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 Multiple Choice QuestionsMultiple Choice Questions

1.

The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. The value of 'g'and 'R'(radius of earth) are 10 m/s2 and 6400km respectively. The required energy for this work will be;

  • 6.4 x1011 J

  • 6.4 x108 J

  • 6.4 x109 J

  • 6.4 x109 J


D.

6.4 x109 J

Potential energy on the earth's surface is -mgR while in free space, it is zero. So, to free the spaceship minimum required energy is
E =mgR
 = 103 x 10 x 6400 x 103 J
 = 6.4 x 1010 J

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2.

A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R ; h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to: (Neglect the effect of atmosphere.)

  • square root of 2 gR end root
  • square root of gR
  • square root of gR divided by 2 end root
  • square root of gR space left parenthesis square root of 2 space minus 1 right parenthesis

D.

square root of gR space left parenthesis square root of 2 space minus 1 right parenthesis

Given, a satellite is revolving in a circular orbit at a height h from the Earth's surface having radius of earth R, i.e h <<R

Orbit velocity of a satellite

straight v equals space square root of fraction numerator GM over denominator straight R plus straight h end fraction end root space equals space square root of GM over straight R end root space left parenthesis as space straight h less than less than straight R right parenthesis
Velocity space required space to space escape
1 half space mv squared space equals space fraction numerator GMn over denominator straight R plus straight h end fraction

straight v apostrophe space equals space square root of fraction numerator 2 GM over denominator straight R plus straight h end fraction end root space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root space left parenthesis straight h less than less than straight R right parenthesis

therefore, the minimum increase in its orbital velocity required to escape from the Earth's Gravitational Field.

straight v apostrophe minus straight v space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root space minus square root of GM over straight R end root
space equals space square root of 2 gR end root minus square root of gR space equals space square root of gR space left parenthesis square root of 2 minus 1 right parenthesis

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3.

From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as shown in the figure. Taking gravitational potential V = 0 at r = , the potential at the centre of the cavity thus formed is (G = gravitational constant)

  • -GM/2R

  • -GM/R

  • -2GM/3R

  • -2GM/R


B.

-GM/R

Consider cavity as negative mass and apply superposition of gravitational potential.
Consider the cavity formed in a solid sphere as shown in figure
V (∞) = 0


According to the question, we can write potential at an internal point P due to the complete solid sphere.
straight V subscript straight s space equals space fraction numerator GM over denominator 2 straight R cubed end fraction open square brackets 3 straight R squared minus open parentheses straight R over 2 close parentheses squared close square brackets
equals negative fraction numerator GM over denominator 2 straight R cubed end fraction open square brackets 3 straight R squared minus straight R squared over 4 close square brackets
equals negative fraction numerator GM over denominator 2 straight R cubed end fraction open square brackets fraction numerator 11 straight R squared over denominator 4 end fraction close square brackets space equals space fraction numerator negative 11 GM over denominator 8 straight R end fraction
Mass of removed part,
=fraction numerator straight M over denominator begin display style 4 over 3 xπR cubed end style end fraction space straight x 4 over 3 straight pi open parentheses straight R over 2 close parentheses cubed space equals space straight M over 8
Potential at point P due to removed part

straight V subscript straight c space equals space fraction numerator negative 3 over denominator 2 end fraction space straight x space fraction numerator GM divided by 8 over denominator straight R end fraction space equals space fraction numerator negative 11 GM over denominator 8 straight R end fraction space minus open parentheses fraction numerator negative 3 GM over denominator 8 straight R end fraction close parentheses
Thus, potential due to remaining part at point P,

straight V subscript straight p space equals space straight V subscript straight s minus straight V subscript straight c space equals space fraction numerator negative 11 space GM over denominator 8 straight R end fraction space minus open parentheses negative fraction numerator 3 GM over denominator 8 straight R end fraction close parentheses
fraction numerator left parenthesis negative 11 space plus 3 right parenthesis GM over denominator 8 straight R end fraction space equals space minus GM over straight R

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4.

The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is

  • 2R

  • fraction numerator straight R over denominator square root of 2 end fraction
  • R/2

  • square root of 2 straight R end root

A.

2R

fraction numerator GM over denominator 9 straight R squared end fraction space equals space fraction numerator GM over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction
rightwards double arrow space 3 straight R space equals space straight R plus straight h
rightwards double arrow space straight h equals space 2 straight R
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5.

From the tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H,u and n is

  • 2gH = n2u2

  • gH = (n-2)2u2

  • 2gH = nu2(n-2)2u2

  • gH = (n-2)2u2


C.

2gH = nu2(n-2)2u2

Time is taken to reach the maximum height t1 = u/g

If t2 is the time taken to hit the ground,
i.e, negative straight H space equals space ut subscript 2 minus 1 half gt subscript 2 superscript 2
straight t subscript 2 space equals space nt subscript 1
minus straight H space equals space straight u nu over straight g minus 1 half straight g fraction numerator straight n squared straight u squared over denominator straight g squared end fraction
minus straight H space equals space nu squared over straight g minus 1 half fraction numerator straight n squared straight u squared over denominator straight g end fraction
2 gH space equals nu squared left parenthesis straight n minus 2 right parenthesis

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6.

Two stones are thrown up simultaneously from the edge of a cliff 240 m high with an initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2 )


C.

The concept of relative motion can be applied to predict the nature of motion of one particle with respect to the other.

Consider the stones thrown up simultaneously as shown in the diagram below.

Considering the motion of the second particle with respect to the first we have relative acceleration
|a21| = |a2-a1| = g-g = 0



Thus, motion of the first particle is straight line with respect to the second particle till the first particle strikes ground at a time given by
negative 240 space equals space 10 straight t space minus space 1 half space straight x space 10 space straight x space straight t squared
or space straight t squared minus 2 straight t minus 48 space equals space 0
or space straight t squared space minus 8 straight t space plus space 6 straight t space minus 48 space equals space 0
or space space straight t squared space equals space 8 comma negative 6 space left parenthesis not space possible right parenthesis
Thus, distance covered by the second particle with respect to the first particle in 8s is
S12 = (v21) t = (40-10)(8s)
 = 30 x 8 = 240m
Similarly, time taken by the second particle to strike the ground is given by
negative 240 space equals space 40 straight t space minus space 1 half space straight x space 10 space straight x space straight t squared
minus 240 space equals space 40 space straight t space minus 5 straight t squared
5 straight t squared minus 40 straight t minus 240 space equals 0
straight t squared minus 8 straight t minus 48 space equals space 0
straight t squared minus 12 straight t space plus space 4 straight t minus 48 space equals 0
straight t left parenthesis straight t minus 12 right parenthesis plus 4 left parenthesis straight t minus 12 right parenthesis space equals 0
straight t space equals space 12 comma negative 4 space left parenthesis not space possible right parenthesis
Thus, after the 8s magnitude of relative velocity will increases up to 12 s when the second particle strikes the ground.

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7.

A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take g = 10 ms−2 )

  • 2 straight pi square root of 2 straight s
  • 2s

  • 2 square root of 2 straight s
  • square root of 2 straight s

C.

2 square root of 2 straight s

A uniform string of length 20 m is suspended from a rigid support. Such that the time taken to reach the support.

straight T equals mgx over straight l
So comma space velocity space at space point space straight P space equals space square root of fraction numerator begin display style mgx over straight l end style over denominator straight m divided by straight l end fraction end root
straight v equals space square root of gx
dx over dt space equals space square root of gx
integral subscript 0 superscript 20 fraction numerator dxt over denominator square root of straight x end fraction space equals space integral subscript 0 superscript straight t square root of straight g space dt
left square bracket 2 square root of straight x right square bracket subscript 0 superscript 20 space equals space square root of 10 straight t
square root of 20 space equals space square root of 10 straight t
straight t space equals space 2 square root of 2 space straight s 

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8.

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

  • 5GmM/6R

  • 2GmM/3R

  • GmM/2R

  • GmM/3R


A.

5GmM/6R

From conservation of energy,
Total energy at the planet = Total energy at the altitude
negative GMm over straight R space plus space left parenthesis KE right parenthesis subscript surface space equals space minus fraction numerator GMm over denominator 3 straight R end fraction space plus space 1 half mv subscript straight A superscript 2 space... space left parenthesis straight i right parenthesis
In its orbit the necessary centripetal force provided by gravitational force.
∴ therefore space fraction numerator mv subscript straight A superscript 2 over denominator left parenthesis straight R space plus 2 straight R right parenthesis end fraction space equals space fraction numerator GMm over denominator left parenthesis straight R space plus 2 straight R right parenthesis squared end fraction
straight v subscript straight A superscript 2 space equals space fraction numerator GM over denominator 3 straight R end fraction space... space left parenthesis ii right parenthesis
From space eq space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space get
left parenthesis KE right parenthesis subscript surface space equals space 5 over 6 GMm over straight R

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9.

Two bodies of masses m and 4 m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

  • -4Gm/r

  • -6Gm/r

  • -9Gm/r

  • zero


C.

-9Gm/r

GM over straight x squared space equals space fraction numerator straight G left parenthesis 4 straight m right parenthesis over denominator left parenthesis straight r minus straight x squared right parenthesis end fraction
1 over straight x space equals space fraction numerator 2 over denominator straight r minus straight x end fraction
straight r minus straight x space equals space 2 straight x
3 straight x space equals space straight r over 3


= -3Gm/r - 6Gm/r = -9Gm/r


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10.

The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius)


B.

straight g space equals space GMx over straight R cubed space inside space the space Earth space
straight g equals space GM over straight r squared space outside space the space Earth
Where M is mass of earth



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