Physics Mechanical Properties of Fluids

A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is___________________ cm.

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Physics Mechanical Properties of Fluids Mechanical Properties of Fluids

Contributor-Level 10

$p_{1} - p_{2} = \frac{4 T}{r_{1}} . . . . . . . (1)$

$p_{2} - p_{0} = \frac{4 T}{r_{2}} . . . . . . . (2)$

(1) + (2), p₁ – p₀ = 4T $(\frac{1}{r_{1}} + \frac{1}{r_{2}})$

$\Rightarrow \frac{4 T}{r_{e q}} = 4 T (\frac{1}{r_{1}} + \frac{1}{r_{2}})$

$r_{e g} = \frac{r_{1} r_{2}}{r_{1} + r_{2}} = \frac{3 * 6}{3 + 6} = 2 c m$

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2 months ago

Two narrow bores of diameter 5.0mm and 8.0 mm are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube.

[Take surface tension of water T = 7.3 * 10^-² Nm^-¹, angle of contact = 0, g = 10 ms^-² and density of water = 1.0 * 10³ kg m^-³]

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Contributor-Level 10

P_B – P_A = $ρ g h$

$\Rightarrow (P_{0} - \frac{2 T}{r^{2}}) - (P_{0} - \frac{2 T}{r_{1}}) = ρ g h$

$h = \frac{2 T (\frac{1}{r_{1}} - \frac{1}{r_{2}})}{ρ g}$

$= \frac{2 * 7.3 * 1 0^{- 2} * (\frac{1}{2.5 * 1 0^{- 3}} - \frac{1}{4 * 1 0^{- 3}})}{1000 * 10}$

$\begin{array}{l} = 2.19 * 1 0^{- 3} m \\ = 2.19 m m \end{array}$

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2 months ago

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

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$ρ = 1 0^{3} k g / m^{3}$

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 * 10^-⁴

= 1J.

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2 months ago

The terminal velocity $(υ_{1})$ of the spherical rain drop depends on the radius (r) of the spherical rain drop as:

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Physics Mechanical Properties of Fluids Mechanical Properties of Fluids

Contributor-Level 10

Let, Rain drop moving with terminal velocity vt in the air Fb (Bnoyancy force) = $\frac{4}{3} π r^{3} ρ_{a i r} g$ $\frac{}{}^{}_{}$

$m g = \frac{4}{3} π r^{3} ρ_{a i r} g$

$F_{v} = (v i s c o n s f o r c e) = 6 π η r v_{T}$

$F_{b} + F_{v} = m g$

$\Rightarrow \frac{4}{3} π r^{3} ρ_{a i r} g + 6 π η r v_{T} = \frac{4}{3} π r^{3} ρ g$

$v_{T} = \frac{2}{9} \frac{r^{2}}{η} (ρ - ρ_{a i r})$

$v_{T} \propto r^{2}$

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The area of cross-section of a large tank is 0.5 m². It has a narrow opening near the bottom having area of cross-section 1 cm². A load of 25 kg is applied on the water, coming out of the opening at the time when the height of water level in the tank is 40cm above the bottom, will be_________cms^-¹. $[T a k e g = 10 m s^{- 2}]$

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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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2 months ago

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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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New answer posted

2 months ago

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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2 months ago

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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