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Gravitation Test - 44

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Gravitation Test - 44
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  • Question 1
    1 / -0
    S.I. Unit of universal gravitational constant GG is-
    Solution
    Gravitational force between two objects of mass m1m_1 and m2m_2 separated by a distance rr, F=Gm1m2r2F = \dfrac{Gm_1 m_2}{r^2}
        \implies G=Fr2m1m2G = \dfrac{F r^2 }{m_1 m_2}
    SI unit of universal gravitational constant GG is Nm2Kg2\dfrac{Nm^2}{Kg^2}.
  • Question 2
    1 / -0
    If the radius of the earth shrinks by 1.51.5% (mass remaining same), then the value of acceleration due to gravity changes by :
    Solution
    Acceleration due to gravity on the earth's surface is g=GMR2g=\dfrac{GM}{R^2}
    When the earth shrinks 1.51.5 %, the radius becomes R=RR(1.5/100)=(9.85/100)RR'=R-R(1.5/100)=(9.85/100)R or R/R=98.5/100R'/R=98.5/100
    After shrinks, the acceleration due to gravity becomes, g=GMR2g'=\dfrac{GM}{R'^2}
    So, gg=(R/R)2=(100/98.5)2=1.03\dfrac{g'}{g}=(R/R')^2=(100/98.5)^2=1.03
    Thus, percentage change in acceleration =ggg×100=[g/g1]100=[1.031]100=3=\dfrac{g'-g}{g}\times 100=[g'/g -1]100=[1.03-1]100=3 %

  • Question 3
    1 / -0
    Variation of acceleration due to gravity (g)\left(g\right) with distance xx from the centre of the earth is best represented by : (RR \rightarrow Radius of the earth)
    Solution
    Acceleration due to gravity at a depth dd from the earth surface     g=go(1dR)g = g_o\bigg(1-\dfrac{d}{R}\bigg)
    Thus distance from earth's centre x=Rdx = R-d  i.e.  d=Rxd = R-x
        \implies    g=goxRg = g_o\dfrac{x}{R}
    Thus acceleration due to gravity increases linearly with the increase in distance from centre of earth.
    Acceleration due to gravity at a height  hh from the earth surface      g=go(RR+h)2g = g_o\bigg(\dfrac{R}{R+h} \bigg)^2
    Thus distance from earth's centre    x=R+hx = R+h    i.e.  h=xRh = x-R
        \implies      g=go(Rx)2g = g_o\bigg(\dfrac{R}{x} \bigg)^2
    Thus acceleration due to gravity decreases as 1x2\dfrac{1}{x^2} with the increase in distance from centre of earth.
    Hence, option D is correct.
  • Question 4
    1 / -0
    At the centre of the earth acceleration due to gravity is:
    Solution
    gravitational  field gg at distance rr where r<Rr<R( radius of earth) is
    g=Gρr3g=\frac{G\rho r}{3}                      (ρ\rho is density of earth)
    So at centre r=0r=0
    hence g=0g=0
    So at the centre of earth acceleration due to gravity is zero 
  • Question 5
    1 / -0
    Dimensional formula of universal gravitational constant GG is-
    Solution

  • Question 6
    1 / -0
    The minimum speed of a particle projected from earth's surface so that it will never return is/are:
    Solution
    An object can be thrown up with a certain minimum initial velocity so that, the object goes beyond the earth's gravitational field and escape from earth, this velocity escape velocity of the earth.
    potential energy at surface is GMmR-\frac{GMm}{R}, so kinetic energy that has to give to reach particle to infinite ( zero energy)  is equal to potential energy.
    12mve2=GMmR=2GMR\frac{1}{2}mv_e^2=\frac{GMm}{R}=\sqrt{\frac{2GM}{R}}.
    so none ofthe above give this expression so the answer is D.
  • Question 7
    1 / -0
    A particle hanging from a massless spring stretches it by 2 cm2\ cm at earth's surface. How much will the same particle stretch the spring at height 2624 km2624\ km from the surface of earth? ( Radius of earth =6400 km=6400\ km)
    Solution
    x=2cm=0.02mx=2cm=0.02m
    h=2624kmh=2624km
    R=6400kmR=6400km
    \therefore x x=g g=64002(6400+2624) 20.5x =0.5×x=0.5×2cm=1cm\cfrac { { x }^{ \prime  } }{ x } =\cfrac { { g }^{ \prime  } }{ g } =\cfrac { { 6400 }^{ 2 } }{ { \left( 6400+2624 \right)  }^{ 2 } } \approx 0.5\\ { x }^{ \prime  }=0.5\times x=0.5\times 2cm=1cm

  • Question 8
    1 / -0
    The SI unit of gravitational potential is
    Solution
    It is a very simple idea. Gravitational potential is the potential energy per kilogram at a point in a field. So the units are joule per kilogram.
    so the best possible answer is option B.

  • Question 9
    1 / -0
    How the gravitational constant will change if a brass plate is introduced between two bodies?
    Solution
    Universal Gravitational Constant (G) remains constant irrespective of the medium between two bodies. It has a constant value of 6.67 × 1011Nm2kg26.67  \times  10^{-11} Nm^{2}kg^{-2}
  • Question 10
    1 / -0
    The minimum energy required to launch a mm kg satellite from earth's surface in a circular orbit at an altitude of 2R2R where RR is the radius of earth, will be:
    Solution

    The kinetic energy at altitude 2R2R is  =GMm6R= \dfrac{G M m} {6 R}
    The gravitational potential energy at altitude 2R2R is =GMm3R=\dfrac{– G M m}{ 3 R}
    Total energy == Kinetic energy ++ potential energy
    =GMm6R+GMm3R= \dfrac{G M m} {6 R}+\dfrac{– G M m}{ 3 R}
                         
    Potential energy at the surfce is  =GMmR= \dfrac{G M m} { R}

    Therefore, required kinetic energy =GMm6R+ GMmR=5GMm6R= -\dfrac{G M m} {6 R}+  \dfrac{G M m} { R}= \dfrac{5G M m} { 6R}
    hence the best possible answer is option B.
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