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Question

Asked in: BITSAT-2018

 From a sphere of mass M and radius R, a  smaller sphere of radius \frac{R}{2} 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

Asked in: BITSAT-2018

 From a sphere of mass M and radius R, a  smaller sphere of radius \frac{R}{2} 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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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...
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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

 
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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
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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 )

 
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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...
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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

 
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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

 
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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
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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

 
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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

 
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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

 
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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}

 
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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

 
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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

 
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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 )}

 
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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}}

 
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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 )}

 
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