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Physics Mechanical Properties of Fluids

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2 weeks ago

Physics Mechanical Properties of Fluids Class 11th

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075N/m. In this process the gain in surface energy will be:

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Raj Pandey

Contributor-Level 9

Gain in surface energy, $Δ U = T Δ A$

from volume centenary, $\frac{4}{3} π R^{3} = 64 * \frac{4}{3} π r^{3}$

$\Rightarrow r = \frac{R}{4}$

Initial surface area, Ai = 4pR²

final surface area, $A_{f} = 64 * 4 π r^{2}$

$∴ Δ U = 12 π R^{2} . T = 0.075 * 12 * 3.14 * 1 0^{- 4} = 2.8 * 1 0^{- 4} J$

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New answer posted

2 weeks ago

Physics Mechanical Properties of Fluids Class 11th

A small spherical ball of radius 0.1mm and density 10⁴kgm^-³ falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity inside water, then the value of h will be_________ m.

(Given g = 10 ms^-², viscosity of water = 1.0 * 10^-⁵ N-sm^-²).

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alok kumar singh

Contributor-Level 10

Since velocity does not change, so acceleration will be zero.

mg = F_B + F_v $\Rightarrow \frac{4 π}{3} r^{3} ρ g = \frac{4 π}{3} r^{3} σ g + 6 π η r v$

$\Rightarrow v = \frac{2 r^{2} (ρ - σ) g}{9 η} = \frac{2 * 0.1 * 0.1 * 1 0^{- 6} * (1 0^{4} - 1 0^{3}) * 10}{9 * 1.0 * 1 0^{- 5}}$

$\Rightarrow h = \frac{400}{2 g} = 20 m$

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New answer posted

3 weeks ago

Physics Mechanical Properties of Fluids Class 11th

A plastic ball is rising in water with terminal speed v. If we use a different plastic whose density is double that of the original ball but still less that density of water, keeping the radius same, the terminal speed would be:

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Vishal Baghel

Contributor-Level 10

$v = \frac{2}{9} \frac{r^{2} g (ρ - σ)}{η}$

$v' = \frac{2}{9} r^{2} g \frac{(ρ - 2 σ)}{η}$

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3 weeks ago

Physics Mechanical Properties of Fluids Class 11th

The electric field line pattern for charges + 2q & - q are shown. Select the most appropriate diagram:-

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Vishal Baghel

Contributor-Level 10

Field lines are proportional to quantity of charge & they originate at positive charge & terminates at -ve charge or at $\infty$

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New answer posted

3 weeks ago

Physics Mechanical Properties of Fluids Class 11th

Two blocks of same mass but different density $ρ_{1} & ρ_{2}$ are floating in water. Block-2 has higher fraction of its volume submerged in water. If buoyant force on them are B₁ & B₂ :-

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Vishal Baghel

Contributor-Level 10

$B = m g \Rightarrow B_{1} = B_{2}$

$ρ_{1} V_{1} g = B = ρ_{w} v_{L D} g$

$\frac{ρ_{2}}{ρ_{W}} = f_{2}$

$\Rightarrow ρ_{1} > ρ_{1}$

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New answer posted

3 weeks ago

Physics Mechanical Properties of Fluids Mechanical Properties of Fluids

Force required to separate two glass plates of area 100cm² with a thin film of water 0.05 cm thick between them is____. (in N). (Surface tension of water is 70x10^-3N / m , consider complete wetting)

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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New answer posted

3 weeks ago

Physics Mechanical Properties of Fluids Mechanical Properties of Fluids

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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Vishal Baghel

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.

Þ 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

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3 weeks ago

Physics Mechanical Properties of Fluids Mechanical Properties of Fluids

Two blocks of masses 3kg and 5kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is $\frac{24}{π} * 1 0^{2} N m^{- 2} .$ What is the minimum radius of the wire? (take g = 10ms^-2)

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Vishal Baghel

Contributor-Level 10

Tension in wire, $T = \frac{2 * 3 * 5 * 10}{3 + 5} = \frac{75}{2} N$

$\frac{T}{π r_{m i n}^{2}} = \frac{24}{π} * 1 0^{2}$

$\Rightarrow r_{m i n}^{2} = \frac{75}{2 * 24} * 1 0^{- 2} = \frac{25}{16} * 1 0^{- 2}$
$\Rightarrow r_{m i n} = \frac{5}{4} * 1 0^{- 1} m = 12.5 c m$

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New answer posted

3 weeks ago

Physics Mechanical Properties of Fluids Mechanical Properties of Fluids

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 Ã— 10³ kg. The inner and outer radii of each column are 50cm and 100cm respectively. Assuming uniform local distribution, calculate the compression strain of each column.

[Use Y = 2.0 Ã— 10¹¹ Pa, g = 9.8 m/s²]

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Vishal Baghel

Contributor-Level 10

Load of mass will be equally distributed among the four colours so force on each columns will be 125 * 10³N.

Cross section area of the column = $π [{(1)}^{2} - {(0.5)}^{2}] = 2.355 m^{2}$

Using young's modulus : $ε = \frac{σ}{Y} = \frac{F}{A Y} = \frac{125 * 1 0^{3}}{2.355 * 2 * 1 0^{11}} = 2.65 * 1 0^{- 7}$

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New answer posted

3 weeks ago

Physics Mechanical Properties of Fluids Class 11th

A bucket of height 20 cm is placed under a tap such that water flows into the bucket at a constant rate of 100 cm³/s. There is a small hole in the bucket of cross-sectional area of 1 cm² at height of 3 cm from the bottom. Area of hole is very small as compared to area of the bucket. Find the maximum height (in cm ) above the bottom upto which the water is filled in the bucket. (Take g = 10 m/s² )

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Vishal Baghel

Contributor-Level 10

Q = a√2gh
h = Q²/ (a² * 2g) = (10? )²/ (10? * 2 * 10) = 5 cm
Height above the ground = 3 cm + 5 cm = 8 cm

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