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NEET-XII-Physics

13: Fluid Mechanics

with Solutions - page 2

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  • Qstn #14
    Water is slowly coming out from a vertical pipe. As the water descends after coming out, its area of cross section reduces. Explain this on the basis of the equation of continuity.
    Ans : Let a be the area of cross section and v be the velocity of water.
    According to the equation of continuity,
    av = Constant
    or, `` v\alpha \frac{1}{a}``
    It means the larger the area of cross section, the smaller will be the velocity of liquid and vice versa.
    Thus, as the water comes out of the vertical pipe, its velocity increases and area of cross section decreases.
    Page No 271:
  • Qstn #15
    While watering a distant plant, a gardener partially closes the exit hole of the pipe by putting his finger on it. Explain why this results in the water stream going to a larger distance.
    Ans : According to the equation of continuity, if the exit hole of the pipe is partially closed, the water stream comes out with more velocity due to decrease in area. This results in the water stream going to a larger distance.
    Page No 271:
  • Qstn #16
    A Gipsy car has a canvass top. When the car runs at high speed, the top bulges out. Explain.
    Ans : This can be explained through Bernoulli's principle, which states that the higher the air speed, the lower the pressure in that area. Because the air inside the car does not move, the pressure in the car is atmospheric. Because air moves outside the car (directly above it), the pressure is low. The canvas top of the Gipsy car is pushed upwards because the pressure inside the car is greater than the pressure directly above the car.
    Page No 271:
  • #
    Section : ii
  • Qstn #1
    A liquid can easily change its shape but a solid can not because
    (a) the density of a liquid is smaller than that of a solid
    (b) the forces between the molecules is stronger in solid than in liquids
    (c) the atoms combine to form bigger molecules in a solid
    (d) the average separation between the molecules is larger in solids
    digAnsr:   b
    Ans : (b) the forces between the molecules is stronger in solids than in liquids
    The forces between the particles of a solid are stronger than those between the particles of a liquid, so the particles cannot move freely but can only vibrate. Thus, a solid has stable, definite shape and volume. A solid can only change its shape by force (when broken or cut), whereas a liquid can easily change its shape because of weak inter-particle forces.
    Page No 271:
  • Qstn #2
    Consider the equations
    P=lim∆s→0F∆S and P1-P2=ρgz.In an elevator accelerating upward
    (a) both the equations are valid
    (b) the first is valid but not the second
    (c) the second is valid but not the first
    (d) both are invalid
    digAnsr:   b
    Ans : (b) the first is valid but not the second
    For a point inside the elevator, pressure can be defined as `` P=\underset{∆s\to 0}{\,\mathrm{\,lim\,}}\frac{F}{∆S}``. It is independent of the acceleration of the elevator.
    The modified form of the second equation, which will be valid in the given case, is given by
    `` {P}_{1}-{P}_{2}=\rho (g+{a}_{0})z``
    `` ``
    Here, acceleration a0 (say) due to elevator accelerating upwards is also taken into account.`` ``
    Page No 271:
  • Qstn #3
    The three vessels shown in figure have same base area. Equal volumes of a liquid are poured in the three vessels. The force on the base will be
    (a) maximum in vessel A
    (b) maximum in vessel B
    (c) maximum in vessel C
    (d) equal in all the vessels

    Figure
    digAnsr:   c
    Ans : (c) maximum in vessel C
    Here, the height of the liquid column is maximum in vessel C. Thus, the force on the base of vessel C, i.e., `` F={P}_{0}+h\,\mathrm{\,\rho \,}g`` where P0 is atmospheric pressure, is maximum.
    Page No 271:
  • Qstn #4
    Equal mass of three liquids are kept in three identical cylindrical vessels A, B and C. The densities are ρA, ρB, ρC with ρA < ρB < ρC. The force on the base will be
    (a) maximum in vessel A
    (b) maximum in vessel B
    (c) maximum in vessel C
    (d) equal in all the vessels
    digAnsr:   d
    Ans : (d) equal in all the vessels
    The force on the base is given by
    `` F=h\rho g\times A``
    `` \Rightarrow F=\left(hA\rho \right)g``
    `` \Rightarrow \,\mathrm{\,F\,}=(\,\mathrm{\,Volume\,}\times \,\mathrm{\,Density\,})\times \,\mathrm{\,g\,}``
    `` \Rightarrow F=\,\mathrm{\,mg\,}``
    In the question, the masses are equal. So, the force on the base is the same in all cases.
    Page No 271:
  • Qstn #5
    Figure shows a siphon. The liquid shown is water. The pressure difference PB - PA between the points A and B is
    (a) 400 Nm-2
    (b) 3000 Nm-2
    (c) 1000 Nm-2
    (d) zero

    figure
    digAnsr:   d
    Ans : (d) zero
    At both points A and B, pressure is equal to atmospheric pressure.
    `` \,\mathrm{\,Thus\,},\,\mathrm{\,we\,}\,\mathrm{\,have\,}:``
    `` {P}_{\,\mathrm{\,A\,}}={P}_{\,\mathrm{\,B\,}}={P}_{\,\mathrm{\,atm\,}}``
    `` \Rightarrow {P}_{\,\mathrm{\,B\,}}-{P}_{\,\mathrm{\,A\,}}=0``
    Page No 271:
  • Qstn #6
    A beaker containing a liquid is kept inside a big closed jar. If the air inside the jar is continuously pumped out, the pressure in the liquid near the bottom of the liquid will
    (a) increase
    (b) decrease
    (c) remain constant
    (d) first decrease and then increase
    digAnsr:   b
    Ans : (b) decrease
    As the air inside the jar is pumped out, the air pressure decreases. Thus, the pressure in the liquid near the bottom of the beaker decreases.
    Page No 271:
  • Qstn #7
    The pressure in a liquid at two points in the same horizontal plane are equal. Consider an elevator accelerating upward and a car accelerating on a horizontal road. The above statement is correct in
    (a) the car only
    (b) the elevator only
    (c) both of them
    (d) neither of them
    digAnsr:   b
    Ans : (b) the elevator only
    The two points in the same horizontal line will not have equal pressure if the liquid is accelerated horizontally. There should be vertical acceleration.
    Page No 271:
  • Qstn #8
    Suppose the pressure at the surface of mercury in a barometer tube is P1 and the pressure at the surface of mercury in the cup is P2.
    (a) P1 = 0, P2 = atmospheric pressure
    (b) P1 = atmospheric pressure P2 = 0
    (c) P1 = P2 = atmospheric pressure
    (d) P1 = P2 = 0
    digAnsr:   a
    Ans : (a) P1 = 0, P2 = atmospheric pressure
    The upper part of the tube contains vacuum as the mercury goes down and no air is allowed in. Thus, the pressure at the upper end, i.e., at the surface of mercury in a barometer tube is zero (P1 = 0). However, the pressure at the surface of mercury in the cup or any another point at the same horizontal plane is equal to the atmospheric pressure.
    Page No 271:
  • Qstn #9
    A barometer kept in an elevator reads 76 cm when it is at rest. If the elevator goes up with increasing speed, the reading will be
    (a) zero
    (b) 76 cm
    (c) < 76 cm
    (d) > 76 cm
    digAnsr:   c
    Ans : (c) < 76 cm
    If the elevator goes up at an increasing speed, then the effective value of g increases.
    We know:
    `` P=\,\mathrm{\,\rho \,}gh``
    So, h will have a lesser value for the same value of P if g increases.
    Page No 272:
  • Qstn #10
    A barometer kept in an elevator accelerating upward reads 76 cm. The air pressure in the elevator is
    (a) 76 cm
    (b) < 76 cm
    (c) > 76 cm
    (d) zero
    digAnsr:   c
    Ans : (c) > 76 cm
    When the elevator is going upwards with acceleration a, the effective acceleration is a' = (g + a).
    Thus, pressure is given by
    `` P=h\,\mathrm{\,\rho \,}\mathit{(}g\mathit{+}a\mathit{)}``
    Air pressure in the elevator = `` P=h\text{'}\,\mathrm{\,\rho \,}g``
    Because the pressure is the same, h' > h.
    ∴ Air pressure > 76 cm
    Page No 272:
  • Qstn #11
    To construct a barometer, a tube of length 1 m is filled completely with mercury and is inverted in a mercury cup. The barometer reading on a particular day is 76 cm. Suppose a 1 m tube is filled with mercury up to 76 cm and then closed by a cork. It is inverted in a mercury cup and the cork is removed. The height of mercury column in the tube over the surface in the cup will be
    (a) zero
    (b) 76 cm
    (c) > 76 cm
    (d) < 76 cm
    digAnsr:   d
    Ans : (d) < 76 cm
    Because of the trapped air, the pressure at the upper end of the mercury column inside the tube is not zero.
    In other words, `` {\,\mathrm{\,P\,}}_{0}>0``.
    Using this relation, we get:
    `` {P}_{\,\mathrm{\,atm\,}}={P}_{0}+\,\mathrm{\,\rho \,}gh``
    `` \,\mathrm{\,Here\,},``
    `` \,\mathrm{\,\rho \,}=\,\mathrm{\,Density\,}\,\mathrm{\,of\,}\,\mathrm{\,mercury\,}``
    `` h=\,\mathrm{\,Height\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,mercury\,}\,\mathrm{\,column\,}``
    `` \because {P}_{0}>0``
    `` \,\mathrm{\,And\,},``
    `` {P}_{\,\mathrm{\,atm\,}}>\,\mathrm{\,\rho \,}gh``
    `` \therefore 76\,\mathrm{\,cm\,}\,\mathrm{\,of\,}\,\mathrm{\,Hg\,}>\,\mathrm{\,\rho \,}gh``
    `` \,\mathrm{\,or\,},h<76\,\mathrm{\,cm\,}``
    Page No 272:
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