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Let the speed of the planet at the perihelion \(P\) in figure shown below be \(v_{_P}\) and the Sun-planet distance \(\mathrm{SP}\) be \(r_{_P}.\) Relation between \((r_{_P},~v_{_P})\) to the corresponding quantities at the aphelion \((r_{_A},~v_{_A})\) is:

1. | \(v_{_P} r_{_P} =v_{_A} r_{_A}\) | 2. | \(v_{_A} r_{_P} =v_{_P} r_{_A}\) |

3. | \(v_{_A} v_{_P} = r_{_A}r_{_P}\) | 4. | none of these |

Subtopic: Kepler's Laws |

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Let the speed of the planet at the perihelion \(P\) in the figure shown below be \(v_p\) and the Sun-planet distance \(SP\) be \(r_p\). Will the planet take equal time to traverse \(BAC\) and \(CPB?\)

1. no

2. yes

3. depends on the mass of the planet

4. we can't say anything

Subtopic: Kepler's Laws |

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Three equal masses of \(m\) kg each are fixed at the vertices of an equilateral triangle \(ABC.\) What is the force acting on a mass \(2m\) placed at the centroid \(G\) of the triangle?

(Take \(AG=BG=CG=1\) m.)

1. \(Gm^2(\hat{i}+\hat{j})\)

2. \(Gm^2(\hat{i}-\hat{j})\)

3. zero

4. \(2Gm^2(\hat{i}+\hat{j})\)

Subtopic: Newton's Law of Gravitation |

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Three equal masses of \(m\) kg each are fixed at the vertices of an equilateral triangle \(ABC.\) What is the force acting on a mass \(2m\) placed at the centroid \(G\) of the triangle if the mass at the vertex \(A\) is doubled?

Take \(AG=BG=CG=1~\text{m}.\)

1. \(Gm^{2} \left(\hat{i} + \hat{j}\right)\)

2. \(Gm^{2} \left(\hat{i} - \hat{j}\right)\)

3. \(0\)

4. \(2Gm^{2} \hat{j}\)$\stackrel{}{}$

Subtopic: Newton's Law of Gravitation |

50%

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The potential energy of a system of four particles placed at the vertices of a square of side \(l\) (as shown in the figure below) and the potential at the centre of the square, respectively, are:

1. \(- 5 . 41 \dfrac{Gm^{2}}{l}\) and \(0\)

2. \(0\) and \(- 5 . 41 \dfrac{Gm^{2}}{l}\)

3. \(- 5 . 41 \dfrac{Gm^{2}}{l}\) and \(- 4 \sqrt{2} \dfrac{Gm}{l}\)

4. \(0\) and \(0\)

Subtopic: Gravitational Potential Energy |

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Two uniform solid spheres of equal radii \({R},\) but mass \({M}\) and \(4M\) have a centre to centre separation \(6R,\) as shown in the figure. The two spheres are held fixed. A projectile of mass \(m\) is projected from the surface of the sphere of mass \(M\) directly towards the centre of the second sphere. The expression for the minimum speed \(v\) of the projectile so that it reaches the surface of the second sphere is:

1. \(\left(\dfrac{3 {GM}}{5 {R}}\right)^{1 / 2}\)2. \(\left(\dfrac{2 {GM}}{5 {R}}\right)^{1 / 2}\)

3. \(\left(\dfrac{3 {GM}}{2 {R}}\right)^{1 / 2}\)

4. \(\left(\dfrac{5 {GM}}{3 {R}}\right)^{1 / 2}\)

Subtopic: Gravitational Potential Energy |

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The planet Mars has two moons, Phobos and Delmos. Phobos has a period of \(7\) hours, \(39\) minutes and an orbital radius of \(9 . 4 \times 10^{3}\) km. The mass of mars is:

1. | \(6 . 48 \times 10^{23} \text{ kg}\) | 2. | \(6 . 48 \times 10^{25} \text{ kg}\) |

3. | \(6 . 48 \times 10^{20} \text{ kg}\) | 4. | \(6 . 48 \times 10^{21} \text{ kg}\) |

Subtopic: Satellite |

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You are given the following data: \(g = 9.81~\text{m/s}^{2}\), \(R_{E} = 6 . 37 \times 10^{6}~\text m\), the distance to the moon, \(R = 3 . 84 \times 10^{8}~\text m\) and the time period of the moon’s revolution is \(27.3\) days. Mass of the Earth \(M_{E}\) in two different ways is:

1. \(5 . 97 \times 10^{24} ~ \text{kg and }6 . 02 \times 10^{24} \text{ kg}\)

2. \(5 . 97 \times 10^{24} \text{ kg and } 6 . 02 \times 10^{23} \text{ kg}\)

3. \(5 . 97 \times 10^{23} ~ \text{kg and }6 . 02 \times 10^{24} \text{ kg}\)

4. \(5 . 97 \times 10^{23} \text{ kg and } 6 . 02 \times 10^{23} \text{ kg}\)

1. \(5 . 97 \times 10^{24} ~ \text{kg and }6 . 02 \times 10^{24} \text{ kg}\)

2. \(5 . 97 \times 10^{24} \text{ kg and } 6 . 02 \times 10^{23} \text{ kg}\)

3. \(5 . 97 \times 10^{23} ~ \text{kg and }6 . 02 \times 10^{24} \text{ kg}\)

4. \(5 . 97 \times 10^{23} \text{ kg and } 6 . 02 \times 10^{23} \text{ kg}\)

Subtopic: Satellite |

55%

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Constant \(k = 10^{- 13} ~ \text s^{2}~ \text m^{- 3}\) in days and kilometres is?

1. \(10^{- 13} ~ \text d^{2} ~\text{km}^{- 3}\)

2. \(1 . 33 \times 10^{14} \text{ dkm}^{- 3}\)

3. \(10^{- 13} ~ \text d^{2} ~\text {km}\)

4. \(1 . 33 \times 10^{- 14} \text{ d}^{2} \text{ km}^{- 3}\)

1. \(10^{- 13} ~ \text d^{2} ~\text{km}^{- 3}\)

2. \(1 . 33 \times 10^{14} \text{ dkm}^{- 3}\)

3. \(10^{- 13} ~ \text d^{2} ~\text {km}\)

4. \(1 . 33 \times 10^{- 14} \text{ d}^{2} \text{ km}^{- 3}\)

Subtopic: Satellite |

56%

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The moon is at a distance of \(3.84\times10^5~\text{km}\) from the earth. Its time period of revolution in days is:

\(\left(\text{Given: }k=\dfrac{4\pi^2}{GM_E}=1.33\times10^{-14}~\text{days}^{2}\text-\text{km}^{-3}\right)\)

1. \(17.3\) days

2. \(33.7\) days

3. \(27.3\) days

4. \(4\) days

\(\left(\text{Given: }k=\dfrac{4\pi^2}{GM_E}=1.33\times10^{-14}~\text{days}^{2}\text-\text{km}^{-3}\right)\)

1. \(17.3\) days

2. \(33.7\) days

3. \(27.3\) days

4. \(4\) days

Subtopic: Satellite |

64%

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