm{W}_{\mathrm{h}}=\frac{\mathrm{W}_{\mathrm{s}}}{4}$ $\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}}=\frac{\mathrm{GMm}}{4 \mathrm{R}^{2}}$ $\frac{1}{\mathrm{R}+\mathrm{h}}=\frac{1}{2 \mathrm{R}}$ $\mathrm{R}+\mathrm{h}=2 \mathrm{R}$ $\mathrm{h}=\mathrm{R}$
AP EAMCET-1999
Gravitation
138265
Mass $M$ is divided into two parts $\mathrm{xm}$ and (1$\mathbf{x}) \mathbf{m}$. For a given separation the value of $x$ for which the gravitational attraction between the two pieces becomes maximum is
1 $\frac{1}{2}$
2 $\frac{3}{5}$
3 1
4 2
Explanation:
A We know that Gravitational force - $F=\frac{\mathrm{Gm}_{1} m_{2}}{\mathrm{r}^{2}}$ $\mathrm{~F}=\frac{\mathrm{Gm}(1-\mathrm{x}) \mathrm{mx}}{\mathrm{r}^{2}}$ $\mathrm{~F}=\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \mathrm{x}(1-\mathrm{x})$ For maximum value of force $\frac{\mathrm{dF}}{\mathrm{dx}}=0$ $\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \mathrm{x}(1-\mathrm{x})\right]=0$ ${\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-\mathrm{x}^{2}\right)\right]=0}$ ${\left[\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}(1-2 \mathrm{x})\right]=0}$ $1-2 \mathrm{x}=0$ $\mathrm{x}=\frac{1}{2}$
AP EAMCET-2001
Gravitation
138266
Two identical objects each of mass $50 \mathrm{~kg}$ are kept at a distance of separation of $50 \mathrm{~cm}$ apart on a horizontal table. The net gravitational force at the mid-point of the line joining their centres is
1 zero
2 $6.6733 \times 10^{-9} \mathrm{~N}$
3 $13.346 \times 10^{-9} \mathrm{~N}$
4 $3.336 \times 10^{-9} \mathrm{~N}$
5 $6.673 \times 10^{6} \mathrm{~N}$
Explanation:
A Given data, $\mathrm{m}=50 \mathrm{~kg}, \mathrm{~d}=50 \mathrm{~cm}=0.5 \mathrm{~m}$ Let $\mathrm{m}_{1}$ be the mass at the midpoint. The net gravitational force at mid-point becomes, $\mathrm{F}_{\mathrm{G}}=\mathrm{F}_{1}-\mathrm{F}_{2}=\frac{50 \times \mathrm{G} \times \mathrm{m}_{1}}{(0.5)^{2}}-\frac{50 \times \mathrm{G} \times \mathrm{m}_{1}}{(0.5)^{2}}=0$ The force exerted by the object on either side of the central object is same and it is opposite in direction. Hence, they cancel out each other and the resultant force is zero.
