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Mechanical Properties of Fluids Test - 52

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Mechanical Properties of Fluids Test - 52
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  • Question 1
    1 / -0
    When water is flowing through a pipe with a speed $$v$$, then its power is proportional to
    Solution

    Using Bernoulli's equation.
    Energy per unit volume before $$=$$ Energy per unit volume after
    $${ p }_{ 1 }+\dfrac { 1 }{ 2 } { pv }_{ 1 }^{ 2 }+{ pgh }_{ 1 }={ p }_{ 2 }+\dfrac { 1 }{ 2 } { pv }_{ 2 }^{ 2 }+{ pgh }_{ 2 }$$
    $${ A }_{ 2 }<{ A }_{ 1 }$$
    $${ V }_{ 2 }>{ V }_{ 1 }$$
    $${ p }_{ 2 }<{ p }_{ 1 }$$
    Hence power $$\alpha { v }^{ 2 }$$.

  • Question 2
    1 / -0
    A container of height $$10\ cm$$ is filled with water. There is a hole at bottom. Find the pressure difference between points $$A$$ & $$B$$.

    Solution
    $$P_{A} - P_{B} = \rho gh$$
    $$= 10^{3} \times 10\times 10\times 10^{-2} Pa$$
    $$= 1000\ Pa$$.
  • Question 3
    1 / -0
    A block of sided $$0.5 m$$ is $$30\%$$ submerged in a liquid of density $$1 \, gm/cc$$. Then find mass of an object placed on block for complete submergence.
    Solution
    Initial condition $$B = mg$$

    $$\rho\dfrac{3}{10}Vg = mg$$.          ...(1)

    $$\rightarrow (1000 \,kg/m^3) \dfrac{3}{10} (0.5m)^3 = m$$

    $$m = 37.5 \,kg$$
    finally $$\rho vg = (m + M)g$$          ...(2)
    From equation (1) & (2)

    $$\dfrac{(m + M)g}{mg} = \dfrac{\rho Vg}{\rho\dfrac{3}{10}Vg}$$

    $$1 + \dfrac{M}{m} = \dfrac{10}{3}$$

    $$\dfrac{M}{m} = \dfrac{7}{3}$$

    $$\dfrac{M}{m} = \dfrac{7}{3}$$

    $$M = \dfrac{7}{3} (37.5)kg = 87.3 \,kg$$
  • Question 4
    1 / -0
    A steel ball is dropped in a viscous liquid. The distance of the steel ball from the top of the liquid is shown below. The terminal velocity of the ball is closest to : 

    Solution
    For terminal velocity, slope should be constant. If we look carefully  slope is constant from $$t=1.6\ sec$$ to $$t=2\ sec$$

    $$V_t=\dfrac{0.4-0.3}{2-1.6}=\dfrac{0.1}{0.4}=0.25\ m/s$$

    So terminal velocity is closest to $$0.26\ m/s$$
  • Question 5
    1 / -0
    The working of venturimeter is based on 
    Solution

  • Question 6
    1 / -0
    The density of Saturn is $$ 687 kg/m^3 $$. It is kept in a lake filled with water. Then:
    Solution

  • Question 7
    1 / -0
    Venturimetre is used to measure
    Solution

  • Question 8
    1 / -0
    A film of water is formed between two straight parallel wires each  $$10\mathrm { cm }$$  long and at separation  $$0.5\mathrm { cm } .$$  Calculate the work required to increase  $$1\mathrm { mm }$$  distance between the wires. Surface tension of water  $$= 72 \times 10 ^ { - 3 } \mathrm { N } / \mathrm { m } .$$
    Solution
    Given:
    Length $$L=10$$ cm
    Width $$W=0.5$$ cm
    Surface tension $$S=72 \times { 10^{ -3 }}$$
    Increase distance by $$\Delta l=1$$ mm
    change in area $$\Delta A=(L\times W)=10(.1)\times 10^{-4}$$

    $$\begin{array}{l} E=S\Delta A \\ =72\times { 10^{ -3 } }\times 2\times 1\times { 10^{ -4 } } \\ =1.44\times { 10^{ -5 } }J \\ Ans.\, \, (C) \end{array}$$
  • Question 9
    1 / -0
    Calculate the energy spent in spraying a drop of mercury of radius $$R$$ into $$n$$ droplets
    all of same size. Given the surface tension of mercury is  $$T .$$
    Solution
    $$\begin{array}{l} n\cdot \frac { 4 }{ 3 } \pi { r^{ 3 } }=\frac { 4 }{ 3 } \pi { R^{ 3 } } &  \\ \Rightarrow n{ r^{ 3 } }={ R^{ 3 } } & ...(i) \\ E=\left( { n\cdot 4\pi { r^{ 2 } }-4\pi { R^{ 2 } } } \right) T &  \\ =4\pi T\left( { n\cdot { { \left( { \frac { { { R^{ 3 } } } }{ n }  } \right)  }^{ \frac { 2 }{ 3 }  } }-{ R^{ 2 } } } \right)  &  \\ =4\pi T{ R^{ 2 } }\left( { { n^{ 1/3 } }-1 } \right)  &  \\ Ans.\, (A) &  \end{array}$$
  • Question 10
    1 / -0
    A flask containing air at $$27^oC$$ at atmospheric pressure is croked up. A pressure of $$2.5\ atm$$, inside the flask would force the rock out. The temperature at which it will happen is :
    Solution

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