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NCERT Exemplar · Class 9 Science Gravitation

26 questions · 26 still being checked

Long Answer Questions 23–26 (part 4 of 4)

  1. Exercise 23

    How does the weight of an object vary with respect to mass and radius of the earth. In a hypothetical case, if the diameter of the earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the earth be affected?

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    NCERT’s answer
    Weight of an object is directly proportional to the mass of the earth and inversely proportional to the square of the radius of the earth. i.e., Weight of a body ∝ M \(\displaystyle R^{2}\) Original weight \(\displaystyle W_{o}\) = m g = m G M \(\displaystyle R^{2}\) When hypothetically M becomes $\displaystyle 4$ M and R becomes R then weight becomes W n= m G M R ( )$\displaystyle 2$ = ($\displaystyle 16$ m G) M R2 = $\displaystyle 16$ × \(\displaystyle W^{o}\) The weight will become $\displaystyle 16$ times.
    Newton's law of gravitation gives weight as the earth's pull on a body: \[F=\frac{GMm}{R^{2}}=mg \] \[g=\frac{GM}{R^{2}} \] so weight \(\displaystyle W=mg\) grows with the earth's mass \(\displaystyle M\) and shrinks with \(\displaystyle R^{2}\). For \(\displaystyle M'=4M\), \(\displaystyle R'=R/2\): \[g'=\frac{G(4M)}{(R/2)^{2}}=\frac{4GM}{R^{2}/4}=16\frac{GM}{R^{2}}=16g \] \[W'=mg'=16\,mg=16\,W \] Answer: weight \(\displaystyle =GMm/R^{2}\); with mass $\displaystyle 4$ times and diameter half, weight becomes $\displaystyle 16$ times the present value.

  2. Exercise 24

    How does the force of attraction between the two bodies depend upon their masses and distance between them? A student thought that two bricks tied together would fall faster than a single one under the action of gravity. Do you agree with his hypothesis or not? Comment.

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    NCERT’s answer
    F ∝ \(\displaystyle m_{1}\)\(\displaystyle m_{2}\) and F ∝ \(\displaystyle d_{2}\) This hypothesis is not correct. The two bricks, like a single body, fall with the same speed to reach the ground at the same time in case of free-fall. This is because acceleration due to gravity is independent of the mass of the falling body.
    Newton's law of gravitation: \[F=\frac{Gm_{1}m_{2}}{r^{2}} \] force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between their centres. Under gravity alone the acceleration of a falling body is \[a=\frac{F}{m}=\frac{mg}{m}=g \] independent of \(\displaystyle m\). For two bricks tied together (mass \(\displaystyle 2m\)): \[a'=\frac{(2m)g}{2m}=g \] Answer: no — a single brick and two tied bricks fall with the same acceleration \(\displaystyle g\) and land together; air resistance apart, mass does not change free-fall acceleration.

  3. Exercise 25

    Two objects of masses m1\displaystyle m_{1} and m2\displaystyle m_{2} having the same size are dropped simultaneously from heights h1\displaystyle h_{1} and h2\displaystyle h_{2} respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if (i) one of the objects is hollow and the other one is solid and (ii) both of them are hollow, size remaining the same in each case. Give reason.

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    NCERT’s answer
    \(\displaystyle h_{1}\) = $\displaystyle 1$ $\displaystyle 2$ gt $\displaystyle 2$ \(\displaystyle h_{2}\) = $\displaystyle 1$ $\displaystyle 2$ gt $\displaystyle 2$ $\displaystyle 2$ , as x = o t t $\displaystyle 2$ = h h $\displaystyle 2$ . Ratio will not change in either case because acceleration remains the same. In case of free-fall acceleration does not depend upon mass and size.
    Under free fall, \[h=\frac{1}{2}gt^{2}\ \Rightarrow\ t=\sqrt{\frac{2h}{g}} \] \[\frac{t_{1}}{t_{2}}=\sqrt{\frac{h_{1}}{h_{2}}} \]This is independent of \(\displaystyle m_1\) and \(\displaystyle m_2\), because \(\displaystyle g\) does not depend on a body's mass or size. Making an object hollow instead of solid only changes its mass, not \(\displaystyle g\); in (i) and (ii) each object still falls with \(\displaystyle a=g\), so the ratio is unchanged.Answer: \(\displaystyle t_1:t_2=\sqrt{h_1}:\sqrt{h_2}\); unaffected by mass, and unaffected by the objects being hollow or solid.

  4. Exercise 26

    (a)
    A cube of side 5\displaystyle 5 cm is immersed in water and then in saturated salt
    solution. In which case will it experience a greater buoyant force. If
    each side of the cube is reduced to 4\displaystyle 4 cm and then immersed in water,
    what will be the effect on the buoyant force experienced by the cube as
    compared to the first case for water. Give reason for each case.
    (b)
    A ball weighing 4\displaystyle 4 kg of density 4000\displaystyle 4000 kg m−3\displaystyle m^{-3} is completely immersed in
    water of density 103\displaystyle 10^{3} kg m−3\displaystyle m^{-3} Find the force of buoyancy on it. (Given g = 10\displaystyle 10 m s−2\displaystyle s^{-2}.)

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    NCERT’s answer
    (i)
    The cube will experience a greater buoyant force in the saturated salt solution because the density of the salt solution is greater than that of water. (ii) The smaller cube will experience lesser buoyant force as its volume is lesser than the initial cube. b) Buoyant force = weight of the liquid displaced = density of water × volume of water displaced ×g = $\displaystyle 1000$ × $\displaystyle 4000$ ×$\displaystyle 10$ = $\displaystyle 10$ N
    Buoyant force equals the weight of fluid displaced: \[F_{B}=\rho_{fluid}\,V\,g \] For the $\displaystyle 5$ cm cube fully immersed, \(\displaystyle V\) is the same in water and in saturated salt solution, but \(\displaystyle \rho_{salt}>\rho_{water}\): \[F_{B}^{salt}=\rho_{salt}Vg>\rho_{water}Vg=F_{B}^{water} \]Shrinking the side to $\displaystyle 4$ cm in water reduces \(\displaystyle V\): \[V'=(0.04\text{ m})^{3}=6.4\times10^{-5}\text{ m}^{3}<(0.05\text{ m})^{3}=1.25\times10^{-4}\text{ m}^{3}=V \] \[F_{B}'=\rho_{water}V'g<\rho_{water}Vg=F_{B} \] For the ball: \[V_{ball}=\frac{m}{\rho_{ball}}=\frac{4\text{ kg}}{4000\text{ kg m}^{-3}}=1\times10^{-3}\text{ m}^{3} \] \[F_{B}=\rho_{water}V_{ball}g=1000\text{ kg m}^{-3}\times1\times10^{-3}\text{ m}^{3}\times10\text{ m s}^{-2}=10\text{ N} \] Answer: (a) greater in salt solution (same \(\displaystyle V\), higher \(\displaystyle \rho\)); smaller for the $\displaystyle 4$ cm cube in water (smaller \(\displaystyle V\)). (b) buoyant force on the ball \(\displaystyle =10\) N.