A spinning cricket ball in air does not follow a parabolic trajectory due to Magnus effect. Let a spinning cricket ball is moving forward with a velocity v and spinning clockwise with velocity u. As ball moves forward, it leaves, a lower pressure region behind it. To fill this region, air moves backward with velocity v. The layers of air in contact with the ball spin with ball with velocity u. Therefore, the resultant velocity of air above the ball is (v - u) and below the ball is (v + u). According to Bernoulli's theorem,
 \(p+\frac { 1 }{ 2 } \rho { v }^{ 2 }\)
Therefore, pressure below the ball becomes lower than above the ball. Due to this pressure difference, a force acts on ball in downward direction. Therefore, the ball follows a curved path in spite of a parabolic trajectory.

Excess pressure inside the soap bubble is 20 Pa;
Pressure inside the air bubble is 1.06 x 10⁵ Pa
Soap bubble is of radius, r = 5.00 mm = 5 x 10^–3 m
Surface tension of the soap solution, S = 2.50 x 10^–2 Nm^–1
Relative density of the soap solution = 1.20
∴ Density of the soap solution, ρ = 1.2 x 10³ kg/m³
Air bubble formed at a depth, h = 40 cm = 0.4 m
Radius of the air bubble, r = 5 mm = 5 x 10^–3 m
1 atmospheric pressure = 1.01 x 10⁵ Pa
Acceleration due to gravity, g = 9.8 m/s²
Hence, the excess pressure inside the soap bubble is given by the relation:
\(P=4 \frac{S}{r}\)
\(=\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\)
= 20 Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by the relation:
\(P=\frac{2 S}{r}\)
\(=\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\)
= 10 Pa
Therefore, the excess pressure inside the air bubble is 10 Pa.
At a depth of 0.4 m, the total pressure inside the air bubble
= Atmospheric pressure + hρg + P’
\(=1.01 \times 10^{5}+0.4 \times 1.2 \times 10^{3} \times 9.8+10\)
\(=1.057 \times 10^{5} \mathrm{~Pa}\)
\(=1.06 \times 10^{5} \mathrm{~Pa}\)
Therefore, the pressure inside the air bubble is \(=1.06 \times 10^{5} \mathrm{~Pa}\)​​​​​​

Here h = 1000 m and ρ = 1.03 x 10 ³ kg m^-3
The pressure outside the submarine is P = P_a + ρgh and the pressure inside it is P_a . Hence, the net pressure acting on the window is gauge pressure, P _g = ρgh. Since the area of the window is A = 0.04 m2 , the force acting on it is 
F = P _g A = 10³ x 10⁵ Pa x 0.04 m² = 4.12 x 10⁵ N

Given, length of the tube (l) = 1.5 m
Radius of the tube(r) = 1.0 cm = 1 x 10^-2 cm
Mass of glycerine flowing per second = 4 x 10^-3 kg/s
Density of glycerine, \(\rho \) = 1.3 x 10³ Kg/ m³
Viscosity of glycerine , \(\eta =0.83\quad Pa-s\)
Volume of glycerine flowing per second, \(V=\frac { m }{ \rho } \)
\(\left[ \because density=\frac { Mass }{ Volume } \right] \)
=  \(\frac { 4\times { 10 }^{ -3 } }{ 1.3\times { 10 }^{ 3 } } { m }^{ 3 }/s=\frac { 4 }{ 1.3 } \times { 10 }^{ -6 }{ m }^{ 3 }/s\)
According to Poiseuille's formula, the rate of flow of liquid through a tube
 \(V=\frac { \pi }{ 8 } \frac { p{ r }^{ 4 } }{ \eta l } \)
Where, p is the pressure difference between the two ends of the tube.
or    \(p=\frac { 8\eta lV }{ \pi { r }^{ 4 } } =\frac { 8\times 0.83\times 1.5\times 4\times { 10 }^{ -6 } }{ 3.14\times (1\times { 10 }^{ -2 })^{ 4 }\times 1.3 } \)
\( =976\ Pa\)
To check the laminar flow in tube, the value of Reynold's number should be less than 2000.
Reynold's number, \({ R }_{ e }=\frac { \rho Dv_{ c } }{ \eta } \)
where, v_c is the critical velocity and D is the diameter of the tube.
Critical velocity,  \({ v }_{ c }=\frac { Volume\ flowing\ out\ per\ second }{ Area\ of\ cross-section } \)
\(=\frac { m/\rho }{ A } =\frac { m }{ \rho \pi { r }^{ 2 } } \quad [\because A=\pi { r }^{ 2 }]\)
\(\therefore \) Reynold's number, \({ R }_{ e }=\frac { \rho D }{ \eta } \times \frac { m }{ \rho \pi { r }^{ 2 } } \)
 \(=\frac { 2r\times m }{ \eta \pi { r }^{ 2 } } \)
\(\\ =\ \frac { 2m }{ \pi r\eta } =\frac { 2\times 4\times 10^{ -3 } }{ 3.14\times { 10 }^{ -2 }\times 0.83 } =0.31\)
As R_e < 2000, therefore. flow of glycerine is laminar.

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

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
\({ p }^{ ' }=p=6.92\times 10^{ 5 }Pa\)

Given, mass of girl (m) = 50kg
Diameter of circular heel (2r) = 1.0cm
\(\therefore \)Radius (r) = 0.5 cm = \(5\times 10^{ -3 }\)m
Area of circular heel (A) = \({ \pi r }^{ 2 }=3.14\times (5\times 10^{ -3 })^{ 2 }m^{ 2 }\)
\(=78.50\times 10^{ -6 }m^{ 2 }\)
\(\therefore \) Pressure exerted on the horizontal floor
\(\rho =\frac { F }{ A } =\frac { mg }{ A } =\frac { 50\times 9.8 }{ 78.50\times 10^{ -6 } } =6.24\times 10^{ 6 }Pa\)

Because water vapour exerts pressure. Thus, the barometric height decreases.

The difference between absolute pressure and atmospheric p[ressure is known as gauge pressure.
As,p_absolute= p_a+ρgh
So,p_absolute− p_a=ρghi.e.p_gauge=ρgh
Here,ρ is the density of a fluid of depth h.

The height of mercury in the inclined case will be 76cm.

The critical speed of a fluid depends on ( a ) diameter of tube,  ( b ) density of fluid, ( c ) coefficient of viscosity of the fluid.

The depth of hole below the upper surface of water is
h  = H/4
= 96/4
= 24 cm
The height of hole from ground is,
h' = 96 - 24
= 72 cm
Horizontal range = 2.\( \sqrt{hh′ } \)
= 2 \( \sqrt{24×72}\)
= 48√3 cm

This is due to equation of continuity, a₁ v_I = a₂ v₂  
∵∵  a₁  > a₂
∴∴     v₂  > v₁

Given, S = 0.03 N/m
Work done =surface area××surface tension
= 2π 4 r² × S
= 24 × 31.4 ×(0.02)² × 0.03
= 3 × 10^-4J

So, they can spread over large area easily.

When a piece of the wood burns, then steam formed and water appears in the form of drops due too surface tension on the other end.

Area of cross-section of tube (A) = 8 cm² = 8\(\times\)10^-4 m²
Number of holes, N = 40
Diameter of each hole, 2r = 1.0 mm   
\(\therefore\) Radius of each hole, r = 0.5 mm = 5\(\times\) 10^{- 4} m
Velocity of liquid flow in tube, = 1.5 m / min
= \( \frac{1.5}{60}\) m/s
Total area of holes = N \(\times\) \(\pi\)r²
= 40\(\times\)3.14\(\times\)(5\(\times\)10^-4)²
= 3.14 \(\times\)10^-5 m²
From equation of continuity,
A₁v₁ = A₂v₂
or v₂ = \(\frac{A_1v_1}{A_2}\)
= \( \frac{8\times10^{-4}\times2.5\times10^{-2}}{3.14\times10^{-5}}\)
= \(\frac{20}{3.14}\times10^{-1}\)
= 0.64 m/s

(a) Given, radius of artery    ( r ) = 2 \(\times\) 10^{- 3} m
\(\therefore \) Diameter of artery   D = 2r = 4 \(\times\) 10^-3m
Density of whole blood ( \(\rho\) ) = 1.06 \(\times\) 10³ kg / m³
Coefficient of viscosity of blood ( \(\eta\) ) = 2.084 \(\times\)10³ Pa - s
For laminar flow, maximum value of Reynold's number
R_e = 2000
Critical velocity  ( v_c ) = \(\frac{R_e\eta}{\rho D} \)
= \(\frac{2000\times2.084\times10^3}{1.06\times10^3\times4\times10^{-3}} \)
= 9.83 \(\times\)10⁵m / s
(b) Flow rate of blood = Volume of blood flowing per second
= Av_c
= \(\pi\)r2\(\times\)vc
= 3.14\(\times\)(2\(\times\)10^-3)²\(\times\)9.83\(\times\)105
= 12.35 m³/s

Let the lower and upper surface of the wings of the aeroplane be at the same height h and speeds of air on the upper and lower surfaces of the wings be V1 and V₂'
Speed of air on the upper surface of the wings
V₁ = 70 m/s
Speed of air on the lower surface of the wings
v₂ = 63 m/s
Density of the air, \(\rho\) = 1.3 kg/ m³ 
Area, A = 2.5 m²
According to Bernoulli's theorem,
p1 +\( \frac{1}{2} \)\(\rho\) \(v_{1}^{2}\) + \(\rho\) gh = p2 + \(\frac{1}{2} \)\(\rho\) \(v_{2}^{2}\) + \(\rho\) gh
or     p2 - p1 = \(\frac{1}{2} \) \(\rho\) \((v_{1}^{2} - v_{2}^{2})\)
\(\therefore\) Lifting force acting on the wings,
F = ( p₂ - p₁ ) \(\times\) A
= \(\frac{1}{2}\) \(\rho\) \((v_{1}^{2} - v_{2}^{2})\)\(\times\) A 
[ \(\because\) force  = pressure\(\times\)area]
= \(\frac{1}{2}\)\(\times\)1.3\(\times\) [ ( 70 )² - ( 63 ) 2 ]\(\times\) 2.5
=\(\frac{1}{2}\)\(\times\)1.3 [ 4900-3969 ]\(\times\) 2.5
= \(\frac{1}{2}\)\(\times\)1.3\(\times\) 931\(\times\) 2.5
= 1.51\(\times\)10³ N

(a) The pressure of liquid column is given by \(p=h\rho g\), where h is depth, \(\rho \)i s density and g is acceleration due to gravity.
Therefore, pressure of liquid column increases with depth. The height of blood column in human body is more at feet than at the brain. Therefore, the blood pressure in humans is greater at the feet than the brain.
(b) The density of air is maximum near the surface of the earth and decreases rapidly with height. At a height of 6 km, the density of air decreases to nearly half its value at the seal level. Beyond 6km height, the density of air decreases very slowly with height.Hence, the atmospheric pressure at a height of about 6km decreases to nearly half of its value at the sea level.
(c) When force is applied on a liquid, the pressure is transmitted equally in all directions inside the liquid. Therefore, hydrostatic pressure has no fixed direction and hence, it is a scalar quantity.

Length pf the slider (i) = 30 cm
As a soap film has two free surfaces, therefore, total length of the film to be supported
l' = 2l = 2\(\times\)30
= 60 cm = 0.06m
Let S be the surface tension of the soap solution 
Total force on the slider due to surface tension.
F = S\(\times\)2l
F = S\(\times\)0.060N
Weight (w) supported by the slider = 1.5 \(\times\)10^-2 N
In equilibrium, 
Force on the slider due to surface tension =weight supported by the slider
\(\therefore\)   F = w
S\(\times\)0.06 = 1.5\(\times\)10^-2
or \(S=\frac { 1.5\times { 10 }^{ -2 } }{ 0.06 } =2.5\times { 10 }^{ -2 }N/m\)