Mechanical Properties Of Fluid

Surface area of drop, A₁ = 4πR²
Surface area of droplets, A₂ = 216(4πr²)
Volume of drops = n x (Volume of droplets)
πR³ = 216 ( πr³)
R = 6r
Now A₂ = 216(4π( )²)
A₂ = 6(4πR²)
E ∝ A
=
=
=
6E₁ = E₂
E₂ = 6E₁
E₂ = 6E {E₁ = E}
Therefore, the correct option is (D) 6E

Given formula for surface tension at critical velocity:

The formula for surface tension at the critical velocity is given by:

where:

• is the surface tension of the liquid,
• is the density of the liquid,
• is the critical velocity of the liquid,
• is the radius of the tube or capillary.

Critical velocity:
The critical velocity is the velocity at which the flow of a liquid through a tube or capillary changes from laminar flow to turbulent flow. At this velocity, the surface tension of the liquid is balanced by the inertial forces of the flowing liquid.

Given values:
- (radius of the tube),
- (density of the liquid),
- (critical velocity).

Step 1: Convert the units to be consistent:
We need to convert the density and velocity into compatible units, such as kg/cm³ for density and m/s for velocity.

• Convert the density: ,
• Convert the velocity: .

Given: Height of water column,
Gravitational acceleration,

Step 1: Use Torricelli's Law According to Torricelli’s law, the speed of a fluid flowing out of a hole is given by: where: - is the acceleration due to gravity, - is the height of the water column.
Step 2: Substitute the given values Substitute the given values into the equation:

Answer: The correct answer is option (b): 10 m/s.

The volume of a cylinder is:

Differentiate both sides with respect to time :

Given:

Substitute into the equation:

Solve for :

Final Answer: