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Engineering
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Question

From a sphere of mass M and radius R, a  smaller sphere of radius  is carved out such that the cavity made in the original sphere is between its centre and the periphery. (See figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is 3R, the gravitational force between the two spheres is :

Question

From a sphere of mass M and radius R, a  smaller sphere of radius  is carved out such that the cavity made in the original sphere is between its centre and the periphery. (See figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is 3R, the gravitational force between the two spheres is :

Solution : Volume of removed sphere, Volume of the sphere (remaining) Therefore mass of sphere carved and remaining sphere are at respectively 1/8M and 7/8M.Therefore, gravitational force between these two sphere,
Engineering
204 Views   |

When body is raised to a height equal to radius of earth, the P.E. change is

• Option 1)

$MgR$

• Option 2)

$\frac{MgR}{2}$

• Option 3)

$2 MgR$

• Option 4)

none of these

As we learnt in  Gravitational Potential energy at a point -  gravitational potential energy   Mass of source body  mass of test body  distance between two - wherein Always negative in the gravitational field because Force is attractive in nature.    Potential energy =   at surface of earth when it is raised at h-R, Option 1) This is incorrect option Option 2) This is correct...
Engineering
797 Views   |

What would be the duration of the year if the distance between the earth and the sun gets doubled?

• Option 1)

1032 days

• Option 2)

129 days

• Option 3)

365 days

• Option 4)

730 days

As we learnt in

Kepler's 3rd law -

$T^{2}\: \alpha\: a^{3}$

From fig.

$AB=AF+FB$

$2a=r_{1}+r_{2}$

$\therefore\; a=\frac{r_{1}+r_{1}}{2}$

$a=$ semi major Axis

$r_{1}=$ Perigee

- wherein

Known as law of periods

$r_{2}=$ apogee

$T^{2}\: \alpha \: \left ( \frac{r_{1}+r_{2}}{2} \right )^{3}$

${r_{1}+r_{2}= 2a$

$T^{2}\propto R^{3}$

If r get double  $\Rightarrow }R_{2}=2R_{1}$

$\frac{T_{2}}{T_{1}}=\left ( \frac{R_{2}}{R_{1}} \right )^{\frac{3}{2}}=2^{\frac{3}{2}}$

$\Rightarrow T_{2}=\left ( 2\sqrt{2} \right ) T_{1}=\left ( 2\sqrt{2} \right ) 365\ days = 1032\ days$

Option 1)

1032 days

This is correct option

Option 2)

129 days

This is incorrect option

Option 3)

365 days

This is incorrect option

Option 4)

730 days

This is incorrect option

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Engineering
593 Views   |

Two planets of radii r1 and r2 are made from the same material. The ratio of the acceleration due to gravity g1/g2 at the surface of the two planets is

• Option 1)

$\frac{r1}{r2}$

• Option 2)

$\frac{r2}{r1}$

• Option 3)

$\left ( \frac{r1}{r2} \right )^{2}$

• Option 4)

$\left ( \frac{r2}{r1} \right )^{2}$

As we learnt in Acceleration due to gravity (g) - Force extended by earth on a body is gravity. Formula:    gravity density of earth Radius of earth   - wherein It's average value is on the surface of earth  Since both are made of same material hence their densities are equal Option 1) This is correct option Option 2) This is incorrect option Option 3) This is incorrect...
Engineering
120 Views   |

The radii of the earth and the moon are in the ratio 10 : 1 while acceleration due to gravity on the earth’s surface and moon’s surface are in the ratio 6 : 1. The ratio of escape velocities from earth’s surface to that of moon surface is

• Option 1)

$10 : 1$

• Option 2)

6 : 1

• Option 3)

1.66 : 1

• Option 4)

7.74 : 1

As we learnt in  Acceleration due to gravity (g) - Force extended by earth on a body is gravity. Formula:    gravity density of earth Radius of earth   - wherein It's average value is on the surface of earth    Escape velocity = Option 1) This is incorrect option Option 2) 6 : 1 This is incorrect option Option 3) 1.66 : 1 This is incorrect option Option 4) 7.74 : 1 This is correct option
Engineering
545 Views   |

The masses of two planets are in the ratio 1 : 2. Their radii are in the ratio 1 : 2. The acceleration due to gravity on the planets are in the ratio.

• Option 1)

1 : 2

• Option 2)

2 : 1

• Option 3)

3 : 5

• Option 4)

5 : 3

As we learnt in  Acceleration due to gravity (g) - Force extended by earth on a body is gravity. Formula:    gravity density of earth Radius of earth   - wherein It's average value is on the surface of earth       Option 1) 1 : 2 This is correct option Option 2) 2 : 1 This is incorrect option Option 3) 3 : 5 This is incorrect option Option 4) 5 : 3 This is incorrect option
Engineering
1321 Views   |

The largest and the shortest distance of the earth from are r1 and r2. It’s distance from the sun when it is perpendicular to the major-axis of the orbit drawn from the sun.

• Option 1)

$\left ( \frac{r1+r2 }{4}\right )$

• Option 2)

$\left ( \frac{r1+r2 }{r1-r2}\right )$

• Option 3)

$\left ( \frac{2r1r2}{r1+r2} \right )$

• Option 4)

$\left ( \frac{r1+r2}{3} \right )$

As we learnt in  Velocity of planet in terms of Eccentricity -  Velocity of planet at apogee - wherein Eccentricity (e) =    The position of a particle moving in an elliptical orbit is represented as is perpendicular distance of particle from focus and e is eccentricity of ellipse Option 1) This is incorrect option Option 2) This is incorrect option Option 3) This is correct...
Engineering
343 Views   |

The figure shows the motion of a planet around the sun in an elliptical orbit with sun at the focus. The shaded areas A and B are also shown in the figure which can be assumed to be equal. If t1 and t2 represent the time for the planet to move from a to b and d to c respectively, then

• Option 1)

t1 > t2

• Option 2)

t1 < t2

• Option 3)

t1 = t2

• Option 4)

none of these

As we learnt in  Kepler's 2nd law - Area of velocity =   Areal velocity   small area traced - wherein Simiar to Law of conservation of momentum  Angular momentum Known  as Law of Area   Areal velocity of planet in elliptical path is constant. This implies equal area will be covered in equal time. Option 1) t1 > t2 This is incorrect option Option 2) t1 < t2 This is incorrect option Option...
Engineering
106 Views   |

The escape velocity from the earth’s surface is 11 km/sec. A certain planet has a radius twice that of the earth but its mean density is the same as that of the earth. The value of the escape velocity from this planet would be

• Option 1)

22 km/sec

• Option 2)

11 km/sec

• Option 3)

5.5 km/sec

• Option 4)

16.5 km/sec

As we learnt in  Escape velocity ( in terms of radius of planet) -  Escape velocity Radius of earth - wherein depends on the reference body greater the value of  or  greater will be the escape velocity   For earth       Option 1) 22 km/sec This is correct option Option 2) 11 km/sec This is incorrect option Option 3) 5.5 km/sec This is incorrect option Option 4) 16.5 km/sec This is...
Engineering
295 Views   |

The escape velocity from earth is 11.2 km per sec. If a body is to be projected in a direction making an angle 45° to the vertical, then the escape velocity is

• Option 1)

11.2 × 2 km/sec

• Option 2)

11.2 km/sec

• Option 3)

$11.2\times \frac{1}{\sqrt{2}}\, \, \, \, \,$  Km/sec

• Option 4)

$11.2\times \sqrt{2} \, \, \, Km/sec$

As we learnt in  Escape velocity - Minimum velocity which is required to escape the body from earth's gravitational pull - wherein It is independent of the mass and direction of projection of body.    Escape velocity is independent of direction of projection. Hence it will remain same for all direction. Option 1) 11.2 × 2 km/sec This is incorrect option Option 2) 11.2 km/sec This is correct...
Engineering
142 Views   |

The distance of two planets from sun are nearly 1014 and 1012 meters. Assuming that they move in circular orbits, their periodic times will be in the ratio

• Option 1)

100

• Option 2)

10

• Option 3)

1000

• Option 4)

$100\sqrt{10}$

As we learnt in  Kepler's 3rd law - From fig.  semi major Axis  Perigee - wherein Known as law of periods  apogee         or      Option 1) 100 This is incorrect option Option 2) 10 This is incorrect option Option 3) 1000 This is correct option Option 4) This is incorrect option
Engineering
486 Views   |

The distance between centre of the earth and moon is 384000 km. If the mass of the earth is 6 × 1024 kg and G = 6.66 × 10–11 Nm2/kg2. The speed of the moon is nearly

• Option 1)

1 km/sec

• Option 2)

4 km/sec

• Option 3)

8 km/sec

• Option 4)

11.2 km/sec

As we learnt in  Orbital velocity of satellite -  Position of satellite from the centre of earth  Orbital velocity - wherein The velocity required to put the satellite into its orbit around the earth.     Option 1) 1 km/sec This is correct option Option 2) 4 km/sec This is incorrect option Option 3) 8 km/sec This is incorrect option Option 4) 11.2 km/sec This is incorrect option
Engineering
121 Views   |

If the radius of the earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth’s surface would

• Option 1)

decrease

• Option 2)

remains unchanged

• Option 3)

increase

• Option 4)

none of these

As we learnt in  Percentage decrease in value of 'g' - Variation in 'g' Radius   - wherein rate of decrease of gravity outside the earth (h<<R) is double of that of inside the earth.   If       g will increase by 2% . Option 1) decrease This is incorrect option Option 2) remains unchanged This is incorrect option Option 3) increase This is correct option Option 4) none of these This is...
Engineering
401 Views   |

If the radius of earth's orbit is made 1/16, the duration of an year will become

• Option 1)

16 times

• Option 2)

1/64 times

• Option 3)

1/16 times

• Option 4)

1/8 times

As we learnt in  Kepler's 3rd law - From figure  semi major Axis  Perigee - wherein Known as law of periods  apogee       Option 1) 16 times This is incorrect option Option 2) 1/64 times This is correct option Option 3) 1/16 times This is incorrect option Option 4) 1/8 times This is incorrect option
Engineering
117 Views   |

If the acceleration due to gravity of a planet is half the acceleration due to gravity of earth’s surface and radius of planet is half the radius of the earth, the mass of planet in terms of mass of earth is

• Option 1)

$\frac{Me}{2}$

• Option 2)

$\frac{Me}{4}$

• Option 3)

$\frac{Me}{6}$

• Option 4)

$\frac{Me}{8}$

As we learnt in  Acceleration due to gravity (g) - Force extended by earth on a body is gravity. Formula:    gravity density of earth Radius of earth   - wherein It's average value is on the surface of earth        and         Option 1) This is incorrect option Option 2) This is incorrect option Option 3) This is incorrect option Option 4) This is correct option
Engineering
138 Views   |

If g is the acceleration due to gravity of the earth’s surface the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is

• Option 1)

$\frac{1}{2} mgR$

• Option 2)

$2 mgR$

• Option 3)

$mgR$

• Option 4)

$\frac{1}{4} mgR$

As we learnt in  Gravitational Potential energy at a point -  gravitational potential energy   Mass of source body  mass of test body  distance between two - wherein Always negative in the gravitational field because Force is attractive in nature.    Potential energy at earth's surface Potential energy at height gain in potential energy   Option 1) This is correct option Option...
Engineering
149 Views   |

If escape velocity from the earth’s surface is 11.2 km/sec. then escape velocity from a planet of mass same as that of earth but radius one fourth as that of earth is

• Option 1)

11.2 km/sec

• Option 2)

22.4 km/sec

• Option 3)

5.65 km/sec

• Option 4)

44.8 km/sec

As we learnt in  Escape velocity ( in terms of radius of planet) -  Escape velocity Radius of earth - wherein depends on the reference body greater the value of  or  greater will be the escape velocity   For earth     Escape velocity   Option 1) 11.2 km/sec This is incorrect option Option 2) 22.4 km/sec This is correct option Option 3) 5.65 km/sec This is incorrect option Option 4) 44.8...
Engineering
283 Views   |

An earth’s satellite of mass m revolves in a circular orbit at a height h from the surface g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by

• Option 1)

$\frac{gR^{2}}{R+h}$

• Option 2)

$gR$

• Option 3)

$\frac{gR}{R+h}$

• Option 4)

$\sqrt{\left (\frac{ gR^{2}}{R+h} \right )}$

As we learnt in  Variation in 'g' with height - gravity at height from surface of earth. Radius of earth height above surface   - wherein   At height h, effect of gravity is  Option 1) This is incorrect option Option 2) This is incorrect option Option 3) This is incorrect option Option 4) This is correct option
Engineering
362 Views   |

A spherical planet far out in space has a mass M0 and diameter D0. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to

• Option 1)

$\frac{GM_{0}}{D^{2}_{0}}$

• Option 2)

$\frac{4mGM_{0}}{D^{2}_{0}}$

• Option 3)

$\frac{4GM_{0}}{D^{2}_{0}}$

• Option 4)

$\frac{GmM_{0}}{D^{2}_{0}}$

As we learnt in  Acceleration due to gravity (g) - Force extended by earth on a body is gravity. Formula:    gravity density of earth Radius of earth   - wherein It's average value is on the surface of earth    Acceleration due to gravity   Option 1) This is incorrect option Option 2) This is incorrect option Option 3) This is correct option Option 4) This is incorrect option
Engineering
692 Views   |

A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is R0 and mass of the earth M, the angular momentum about the centre of the earth is

• Option 1)

$M\sqrt{\left ( GmRo \right )}$

• Option 2)

$M\sqrt{\left ( \frac{Gm}{Ro }\right )}$

• Option 3)

$m\sqrt{\left ( \frac{GM}{Ro }\right )}$

• Option 4)

$m\sqrt{\left ( GMRo\right )}$

As we learnt in  Angular momentum of satellite -  Angular momentum  mass of satellite - wherein depends on both the masses , mass of centre of body and mass of planet as well as radius of earth.     Centripetal force = Angular momentum =   Option 1) This is incorrect option Option 2) This is incorrect option Option 3) This is incorrect option Option 4) This is correct option
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