Physics Mechanical Properties of Fluids

Contributor-Level 10

Bernoulli's equation between (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{z g} + z_{2}$

$\frac{P_{a t m} + m g / A}{ρ g} + 0 + 40 + 1 0^{- 2}$

$\frac{P_{a t m}}{ρ g} + \frac{v_{2}^{2}}{2 g} + 0 [v_{1} \approx 0]$

$\frac{m g}{A ρ g} + 40 * 1 0^{- 2} = \frac{v_{2}^{2}}{2 g}$

$\frac{m}{A ρ} + 40 * 1 0^{- 2} = \frac{v_{2}^{2}}{2 g}$

$\Rightarrow \frac{25}{0.5 * 1 0^{3}} + 40 * 1 0^{2} = \frac{V_{2}^{2}}{2 * 10}$

$\Rightarrow V_{2}^{2} = (5 * 1 0^{- 2} + 40 * 1 0^{- 2}) 20$

$V_{2}^{2} = 45 * 1 0^{- 2} * 20$

$V_{2}^{2} = 90 * 1 0^{- 1}$

$V_{2} = 3 m / s e c$

$= 300 c m / s e c$

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An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are $6.4 c m$ and $4.8 c m$ , respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is

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Contributor-Level 10

Using equation of continuity

$\begin{array}{r} A_{1} V_{1} = A_{2} V_{2} \\ \frac{V_{1}}{V_{2}} = \frac{A_{2}}{A_{1}} = {(\frac{d_{2}}{d_{1}})}^{2} = {(\frac{4.8}{6.4})}^{2} = \frac{9}{16} \end{array}$

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Two cylindrical of equal cross-sectional are 16 cm² contain water upto height 100cm and 150 cm respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, it [Take, density of water = 10³ kg/m³ and g = 10 ms^-2]:

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Physics Mechanical Properties of Fluids Mechanical Properties of Fluids

Contributor-Level 10

$ρ = 1 0^{3} k g / m^{3}$

g = 10 m/s²

Final height in both vessels

$= \frac{100 + 150}{2} = 125 c m$

So, less in U = $\int A * 0.25 * g * 0.25$

$1 0^{3} * 625 * 1 0^{- 4} * 10 * 16 * 1 0^{- 4}$

= 625 * 16 * 104

= 1J.

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A spherical soap bubble of radius 3cm is formed inside another spherical soap bubble of radius 6cm. If the internal pressure of the smaller bubble of radius 3cm in the above system is equal to the internal pressure of the another single soap bubble of radius r cm.

The value of r is___________.

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Contributor-Level 10

$P_{2} - P_{06} = \frac{4 T}{6} & P_{1} - P_{2} = \frac{4 T}{3}$

$P_{1} - P_{0} = \frac{4 T}{2} = \frac{4 T}{r}$

r = 2cm

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The velocity of a small ball of mass 0.3g and density 8 g/cc when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is 1.3 g/cc, then the value of viscous force acting on the ball will be x * 10^-4 N, value of x is________. [use g = 10 m/s²]

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Contributor-Level 10

m = 0.3g density, $ρ_{(b a l l)} = 8 g / c c$ $∴ V = \frac{m}{ρ} = \frac{0.3}{8} c c$

$ρ_{g l y c e r e n e} = 1.3 g / c c = 1.3 * 1 0^{3} k g / m^{3}$

F_v + F_B = mg.

=F_v = mg – F_B = .3 * 10^-3 * 10 – 1.3 * $\frac{0.3}{8} * 1 0^{- 6} * 10 * 1 0^{3}$

$= 3 * 1 0^{- 3} - \frac{. 39}{8} * 1 0^{- 2} = 3 * 1 0^{- 3} - . 5 * 1 0^{- 3}$

= 2.5 * 10^-3N

= 25 * 10^-4N

x = 25

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A tube of length 50 cm is filled completely with an incompressible liquid of mass 250g closed both ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity $z \sqrt[]{F} r a d s^{- 1} .$ If F be the force exertaed by the liquid at the other end then the value of x will be __________.

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Physics Mechanical Properties of Fluids Mechanical Properties of Fluids

Contributor-Level 10

$ω = x \sqrt[]{F} r a d / 5$

As, F = $m ω^{2} \frac{L}{2}$

$\Rightarrow F = \frac{m L}{2} (x^{2} F) \Rightarrow x^{2} = \frac{2}{m L} = \frac{2}{\frac{1}{4} * \frac{1}{2}} = 16$

$∴ x = 4$

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A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force F, its length increases by 5cm. Another wire of the same material of length 4L and radius 4r pulled by a force 4F under same conditions. The increases in length of this wire is _________cm.

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Contributor-Level 10

$Δ L_{1} = \frac{F L}{A Y} = \frac{F L}{π r^{2} Y} = 5 c m$

$Δ L_{2} = \frac{4 F 4 L}{π 16 r^{2} y}$ = $\frac{F L}{π r^{2} Y} = 5 c m$

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Payal Gupta

Contributor-Level 10

m = 0.3g density, $ρ_{(b a l l)} = 8 g / c c$ $∴ V = \frac{m}{ρ} = \frac{0.3}{8} c c$

$ρ_{g l y c e r e n e} = 1.3 g / c c = 1.3 * 1 0^{3} k g / m^{3}$

F_v + F_B = mg.

F_v = mg – F_B = .3 * 103 * 10 – 1.3 * $\frac{0.3}{8} * 1 0^{- 6} * 10 * 1 0^{3}$

$= 3 * 1 0^{- 3} - \frac{. 39}{8} * 1 0^{- 2} = 3 * 1 0^{- 3} - . 5 * 1 0^{- 3}$

= 2.5 * 103N

= 25 * 104N

x = 25

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Physics Mechanical Properties of Fluids Mechanical Properties of Fluids

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Payal Gupta

Contributor-Level 10

$ω = x \sqrt[]{F} r a d / 5$

As, F = $m ω^{2} \frac{L}{2}$

$\Rightarrow F = \frac{m L}{2} (x^{2} F) \Rightarrow x^{2} = \frac{2}{m L} = \frac{2}{\frac{1}{4} * \frac{1}{2}} = 16$

$∴ x = 4$

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3 months ago

The diameter of an air bubble which was initially 2 mm, rises steadily through a solution of density 1750 kg m^-³ at the rate of 0.35 cms^-¹. The coefficient of viscosity of the solution is_________ poise (in nearest integer). (the density of air is negligible).

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