What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 km

 (a)  7.4×10-4 rad/sac             

 (b)  6.4×10-4 rad/sac

 (c)  7.8×10-4 rad/sac               

(d)  8.7×10-4 rad/sac 

Subtopic:  Acceleration due to Gravity |
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If g is the acceleration due to gravity at the earth's surface and r is the radius of the earth, the escape velocity for the body to escape out of the earth's gravitational field is:

(1) gr                                       

(2) 2gr

(3) g/r                                       

(4) r/g

Subtopic:  Escape velocity |
 95%
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The escape velocity of a projectile from the earth is approximately

(1)11.2 m/sec                            (2)112 km/sec

(3)11.2 km/sec                           (4)11200 km/sec

Subtopic:  Escape velocity |
 83%
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The escape velocity of a particle of mass m varies as:

(1) m2                                   

(2) m

(3) m0                                   

(4) m-1

Subtopic:  Escape velocity |
 80%
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Acceleration due to gravity is ‘g’ on the surface of the earth. The value of acceleration due to gravity at a height of 32 km above earth’s surface is (Radius of the earth = 6400 km)

(1) 0.9 g           

(2) 0.99g

(3) 0.8 g           

(4) 1.01 g

Subtopic:  Acceleration due to Gravity |
 80%
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The time period of a simple pendulum on a freely moving artificial satellite is

(1) Zero                           

(2) 2 sec

(3)  3 sec                          

(4)  Infinite

Subtopic:  Satellite |
 74%
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For the moon to cease as the earth's satellite, its orbital velocity has to be increased by a factor of -

1. 2 2. \(\sqrt{2}\)
3. \(1/\sqrt{2}\) 4. 4
Subtopic:  Orbital velocity |
 77%
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The height of a point vertically above the earth’s surface, at which the acceleration due to gravity becomes 1% of its value at the surface is: (Radius of the earth = R)

1. 8R           

2. 9R

3. 10R         

4. 20R

Subtopic:  Acceleration due to Gravity |
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The escape velocity of an object from the earth depends upon the mass of the earth (M), its mean density, its radius (R) and the gravitational constant (G). Thus the formula for escape velocity is:

(1) v=R8π3Gp                                         

(2) v=M8π3GR

(3) v=2GMR                                             

(4) v=2GMR2

Subtopic:  Escape velocity |
 77%
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Escape velocity on a planet is ve. If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes

(1)4ve                                               

(2)2ve

(3)ve                                                 

(4)12ve

Subtopic:  Escape velocity |
 77%
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