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Physics Mechanical Properties of Fluids

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2 months 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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2 months 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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2 months 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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2 months 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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2 months 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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2 months 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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2 months ago

Physics Mechanical Properties of Fluids Class 11th

A massless metal plate is placed on a horizontal tabletop lubricated with oil. The sheet is a square of side length $l$ = 1.0m and the oil layer has thickness h = 1.0mm. Initially one edge of the sheet coincides with one edge of the table. The sheet is pulled outwards without rotation with a constant force F = 15 N. If coefficient of viscosity of the oil is $η$ = 0.2 N-s/m², how long (in second) will it take to pull half of the sheet out of the table?

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alok kumar singh

Contributor-Level 10

In this case,

$F = η \frac{v}{h} l (l - x)$

$∴ v = \frac{F h}{η l (l - x)} = \frac{d x}{d t}$

So, $t = \frac{3 η l^{3}}{8 F h} = 5 s e c$

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2 months ago

Physics Mechanical Properties of Fluids Class 11th

A plastic circular disc of radius 10cm is placed on a thin oil film of thickness 2mm, spread over a flat horizontal surface. The torque (N-m) required to spin the disc about its central vertical axis with a constant angular velocity 8 rad/sec is (coefficient of viscosity of oil is 1.0 kg/m-s) (take p = 3.14)

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alok kumar singh

Contributor-Level 10

$τ = \int F . r = \int_{0}^{R} η (2 π r d r) (\frac{r w}{t}) . r$

$= \frac{2 π η W R^{4}}{4 t} = \frac{2 * 3.14 * 1 * 8 * 1 0^{- 4}}{4 * 2 * 1 0^{- 3}}$ f

$= 0.625 N - m ? 0.63 N - m$

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2 months ago

Physics Mechanical Properties of Fluids Class 11th

A uniform rod of length $b$ capable of turning about its end which is out of water, rests inclined to the vertical. If the specific gravity of the rod is $\frac{5}{9}$ , find the length of the rod immersed in water.

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Raj Pandey

Contributor-Level 9

Let $2 x$ length of rod is immersed in water.

$τ_{Hinge} ($ net $) = 0$

$\Rightarrow m g ? \frac{b}{2} s i n ? θ - F_{B} (b - x) s i n ? θ = 0$

$\Rightarrow m g b = 2 (b - x) F_{B}$

$∴ F_{B} = \frac{m g b}{2 (b - x)} = \frac{(A b d) b}{2 (b - x)} g$ , where $d =$ density of material of rod

$_{}$ $F_{B} =$ Buoyant force $\frac{}{} \frac{}{}$

Equate $F_{B}, \frac{A b^{2} d}{2 (b - x)} = 2 A x ρ$ $_{} \frac{^{}}{}$

$\Rightarrow b^{2} = 4 x (b - x) \frac{9}{5}$

$^{} \frac{^{}}{} \frac{}{}$ $\Rightarrow x^{2} - b x + \frac{5 b^{2}}{36} = 0 \Rightarrow x = \frac{b}{6} ∴$ immersed $= \frac{b}{3}$

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New answer posted

2 months ago

Physics Mechanical Properties of Fluids Class 11th

If temperature of a liquid is increased, choose the correct option regarding change in its surface tension and viscosity.

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Vishal Baghel

Contributor-Level 10

Fact. Due to decrease in intermolecular forces

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