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NCERT Exemplar · Class 9 Science Gravitation

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Multiple Choice Questions 1–10 (part 1 of 4)

  1. Exercise 1

    Two objects of different masses falling freely near the surface of moon would
    (a)
    have same velocities at any instant
    (b)
    have different accelerations
    (c)
    experience forces of same magnitude
    (d)
    undergo a change in their inertia

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    NCERT’s answer
    (a)
    (a) Same velocities at any instant.\[a = g_{moon} = \frac{GM_{moon}}{R_{moon}^{2}} \]independent of the falling body's mass, so both objects share the same acceleration.\[v = g_{moon}\,t \]Equal \(\displaystyle a\) and equal \(\displaystyle t\) give equal \(\displaystyle v\).

  2. Exercise 2

    The value of acceleration due to gravity
    (a)
    is same on equator and poles
    (b)
    is least on poles
    (c)
    is least on equator
    (d)
    increases from pole to equator

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    NCERT’s answer
    (c)
    (c) Least on the equator.\[g = \frac{GM_e}{R^{2}} \]Earth bulges at the equator, so \(\displaystyle R_{equator} > R_{pole}\), and \(\displaystyle g\), which falls as \(\displaystyle R\) grows, is smallest there.

  3. Exercise 3

    The gravitational force between two objects is F. If masses of both objects
    are halved without changing distance between them, then the gravitational
    force would become
    (a)
    F/4\displaystyle 4 (b) F/2\displaystyle 2
    (c)
    F (d) 2\displaystyle 2 F

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    NCERT’s answer
    (a)
    (a) \(\displaystyle F/4\).\[F = \frac{G\,m_1 m_2}{d^{2}} \] \[F' = \frac{G\,(m_1/2)(m_2/2)}{d^{2}} = \frac{F}{4} \]Halving each mass halves each factor in the product, so the force falls to one-quarter.

  4. Exercise 4

    A boy is whirling a stone tied with a string in an horizontal circular path. If
    the string breaks, the stone
    (a)
    will continue to move in the circular path
    (b)
    will move along a straight line towards the centre of the circular path
    (c)
    will move along a straight line tangential to the circular path
    (d)
    will move along a straight line perpendicular to the circular path away
    from the boy

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    NCERT’s answer
    (c)
    (c) Along a straight line tangential to the circular path.\[T=0 \;\Rightarrow\; F_{\text{horizontal}}=0 \]Weight acts vertically only, so by Newton's first law the horizontal velocity stays constant — tangent to the circle at the break point.NCERT_Solution_Class9_Science_Exemplar_Ch10_Q4

  5. Exercise 5

    An object is put one by one in three liquids having different densities. The object floats with 19\displaystyle \frac{1}{9}, 211\displaystyle \frac{2}{11} and 37\displaystyle \frac{3}{7} parts of their volumes outside the liquid surface in liquids of densities d1\displaystyle d_{1}, d2\displaystyle d_{2} and d3\displaystyle d_{3} respectively. Which of the following statement is correct?
    (a)
    d1\displaystyle d_{1}> d2\displaystyle d_{2}> d3\displaystyle d_{3}
    (b)
    d1\displaystyle d_{1}> d2\displaystyle d_{2}< d3\displaystyle d_{3}
    (c)
    d1\displaystyle d_{1}< d2\displaystyle d_{2}> d3\displaystyle d_{3}
    (d)
    d1\displaystyle d_{1}< d2\displaystyle d_{2}< d3\displaystyle d_{3}

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    NCERT’s answer
    (d)
    (d) \(\displaystyle d_1<d_2<d_3\).For a floating object of density \(\displaystyle \rho\) and volume \(\displaystyle V\), weight equals upthrust: \[\rho V g = d_i\,(1-f_i)\,V g \ \Rightarrow\ d_i=\frac{\rho}{1-f_i} \]The fractions outside are \(\displaystyle f_1=\tfrac{1}{9}\), \(\displaystyle f_2=\tfrac{2}{11}\), \(\displaystyle f_3=\tfrac{3}{7}\), so the submerged fractions are \(\displaystyle \tfrac{8}{9}\), \(\displaystyle \tfrac{9}{11}\), \(\displaystyle \tfrac{4}{7}\), giving \[d_1=\tfrac{9}{8}\rho,\qquad d_2=\tfrac{11}{9}\rho,\qquad d_3=\tfrac{7}{4}\rho \]Answer: (d) \(\displaystyle d_1<d_2<d_3\).

  6. Exercise 6

    In the relation F = G M m/d2\displaystyle d^{2}, the quantity G
    (a)
    depends on the value of g at the place of observation
    (b)
    is used only when the earth is one of the two masses
    (c)
    is greatest at the surface of the earth
    (d)
    is universal constant of nature

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    NCERT’s answer
    (d)
    (d) Universal constant of nature.\[F = \frac{G\,Mm}{d^{2}} \]\(\displaystyle G\) is the same for every pair of masses, everywhere, independent of \(\displaystyle g\), of whether earth is involved, and of location.

  7. Exercise 7

    Law of gravitation gives the gravitational force between
    (a)
    the earth and a point mass only
    (b)
    the earth and Sun only
    (c)
    any two bodies having some mass
    (d)
    two charged bodies only

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    NCERT’s answer
    (c)
    (c) Any two bodies having some mass.\[F = \frac{G\,m_1 m_2}{d^{2}} \]Newton's law applies to every pair of masses, not only the earth, and not to charge, which follows Coulomb's law instead.

  8. Exercise 8

    The value of quantity G in the law of gravitation
    (a)
    depends on mass of earth only
    (b)
    depends on radius of earth only
    (c)
    depends on both mass and radius of earth
    (d)
    is independent of mass and radius of the earth

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    NCERT’s answer
    (d)
    (d) Independent of the mass and radius of the earth.\[F = \frac{G\,Mm}{d^{2}} \]\(\displaystyle G\) is a fixed constant of nature; only \(\displaystyle M\) and \(\displaystyle d\) depend on the earth, and \(\displaystyle G\) itself does not change if either does.

  9. Exercise 9

    Two particles are placed at some distance. If the mass of each of the two
    particles is doubled, keeping the distance between them unchanged, the
    value of gravitational force between them will be
    (a)
    4\displaystyle 4 times
    (b)
    4\displaystyle 4 times
    (c)
    2\displaystyle 2 times
    (d)
    unchanged

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    NCERT’s answer
    (b)
    (b) $\displaystyle 4$ times.\[F = \frac{G\,m_1 m_2}{d^{2}} \] \[F' = \frac{G\,(2m_1)(2m_2)}{d^{2}} = 4F \]Doubling both masses multiplies the product \(\displaystyle m_1 m_2\) by $\displaystyle 4$, at the same \(\displaystyle d\).

  10. Exercise 10

    The atmosphere is held to the earth by
    (a)
    gravity
    (b)
    wind
    (c)
    clouds
    (d)
    earth’s magnetic field

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    NCERT’s answer
    (a)
    (a) Gravity.\[F_g=\frac{GM_em_{air}}{R_e^{2}} \]Earth's gravitational pull acts on every air molecule, holding the gaseous envelope around the planet instead of letting it escape into space.Answer: (a) Gravity.