138270
Suppose the force of gravitation between two bodies of equal masses is $F$. If each mass is doubled keeping the distance of separation between them unchanged, the force would become
1 $\mathrm{F}$
2 $2 \mathrm{~F}$
3 $4 \mathrm{~F}$
4 $\frac{1}{4} F$
Explanation:
C Two bodies are of equal mass. So, $\quad \mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}$ $\mathrm{m}_{1}=$ mass of $1^{\text {st }}$ body, $\mathrm{m}_{2}=$ mass of $2^{\text {nd }}$ body and $\mathrm{r}=$ distance between the two bodies Then, $\quad \mathrm{F}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}$ $\mathrm{F}=\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}$ Now masses are doubled and distance is the same $\mathrm{m}_{1}=\mathrm{m}_{2}=2 \mathrm{~m}$ The magnitude of the gravitational force $\mathrm{F}$ using equation (i) $\mathrm{F}=\mathrm{G} \frac{2 \mathrm{~m} \times 2 \mathrm{~m}}{\mathrm{r}^{2}}$ $\mathrm{~F}^{1}=4 \times \frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}=4 \mathrm{~F}$
NDA (II) 2019
Gravitation
138271
Which one of the following statements is true for the relation $F=\frac{G m_{1} m_{2}}{r^{2}}$ ? (All symbols have their usual meanings)
1 The quantity $\mathrm{G}$ depends on the local value of $\mathrm{g}$, acceleration due to gravity
2 The quantity $G$ is greatest at the surface of the Earth
3 The quantity G is used only when earth is one of the two masses
4 The quantity $\mathrm{G}$ is a universal constant
Explanation:
D Gravitational Force $(F)=G \frac{m_{1} m_{2}}{r^{2}}$ Where, $\mathrm{m}_{1}=$ mass of first body $\mathrm{m}_{2}=$ mass of second body $\mathrm{r}=$ distance between two body $\mathrm{G}=$ Gravitational constant $G$ is the universal gravitational constant which remain constant at all places in the universe $\mathrm{G}$ is equivalent to the force of all reaction between two bodies of unit mass and unit distance apart. The value of $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}$
NDA (I) 2017
Gravitation
138273
Three particles, two with masses $m$ and one with mass $M$, might be arranged in any of the four configurations shown below. Rank the configuration according to the magnitude of the gravitational force on $M$, least to greatest.
1 $1,2,3,4$
2 $2,1,3,4$
3 $2,1,4,3$
4 2, 3, 4, 1
Explanation:
B From figure (i), $\because \quad \mathrm{F}_{\text {net }}=\mathrm{F}_{1}+\mathrm{F}_{2}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}+\frac{\mathrm{GMm}}{(2 \mathrm{~d})^{2}}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\left(1+\frac{1}{4}\right)$ $=\frac{5}{4} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}=1.25 \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ From figure (ii), $\because \quad \mathrm{F}_{\text {net }}=\mathrm{F}_{4}-\mathrm{F}_{3}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}-\frac{\mathrm{GMm}}{\mathrm{d}^{2}}=0$ From figure (iii), $\because \quad \mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_{5}^{2}+\mathrm{F}_{6}^{2}+2 \mathrm{~F}_{5} \mathrm{~F}_{6} \cos \theta}$ $=\sqrt{\mathrm{F}_{5}^{2}+\mathrm{F}_{6}^{2}+2 \mathrm{~F}_{5} \mathrm{~F}_{6} \cos 90^{\circ}} \quad\left(\because \cos 90^{\circ}=0\right)$ $=\sqrt{\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}} \quad$ $\mathrm{F}_{\text {net }}=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}=1.414 \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ From equation (iv), $\because \mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_{7}^{2}+\mathrm{F}_{8}^{2}+2 \mathrm{~F}_{7} \mathrm{~F}_{8} \cos \theta}$ $=\sqrt{\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+2 \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \times \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \cos \theta}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{1+1+2 \cos \theta}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}(\sqrt{2(1+\cos \theta)})$ $=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{1+\cos \theta}=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{2} \cos \frac{\theta}{2}$ $=2 \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \cos \frac{\theta}{2}$ For, $0 \lt \theta \lt 90^{\circ}$ then, $\therefore \quad \mathrm{F}_{\text {net }}>\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ Thus, the rank of gravitational force from least to greatest, $2 \lt 1 \lt 3 \lt 4$
AMU-2018
Gravitation
138274
The figure shows four arrangements of three particles of equal masses. Arrange them according to the magnitude of the net gravitational force on the particle labeled $\mathbf{m}$, in decreasing order.
1 (i), (iii) = (iv), (ii)
2 (i) = (iii), (ii), (iv)
3 (i), (ii), (iii), (iv)
4 (iv), (iii), (ii), (ii)
Explanation:
A We know that, Gravitational force $(\mathrm{F})=\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}$ From figure (i), $\mathrm{F}_{1}=\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}+\frac{\mathrm{Gmm}}{\mathrm{D}^{2}}=\mathrm{Gm}^{2}\left(\frac{1}{\mathrm{~d}^{2}}+\frac{1}{\mathrm{D}^{2}}\right)$ From figure (ii), $\mathrm{F}_{2}=\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}-\frac{\mathrm{Gmm}}{(\mathrm{D}-\mathrm{d})^{2}}=\mathrm{Gm}^{2}\left(\frac{1}{\mathrm{~d}^{2}}-\frac{1}{\mathrm{D}^{2}}\right)$ From figure (iii), $F_{3}=\sqrt{\left(\frac{G m m}{d^{2}}\right)^{2}+\left(\frac{G m m}{D^{2}}\right)^{2}}=\mathrm{Gm}^{2} \sqrt{\left(\frac{1}{\mathrm{~d}^{2}}\right)^{2}+\left(\frac{1}{(D)^{2}}\right)^{2}}$ From figure (iv), $\mathrm{F}_{4}=\sqrt{\left(\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}\right)^{2}-\left(\frac{\mathrm{Gmm}}{\mathrm{D}^{2}}\right)^{2}}=\mathrm{Gm}^{2} \sqrt{\left(\frac{1}{\mathrm{~d}^{2}}\right)^{2}-\left(\frac{1}{\mathrm{D}^{2}}\right)^{2}}$ Hence, the magnitude of the net gravitational force on the particle labeled $\mathrm{m}$, in decreasing order- $\text { (i), (iii) = (iv), (ii) }$
138270
Suppose the force of gravitation between two bodies of equal masses is $F$. If each mass is doubled keeping the distance of separation between them unchanged, the force would become
1 $\mathrm{F}$
2 $2 \mathrm{~F}$
3 $4 \mathrm{~F}$
4 $\frac{1}{4} F$
Explanation:
C Two bodies are of equal mass. So, $\quad \mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}$ $\mathrm{m}_{1}=$ mass of $1^{\text {st }}$ body, $\mathrm{m}_{2}=$ mass of $2^{\text {nd }}$ body and $\mathrm{r}=$ distance between the two bodies Then, $\quad \mathrm{F}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}$ $\mathrm{F}=\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}$ Now masses are doubled and distance is the same $\mathrm{m}_{1}=\mathrm{m}_{2}=2 \mathrm{~m}$ The magnitude of the gravitational force $\mathrm{F}$ using equation (i) $\mathrm{F}=\mathrm{G} \frac{2 \mathrm{~m} \times 2 \mathrm{~m}}{\mathrm{r}^{2}}$ $\mathrm{~F}^{1}=4 \times \frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}=4 \mathrm{~F}$
NDA (II) 2019
Gravitation
138271
Which one of the following statements is true for the relation $F=\frac{G m_{1} m_{2}}{r^{2}}$ ? (All symbols have their usual meanings)
1 The quantity $\mathrm{G}$ depends on the local value of $\mathrm{g}$, acceleration due to gravity
2 The quantity $G$ is greatest at the surface of the Earth
3 The quantity G is used only when earth is one of the two masses
4 The quantity $\mathrm{G}$ is a universal constant
Explanation:
D Gravitational Force $(F)=G \frac{m_{1} m_{2}}{r^{2}}$ Where, $\mathrm{m}_{1}=$ mass of first body $\mathrm{m}_{2}=$ mass of second body $\mathrm{r}=$ distance between two body $\mathrm{G}=$ Gravitational constant $G$ is the universal gravitational constant which remain constant at all places in the universe $\mathrm{G}$ is equivalent to the force of all reaction between two bodies of unit mass and unit distance apart. The value of $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}$
NDA (I) 2017
Gravitation
138273
Three particles, two with masses $m$ and one with mass $M$, might be arranged in any of the four configurations shown below. Rank the configuration according to the magnitude of the gravitational force on $M$, least to greatest.
1 $1,2,3,4$
2 $2,1,3,4$
3 $2,1,4,3$
4 2, 3, 4, 1
Explanation:
B From figure (i), $\because \quad \mathrm{F}_{\text {net }}=\mathrm{F}_{1}+\mathrm{F}_{2}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}+\frac{\mathrm{GMm}}{(2 \mathrm{~d})^{2}}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\left(1+\frac{1}{4}\right)$ $=\frac{5}{4} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}=1.25 \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ From figure (ii), $\because \quad \mathrm{F}_{\text {net }}=\mathrm{F}_{4}-\mathrm{F}_{3}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}-\frac{\mathrm{GMm}}{\mathrm{d}^{2}}=0$ From figure (iii), $\because \quad \mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_{5}^{2}+\mathrm{F}_{6}^{2}+2 \mathrm{~F}_{5} \mathrm{~F}_{6} \cos \theta}$ $=\sqrt{\mathrm{F}_{5}^{2}+\mathrm{F}_{6}^{2}+2 \mathrm{~F}_{5} \mathrm{~F}_{6} \cos 90^{\circ}} \quad\left(\because \cos 90^{\circ}=0\right)$ $=\sqrt{\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}} \quad$ $\mathrm{F}_{\text {net }}=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}=1.414 \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ From equation (iv), $\because \mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_{7}^{2}+\mathrm{F}_{8}^{2}+2 \mathrm{~F}_{7} \mathrm{~F}_{8} \cos \theta}$ $=\sqrt{\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+2 \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \times \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \cos \theta}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{1+1+2 \cos \theta}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}(\sqrt{2(1+\cos \theta)})$ $=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{1+\cos \theta}=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{2} \cos \frac{\theta}{2}$ $=2 \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \cos \frac{\theta}{2}$ For, $0 \lt \theta \lt 90^{\circ}$ then, $\therefore \quad \mathrm{F}_{\text {net }}>\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ Thus, the rank of gravitational force from least to greatest, $2 \lt 1 \lt 3 \lt 4$
AMU-2018
Gravitation
138274
The figure shows four arrangements of three particles of equal masses. Arrange them according to the magnitude of the net gravitational force on the particle labeled $\mathbf{m}$, in decreasing order.
1 (i), (iii) = (iv), (ii)
2 (i) = (iii), (ii), (iv)
3 (i), (ii), (iii), (iv)
4 (iv), (iii), (ii), (ii)
Explanation:
A We know that, Gravitational force $(\mathrm{F})=\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}$ From figure (i), $\mathrm{F}_{1}=\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}+\frac{\mathrm{Gmm}}{\mathrm{D}^{2}}=\mathrm{Gm}^{2}\left(\frac{1}{\mathrm{~d}^{2}}+\frac{1}{\mathrm{D}^{2}}\right)$ From figure (ii), $\mathrm{F}_{2}=\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}-\frac{\mathrm{Gmm}}{(\mathrm{D}-\mathrm{d})^{2}}=\mathrm{Gm}^{2}\left(\frac{1}{\mathrm{~d}^{2}}-\frac{1}{\mathrm{D}^{2}}\right)$ From figure (iii), $F_{3}=\sqrt{\left(\frac{G m m}{d^{2}}\right)^{2}+\left(\frac{G m m}{D^{2}}\right)^{2}}=\mathrm{Gm}^{2} \sqrt{\left(\frac{1}{\mathrm{~d}^{2}}\right)^{2}+\left(\frac{1}{(D)^{2}}\right)^{2}}$ From figure (iv), $\mathrm{F}_{4}=\sqrt{\left(\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}\right)^{2}-\left(\frac{\mathrm{Gmm}}{\mathrm{D}^{2}}\right)^{2}}=\mathrm{Gm}^{2} \sqrt{\left(\frac{1}{\mathrm{~d}^{2}}\right)^{2}-\left(\frac{1}{\mathrm{D}^{2}}\right)^{2}}$ Hence, the magnitude of the net gravitational force on the particle labeled $\mathrm{m}$, in decreasing order- $\text { (i), (iii) = (iv), (ii) }$
138270
Suppose the force of gravitation between two bodies of equal masses is $F$. If each mass is doubled keeping the distance of separation between them unchanged, the force would become
1 $\mathrm{F}$
2 $2 \mathrm{~F}$
3 $4 \mathrm{~F}$
4 $\frac{1}{4} F$
Explanation:
C Two bodies are of equal mass. So, $\quad \mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}$ $\mathrm{m}_{1}=$ mass of $1^{\text {st }}$ body, $\mathrm{m}_{2}=$ mass of $2^{\text {nd }}$ body and $\mathrm{r}=$ distance between the two bodies Then, $\quad \mathrm{F}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}$ $\mathrm{F}=\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}$ Now masses are doubled and distance is the same $\mathrm{m}_{1}=\mathrm{m}_{2}=2 \mathrm{~m}$ The magnitude of the gravitational force $\mathrm{F}$ using equation (i) $\mathrm{F}=\mathrm{G} \frac{2 \mathrm{~m} \times 2 \mathrm{~m}}{\mathrm{r}^{2}}$ $\mathrm{~F}^{1}=4 \times \frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}=4 \mathrm{~F}$
NDA (II) 2019
Gravitation
138271
Which one of the following statements is true for the relation $F=\frac{G m_{1} m_{2}}{r^{2}}$ ? (All symbols have their usual meanings)
1 The quantity $\mathrm{G}$ depends on the local value of $\mathrm{g}$, acceleration due to gravity
2 The quantity $G$ is greatest at the surface of the Earth
3 The quantity G is used only when earth is one of the two masses
4 The quantity $\mathrm{G}$ is a universal constant
Explanation:
D Gravitational Force $(F)=G \frac{m_{1} m_{2}}{r^{2}}$ Where, $\mathrm{m}_{1}=$ mass of first body $\mathrm{m}_{2}=$ mass of second body $\mathrm{r}=$ distance between two body $\mathrm{G}=$ Gravitational constant $G$ is the universal gravitational constant which remain constant at all places in the universe $\mathrm{G}$ is equivalent to the force of all reaction between two bodies of unit mass and unit distance apart. The value of $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}$
NDA (I) 2017
Gravitation
138273
Three particles, two with masses $m$ and one with mass $M$, might be arranged in any of the four configurations shown below. Rank the configuration according to the magnitude of the gravitational force on $M$, least to greatest.
1 $1,2,3,4$
2 $2,1,3,4$
3 $2,1,4,3$
4 2, 3, 4, 1
Explanation:
B From figure (i), $\because \quad \mathrm{F}_{\text {net }}=\mathrm{F}_{1}+\mathrm{F}_{2}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}+\frac{\mathrm{GMm}}{(2 \mathrm{~d})^{2}}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\left(1+\frac{1}{4}\right)$ $=\frac{5}{4} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}=1.25 \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ From figure (ii), $\because \quad \mathrm{F}_{\text {net }}=\mathrm{F}_{4}-\mathrm{F}_{3}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}-\frac{\mathrm{GMm}}{\mathrm{d}^{2}}=0$ From figure (iii), $\because \quad \mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_{5}^{2}+\mathrm{F}_{6}^{2}+2 \mathrm{~F}_{5} \mathrm{~F}_{6} \cos \theta}$ $=\sqrt{\mathrm{F}_{5}^{2}+\mathrm{F}_{6}^{2}+2 \mathrm{~F}_{5} \mathrm{~F}_{6} \cos 90^{\circ}} \quad\left(\because \cos 90^{\circ}=0\right)$ $=\sqrt{\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}} \quad$ $\mathrm{F}_{\text {net }}=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}=1.414 \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ From equation (iv), $\because \mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_{7}^{2}+\mathrm{F}_{8}^{2}+2 \mathrm{~F}_{7} \mathrm{~F}_{8} \cos \theta}$ $=\sqrt{\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+2 \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \times \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \cos \theta}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{1+1+2 \cos \theta}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}(\sqrt{2(1+\cos \theta)})$ $=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{1+\cos \theta}=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{2} \cos \frac{\theta}{2}$ $=2 \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \cos \frac{\theta}{2}$ For, $0 \lt \theta \lt 90^{\circ}$ then, $\therefore \quad \mathrm{F}_{\text {net }}>\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ Thus, the rank of gravitational force from least to greatest, $2 \lt 1 \lt 3 \lt 4$
AMU-2018
Gravitation
138274
The figure shows four arrangements of three particles of equal masses. Arrange them according to the magnitude of the net gravitational force on the particle labeled $\mathbf{m}$, in decreasing order.
1 (i), (iii) = (iv), (ii)
2 (i) = (iii), (ii), (iv)
3 (i), (ii), (iii), (iv)
4 (iv), (iii), (ii), (ii)
Explanation:
A We know that, Gravitational force $(\mathrm{F})=\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}$ From figure (i), $\mathrm{F}_{1}=\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}+\frac{\mathrm{Gmm}}{\mathrm{D}^{2}}=\mathrm{Gm}^{2}\left(\frac{1}{\mathrm{~d}^{2}}+\frac{1}{\mathrm{D}^{2}}\right)$ From figure (ii), $\mathrm{F}_{2}=\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}-\frac{\mathrm{Gmm}}{(\mathrm{D}-\mathrm{d})^{2}}=\mathrm{Gm}^{2}\left(\frac{1}{\mathrm{~d}^{2}}-\frac{1}{\mathrm{D}^{2}}\right)$ From figure (iii), $F_{3}=\sqrt{\left(\frac{G m m}{d^{2}}\right)^{2}+\left(\frac{G m m}{D^{2}}\right)^{2}}=\mathrm{Gm}^{2} \sqrt{\left(\frac{1}{\mathrm{~d}^{2}}\right)^{2}+\left(\frac{1}{(D)^{2}}\right)^{2}}$ From figure (iv), $\mathrm{F}_{4}=\sqrt{\left(\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}\right)^{2}-\left(\frac{\mathrm{Gmm}}{\mathrm{D}^{2}}\right)^{2}}=\mathrm{Gm}^{2} \sqrt{\left(\frac{1}{\mathrm{~d}^{2}}\right)^{2}-\left(\frac{1}{\mathrm{D}^{2}}\right)^{2}}$ Hence, the magnitude of the net gravitational force on the particle labeled $\mathrm{m}$, in decreasing order- $\text { (i), (iii) = (iv), (ii) }$
138270
Suppose the force of gravitation between two bodies of equal masses is $F$. If each mass is doubled keeping the distance of separation between them unchanged, the force would become
1 $\mathrm{F}$
2 $2 \mathrm{~F}$
3 $4 \mathrm{~F}$
4 $\frac{1}{4} F$
Explanation:
C Two bodies are of equal mass. So, $\quad \mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}$ $\mathrm{m}_{1}=$ mass of $1^{\text {st }}$ body, $\mathrm{m}_{2}=$ mass of $2^{\text {nd }}$ body and $\mathrm{r}=$ distance between the two bodies Then, $\quad \mathrm{F}=\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}$ $\mathrm{F}=\frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}$ Now masses are doubled and distance is the same $\mathrm{m}_{1}=\mathrm{m}_{2}=2 \mathrm{~m}$ The magnitude of the gravitational force $\mathrm{F}$ using equation (i) $\mathrm{F}=\mathrm{G} \frac{2 \mathrm{~m} \times 2 \mathrm{~m}}{\mathrm{r}^{2}}$ $\mathrm{~F}^{1}=4 \times \frac{\mathrm{Gm}^{2}}{\mathrm{r}^{2}}=4 \mathrm{~F}$
NDA (II) 2019
Gravitation
138271
Which one of the following statements is true for the relation $F=\frac{G m_{1} m_{2}}{r^{2}}$ ? (All symbols have their usual meanings)
1 The quantity $\mathrm{G}$ depends on the local value of $\mathrm{g}$, acceleration due to gravity
2 The quantity $G$ is greatest at the surface of the Earth
3 The quantity G is used only when earth is one of the two masses
4 The quantity $\mathrm{G}$ is a universal constant
Explanation:
D Gravitational Force $(F)=G \frac{m_{1} m_{2}}{r^{2}}$ Where, $\mathrm{m}_{1}=$ mass of first body $\mathrm{m}_{2}=$ mass of second body $\mathrm{r}=$ distance between two body $\mathrm{G}=$ Gravitational constant $G$ is the universal gravitational constant which remain constant at all places in the universe $\mathrm{G}$ is equivalent to the force of all reaction between two bodies of unit mass and unit distance apart. The value of $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}$
NDA (I) 2017
Gravitation
138273
Three particles, two with masses $m$ and one with mass $M$, might be arranged in any of the four configurations shown below. Rank the configuration according to the magnitude of the gravitational force on $M$, least to greatest.
1 $1,2,3,4$
2 $2,1,3,4$
3 $2,1,4,3$
4 2, 3, 4, 1
Explanation:
B From figure (i), $\because \quad \mathrm{F}_{\text {net }}=\mathrm{F}_{1}+\mathrm{F}_{2}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}+\frac{\mathrm{GMm}}{(2 \mathrm{~d})^{2}}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\left(1+\frac{1}{4}\right)$ $=\frac{5}{4} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}=1.25 \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ From figure (ii), $\because \quad \mathrm{F}_{\text {net }}=\mathrm{F}_{4}-\mathrm{F}_{3}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}-\frac{\mathrm{GMm}}{\mathrm{d}^{2}}=0$ From figure (iii), $\because \quad \mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_{5}^{2}+\mathrm{F}_{6}^{2}+2 \mathrm{~F}_{5} \mathrm{~F}_{6} \cos \theta}$ $=\sqrt{\mathrm{F}_{5}^{2}+\mathrm{F}_{6}^{2}+2 \mathrm{~F}_{5} \mathrm{~F}_{6} \cos 90^{\circ}} \quad\left(\because \cos 90^{\circ}=0\right)$ $=\sqrt{\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}} \quad$ $\mathrm{F}_{\text {net }}=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}=1.414 \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ From equation (iv), $\because \mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_{7}^{2}+\mathrm{F}_{8}^{2}+2 \mathrm{~F}_{7} \mathrm{~F}_{8} \cos \theta}$ $=\sqrt{\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+\left(\frac{\mathrm{GMm}}{\mathrm{d}^{2}}\right)^{2}+2 \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \times \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \cos \theta}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{1+1+2 \cos \theta}$ $=\frac{\mathrm{GMm}}{\mathrm{d}^{2}}(\sqrt{2(1+\cos \theta)})$ $=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{1+\cos \theta}=\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \sqrt{2} \cos \frac{\theta}{2}$ $=2 \frac{\mathrm{GMm}}{\mathrm{d}^{2}} \cos \frac{\theta}{2}$ For, $0 \lt \theta \lt 90^{\circ}$ then, $\therefore \quad \mathrm{F}_{\text {net }}>\sqrt{2} \frac{\mathrm{GMm}}{\mathrm{d}^{2}}$ Thus, the rank of gravitational force from least to greatest, $2 \lt 1 \lt 3 \lt 4$
AMU-2018
Gravitation
138274
The figure shows four arrangements of three particles of equal masses. Arrange them according to the magnitude of the net gravitational force on the particle labeled $\mathbf{m}$, in decreasing order.
1 (i), (iii) = (iv), (ii)
2 (i) = (iii), (ii), (iv)
3 (i), (ii), (iii), (iv)
4 (iv), (iii), (ii), (ii)
Explanation:
A We know that, Gravitational force $(\mathrm{F})=\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}$ From figure (i), $\mathrm{F}_{1}=\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}+\frac{\mathrm{Gmm}}{\mathrm{D}^{2}}=\mathrm{Gm}^{2}\left(\frac{1}{\mathrm{~d}^{2}}+\frac{1}{\mathrm{D}^{2}}\right)$ From figure (ii), $\mathrm{F}_{2}=\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}-\frac{\mathrm{Gmm}}{(\mathrm{D}-\mathrm{d})^{2}}=\mathrm{Gm}^{2}\left(\frac{1}{\mathrm{~d}^{2}}-\frac{1}{\mathrm{D}^{2}}\right)$ From figure (iii), $F_{3}=\sqrt{\left(\frac{G m m}{d^{2}}\right)^{2}+\left(\frac{G m m}{D^{2}}\right)^{2}}=\mathrm{Gm}^{2} \sqrt{\left(\frac{1}{\mathrm{~d}^{2}}\right)^{2}+\left(\frac{1}{(D)^{2}}\right)^{2}}$ From figure (iv), $\mathrm{F}_{4}=\sqrt{\left(\frac{\mathrm{Gmm}}{\mathrm{d}^{2}}\right)^{2}-\left(\frac{\mathrm{Gmm}}{\mathrm{D}^{2}}\right)^{2}}=\mathrm{Gm}^{2} \sqrt{\left(\frac{1}{\mathrm{~d}^{2}}\right)^{2}-\left(\frac{1}{\mathrm{D}^{2}}\right)^{2}}$ Hence, the magnitude of the net gravitational force on the particle labeled $\mathrm{m}$, in decreasing order- $\text { (i), (iii) = (iv), (ii) }$
