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Mechanical Properties of Fluids Test - 42

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Mechanical Properties of Fluids Test - 42
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  • Question 1
    1 / -0
    Depth of sea is maximum at Mariana Trench in West Pacific Ocean. Trench has maximum depth of about $$11km$$. At bottom of trench water column above it exerts $$1000$$ atm pressure. Percentage change in density of sea water at such depth will be around
    (Given, $$B=2\times {10}^{9}N{m}^{-2}$$ and $${p}_{atm}=1\times {10}^{5}N{m}^{-2}$$)
    Solution

  • Question 2
    1 / -0
    A $$20$$cm long capillary tube is dipped in water. The water rises up to $$8$$cm. If the entire arrangement is put in a freely falling elevator, the length of water column in the capillary tube will be:
    Solution
    In a freely falling lift, gravitational pull is zero hence the capillary tube will be filled completely.
  • Question 3
    1 / -0
    Water rises in a vertical capillary tube up to a length of $$10$$cm. If the tube is inclined at $$45^o$$, the length of water risen in the tube will be:
    Solution

  • Question 4
    1 / -0
    A vertical tank, open at the top, is filled with a liquid and rests on a smooth horizontal surface. A small hole is opened at the center of one side of the tank. The area of a cross-section of the tank is $$N$$ times the area of the hole, where $$N$$ is a large number. Neglect mass of the tank itself. The initial acceleration of the tank is:
    Solution
    Let the height of the tank is L, density \rho and area is A
    velocity of efflux=v
    $$\Rightarrow v=\sqrt{2g\dfrac{L}{2}}=\sqrt{gL}$$...eq(1)
    when the liquid flows out of the container horizontally, a force is exerted on the container.
    force $$F=\rho\times$$ (area of the hole)$$\times v^2$$
    or $$m_{0}a=\rho \times \dfrac{A}{N}\times v^2$$     \therefore a is acceleration 
    $$\rho = \dfrac{m_{0}}{AL}$$...eq(2)
    from eq(1) and eq(2)
    $$\Rightarrow m_{0}a=\dfrac{m_{0}}{AL}\times\dfrac{A}{N}\times gL $$
    $$\Rightarrow a=\dfrac{g}{N}$$
    Hence C option is correct .
  • Question 5
    1 / -0
    The pressure of water in a water pipe when tap is opened and closed in respectively $$3\times 10^{5}N/m^{2}$$ and $$3.5\times 10^{5} N/m^{2}$$. Determine the velocity of flow when tap is open?
    Solution
     $$\Delta P=\cfrac { 1 }{ 2 } \rho { v }^{ 2 }\\ \left( 0.5\times { 10 }^{ 5 } \right) =\cfrac { 1 }{ 2 } \times 1000\times { v }^{ 2 }\\ v=10m/s$$
  • Question 6
    1 / -0
    A ball of mass $$m$$ and radius $$r$$ is released in viscous liquid. The value of its terminal velocity is proportional to
    Solution

  • Question 7
    1 / -0
    The two femurs each of the cross-sectional area 10 cm$$^2$$ support the upper part of a human body of mass 40 kg. The average pressure sustained by the femurs is then (Takes g= 10 m s$$^{-2}$$)
    Solution
    Given that,
    $$Area=10 \ cm^{2}$$
    $$Mass=40 \ kg$$
    $$g=10 \ m/s^{2}$$

    Total cross-sectional area of the femurs is,
    $$A = 2 \times 10 \ cm^2$$

    $$ = 2 \times 10\times 10^{-4} m^2 $$

     $$ = 20 \times 10^{-4} m^2$$

    Force acting on them is

    $$F= mg =40\,kg \times 10\,m\,s^{-2} =400\,N$$

    $$\therefore$$ Average pressure sustained by them is
    $$\displaystyle P=\dfrac{F}{A}= \dfrac {400\,N}{20 \times 10^{-4}\,m^2}$$

                   $$ = 2 \times 10^5\,N\,m^{-2}$$
  • Question 8
    1 / -0
    Two syringes of different cross section (without needle) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1 cm and 3 cm respectively. If a force of 10 N is applied to the smaller piston then the force exerted on the larger piston is
    Solution
    Since pressure is transmitted undiminished throughout the water
    $$\therefore \displaystyle \frac{F_1}{A_1} =\frac{F_2}{A_2}$$
    where $$F_1$$ and $$F_2$$ are the forces on the smaller and on the larger pistons respectively and $$A_1$$ and $$A_2$$ are the respective areas.
    $$\therefore \displaystyle F_2=\frac{A_2}{A_1}F_1 = \frac{\pi (D_2/2)^2}{\pi (D_1/2)^2} F_1 = \left ( \frac{D_2}{D_1} \right )^2 F_1$$
    $$\displaystyle = \frac {(3 \times 10^{-2}\,m)^2}{(1 \times 10^{-2}\,m)^2} \times 10\,N = 90\,N$$
  • Question 9
    1 / -0
    When the flow parameters of any given instant remain same at every point, then flow is said to be
    Solution
    When the flow parameters (like pressure, velocity etc) doesn't change w.r.t time i.e remains constant through out the time period, then this state is called as STEADY STATE.
  • Question 10
    1 / -0
    A sealed tank contains water to a height of 11 m and air at 3 atm. Water  flower out from the bottom of a tank through a small hole. The velocity of efflux is (g=$$10{ ms }^{ -2 }$$)
    Solution
    According to Bernaulis theoram
    $$0.01\times { 10 }^{ 5 }\times 3+\rho gh=\cfrac { 1 }{ 2 } \rho { v }^{ 2 }\\ 3.03\times { 10 }^{ 5 }+1000\times 10\times 11=\cfrac { 1 }{ 2 } \times 1000{ v }^{ 2 }\\ v=\sqrt { \cfrac { 4.13\times { 10 }^{ 5 }\times 2 }{ 1000 }  } \\ v=28.6m/s$$

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