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Mechanical Properties of Fluids Test - 29

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Mechanical Properties of Fluids Test - 29
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  • Question 1
    1 / -0
    In a hydraulic machine, a force of 2 N2  N is applied on the piston of area of cross section 10 cm210  c{m}^{2}. What force is obtained on its piston of area of cross section 100 cm2100  c{m}^{2}?
    Solution
    We know that, Pressure=ForceAreaPressure = \dfrac{Force}{Area}
    Pressure =210=0.2Ncm2\Rightarrow Pressure  = \dfrac{2}{10} = 0.2Ncm^{-2}
    Now, Pressure=0.2Pressure = 0.2
    Again, Force=Pressure× AreaForce = Pressure \times  Area
    Force=0.2× 100=20NForce = 0.2 \times  100 = 20 N
  • Question 2
    1 / -0
    The diameter of neck and bottom of a bottle are 2 cm2   cm and 10 cm10  cm respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of 1.2 kgf1.2  kgf, the amount of force applied on the bottom is

    Solution
    Radius of the cork at the neck is 1 cm1  cm.
    Area of the cork, Ac=π×12 cm2A_{c}= \pi \times 1^2  cm^2
    So, pressure applied through the cork, Pc=1.2π×12 kgf.cm2P_{c}=\dfrac{1.2}{\pi \times 1^2}  kgf.{cm}^{-2}
    Same pressure is applied on the bottom of the bottle as per Pascal's law.
    Radius of the bottom of bottle is 5 cm5  cm.
    Area of the bottom, Ab=π×52cm2A_{b}= \pi \times 5^2 cm^2
    So, force applied at the bottom = Pc×Ab=1.2π×12×π×52=1.2×25=30 kgfP_{c} \times A_{b}=\dfrac{1.2}{\pi \times 1^2} \times \pi \times 5^2=1.2 \times 25=30  kgf

  • Question 3
    1 / -0
    A water tank of height 10m10m, completely filled with water is placed on a level ground. It has two holes one at 3m3 m and the other at 7m7 m from its base. The water ejecting from:
    Solution
    Let hh be the height of the hole.
    From Bernoulli's Theorem,
    v=2g(10h)v=\sqrt{2g(10-h)}
    This is the velocity of water from a hole at height hh from bottom.
    Thus the range of the water from the hole=vt=2g(10h)2hg=2h(10h)vt=\sqrt{2g(10-h)}\sqrt{\dfrac{2h}{g}}=2\sqrt{h(10-h)}
    Thus it is same for h=3mh=3m and h=7m.h=7m.
  • Question 4
    1 / -0
    In a surface tension experiment with a capillary tube, water rises upto 0.1 m.  If the same experiment is repeated in an artificial satellite revolving around the earth, then the water will rise in the capillary upto a height of
    Solution

  • Question 5
    1 / -0
    A tank is filled up to a height 2 H2\ H with a liquid and is placed on a platform of height HH from the ground.The distance xx from the ground where a small hole is punched to get the maximum range RR is

    Solution
    Pressure at a height x=ρg(3Hx)x=\rho g(3H-x)
    Using Bernoulli's Equation inside and outside the hole,
    12ρv2=ρg(3Hx)\dfrac{1}{2}\rho v^2=\rho g(3H-x)
        v=2(3Hx)g\implies v=\sqrt{2(3H-x)g}
    Range of the liquid=vt=v2xgvt=v\sqrt{\dfrac{2x}{g}}
    =2x(3Hx)=2\sqrt{x(3H-x)}
    This is maximum for x=1.5Hx=1.5H
  • Question 6
    1 / -0
    A spherical ball of density ρ\rho and radius 0.003m0.003 m is dropped into a tube containing a viscous fluid up to the 0 cm0 \ cm mark as shown in the figure. Viscosity of the fluid =1.26Ns/m2=1.26 N-s/m^{2} and its density ρL=ρ2=1260kg/m3 \displaystyle \rho_{L} =\frac{\rho}{2}= 1260 kg/ m^{3} Assume the ball reaches a terminal speed at 10cm10 cm mark. The time taken by the ball to travel the distance between the 10cm10 cm and 20cm20 cm mark is (g=10m/s2)(g = 10m/s ^{2} )

    Solution
    vT=29r2(ρσ )gη\displaystyle v_{T}=\frac{2}{9}\frac{r^{2}\left ( \rho -\sigma  \right )g}{\eta }
       =2×(0.003)2(1260×21260)×109×1.26\displaystyle =\frac{2\times \left ( 0.003 \right )^{2}\left ( 1260\times 2-1260 \right )\times 10}{9\times 1.26}
       =0.02=0.02 m/s
    t=dvT=10×1020.02=5\displaystyle t=\frac{d}{v_{T}}=\frac{10\times 10^{-2}}{0.02}=5 s
  • Question 7
    1 / -0

    Directions For Questions

    The spouting can is something used to demonstrate the variation of pressure with depth. When the corks are removed from the tubes in the side of the can, water flows out with a speed that depends on the depth. In a certain can, three tubes T1,T2T_{1},T_{2} and T3T_{3} are set at equal distances a above the base of the can. When water contained in this can is allowed to come out of the tubes the distances on the horizontal surface are measured as x1,x2x_{1},x_{2} and x3x_{3}

    ...view full instructions

    Speed of efflux is

    Solution
    Applying Bernoulli's Theorem inside and outside the hole,
    Pinside+12ρ(0)2+ρgH=Poutside+12ρv2+ρgHP_{inside}+\dfrac{1}{2}\rho (0)^2+\rho gH=P_{outside}+\dfrac{1}{2}\rho v ^2+\rho gH
    Thus P0+ρgh=P0+12ρv2P_0+\rho gh=P_0+\dfrac{1}{2}\rho v^2
        v=2gh\implies v=\sqrt{2gh}
  • Question 8
    1 / -0
    A large open tank has two holes in the wall. One is a square hole of side LL at a depth h from the top and the other is a circular hole of radius RR at a depth 4h4 h from the top, When the tank is completely filled with water, quantities of water flowing out per second from both holes are the same. Then RR is equal to
    Solution
    Since the water quantity from both the holes are equal, we have
    a1v1=a2v2a_1v_1=a_2v_2
    L22gy=πR22g(4y)  \Rightarrow L^2\sqrt{2gy}=\pi R^2\sqrt{2g(4y)}    \because velocity of efflux at depth h is given by v=2ghv =\sqrt{2gh}
    R=L2π\Rightarrow R=\frac{L}{\sqrt{2\pi}}
  • Question 9
    1 / -0
    Three points A,BA, B and CC on a steady flow of a non viscous and incompressible fluid are observed. The pressure, velocity and height of the points A,BA, B and CC are (2,3,1).(1,2,2)(2,3, 1). (1, 2, 2)and (4,1,2)(4, 1, 2) respectively. Density of the fluid is 1kgm31 kgm^{-3} and all otner parameters are given in SI units. Then which of the following is correct? (g=10ms2)(g = 10 ms ^{-2})
    Solution
    Given :     ρ =1 kg/m3\rho  = 1  kg/m^3
    Using Bernoulli's theorem,           P+ρgh+12ρv2 =constantP + \rho gh + \dfrac{1}{2} \rho v^2   =constant
    For point A :    2+1(10)(1)+12×(1)(32) =16.52 + 1 (10) (1) + \dfrac{1}{2} \times (1) (3^2)   =16.5

    For point B :    1+1(10)(2)+12×(1)(22) =231 + 1 (10) (2) + \dfrac{1}{2} \times (1) (2^2)   =23

    For point C :    4+1(10)(2)+12×(1)(12) =24.54 + 1 (10) (2) + \dfrac{1}{2} \times (1) (1^2)   = 24.5
    As the value of the constants are different in each case, thus all these points do not lie on same stream line.
  • Question 10
    1 / -0
    Two cylindrical vessels fitted with pistons A and B of area of cross section 8 cm28  c{m}^{2} and 320 cm2320  c{m}^{2} respectively, are joined at their bottom by a tube and they are completely filled with water. Find : (i) the pressure on piston A, (ii) the pressure on piston B, and (iii) the thrust on piston B.(consider effort on piston A E=4kgfE=4kgf)
    Solution
    Let EE be the effort applied on the smaller piston AA and L L be the load required on the larger piston BB
    Area of cross section of pistion A=8cm2A = 8 cm^{2}
    Area of cross section of pistion A=320cm2A = 320 cm^{2}
    (i) Pressure acting on piston AA in the downward direction =Thrust/Area=E/8=4/8=0.5= Thrust/Area = E/8 = 4/8 = 0.5
    \thereforePressure acting on water = E/8=0.5= E/8 =0.5 (according to Pascal's Law)
    (ii) \therefore Pressure acting on piston B= E/8=0.5B =  E/8 =0.5 (according to Pascal's Law)
    (iii) Thrust acting on piston B in the upward direction =Pressure  ×Area  ofB= Pressure \ \ \times Area \ \ of B

    =E× 3208=160 = \dfrac{E \times  320}{8} = 160
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