Physics Mechanical Properties of Fluids

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A
alok kumar singh

Contributor-Level 10

k x = m g - F u p

k x = m g - m d D g

x = m g k 1 - d D

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V
Vishal Baghel

Contributor-Level 10

v T r 2

v T 1 v T 2 = r 1 2 r 2 2 = r 2 9 r 2

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V
Vishal Baghel

Contributor-Level 10

Applying Bernoulli's theorem

P 0 + ρ g 2 h + 2 ρ g h + 0 = P 0 + 1 2 2 ρ v 2

V 2 = 4 g h

v = 2 g h

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A
alok kumar singh

Contributor-Level 10

Cohesive force in mercury molecules are greater than adhesive force

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Raj Pandey

Contributor-Level 9

ρ = 8 0 0 k g / m 3

P1 – P2 = 4100 Pa

 Bernoulli's question b/w (1) & (2)

P 1 ρ g + v 1 2 2 g + z 1 = P 2 ρ g + v 2 2 2 g + z 2  

⇒  P 1 P 2 ρ g + v 1 2 2 g + 1 = v 2 2 2 g + 0            -(1)

Equation of continuity

A1v1 = A2v2

v2 = 2v1                -(2)

from (1) & (2)

P 1 P 2 ρ g + v 1 2 2 g + 1 = ( 2 v 1 ) 2 2 g  

4 1 0 0 8 0 0 * 1 0 + v 1 2 2 g + 1 = 4 v 1 2 2 g  

1 2 1 0 0 8 0 0 0 = 3 v 1 2 2 g

v 1 2 = 2 * 1 0 3 * 1 2 1 0 0 8 0 0 0  

  = 2 3 * 1 2 1 8  

                 x = 363

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2 months ago

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Vishal Baghel

Contributor-Level 10

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

mg = FB + Fv 4 π 3 r 3 ρ g = 4 π 3 r 3 σ g + 6 π η r v

v = 2 r 2 ( ρ σ ) g 9 η = 2 * 0 . 1 * 0 . 1 * 1 0 6 * ( 1 0 4 1 0 3 ) * 1 0 9 * 1 . 0 * 1 0 5

h = 4 0 0 2 g = 2 0 m

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2 months ago

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R
Raj Pandey

Contributor-Level 9

Gain in surface energy, Δ U = T Δ A  

from volume centenary,   4 3 π R 3 = 6 4 * 4 3 π r 3  

r = R 4  

Initial surface area, Ai = 4pR2

final surface area,   A f = 6 4 * 4 π r 2  

  Δ U = 1 2 π R 2 . T = 0 . 0 7 5 * 1 2 * 3 . 1 4 * 1 0 4 = 2 . 8 * 1 0 4 J  

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

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

mg = FB + Fv 4 π 3 r 3 ρ g = 4 π 3 r 3 σ g + 6 π η r v  

v = 2 r 2 ( ρ σ ) g 9 η = 2 * 0 . 1 * 0 . 1 * 1 0 6 * ( 1 0 4 1 0 3 ) * 1 0 9 * 1 . 0 * 1 0 5  

h = 4 0 0 2 g = 2 0 m  

 

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

v = 2 9 r 2 g ( ρ σ ) η

v ' = 2 9 r 2 g ( ρ 2 σ ) η

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

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

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