greater

conservation of mass, Bernoulli's principle

The liquid comes to rest due to the viscous force i.e., due to internal fluid friction between its different layers.

Because terminal velocity of air bubble is negative.

Rate of flow of a liquid through a tube is directly proportional to the fourth power of the radius of its bore i.e., V \(\propto\) r⁴.

It is so because the pressure of water in reservoir increases with depth.

For capillary action.

(i) \(\frac{v^2}{2g}\)
(ii) \(\frac{p}{\rho g}\)

\(p\propto\frac {1}{r}\)

The net force acting on the base
\(=\left( \rho gh \right) A+mg\)
\(F=1000\times 9.8\times 1\times 0.02+10\times 9.8=294N\)
Pressure =\(\quad \frac { thrust }{ area } =\frac { 294N }{ 0.02 } =14700N\) 

Here, r = 0.005 m , diameter D = 2 r = 0.010m
\(\eta\) = 0.001 Pa-s, \(\rho\) = 1000 kgm^-3
For flow to be just turbulent , Re = 3000
\(\therefore\) v = \( \frac{Re\eta}{\rho D}\)
=  \(\frac{3000\times0.001}{1000\times0.010} \)
= 0.3 ms^-1

The density of hydrogen is less than air. So, the buoyant force on the balloon will be more than its weight in case of the hydrogen. So, in this case the balloon is more than the buoyant force acting on it, so balloon will come down.

Given, depth of ocean (h) = 3km = 3000m
Density of water\((\rho )\) = \({ 10 }^{ 3 }kg/m^{ 3 }\)
Pressure exerted by water column
\(p=h\rho g=3000\times 10^{ 3 }9.8\)
\(\\=29.4\times 10^{ 6 }Pa=2.94\times 10^{ 7 }Pa\)
Maxim stress which can be withstand by the vertical off-shore structure = \({ 10 }^{ 9 }Pa\)
As, \({ 10 }^{ 9 }Pa>2.9\times { 10 }^{ 7 }Pa\)
Therefore, the vertical structure is suitable for putting up in top of an oil well in the ocean.

Given, maximum mass that can be lifted (m) = 3000 kg
Area of cross-section (A) = 425\({ cm }^{ 2 }\) = 4.25\(\times 10^{ -2 }m^{ 2 }\)
\(\therefore \)Maximum pressure on the bigger piston
\(p=\frac { F }{ A } =\frac { mg }{ A } =\frac { 3000\times 9.8 }{ 4.25\times 10^{ -2 } } =6.92\times 10^{ 5 }Pa\)
According to Pascal's law, the pressure applied on an enclosed liquid is transmitted equally in all directions.
\(\therefore \)Maximum pressure on smaller piston = maximum pressure on bigger piston
\({ p }^{ ' }=p=6.92\times 10^{ 5 }Pa\)

Absolute pressur
\(p={ p }_{ a }+{ p }_{ g }\ \ [\because \ from\ Eq.(I)]\)
\(\\ =101\ kPa+199kPa=300kPa\)