The situation is shown in fig. Here, we have
\(v_A=\sqrt{2gh}\) .............(1)
and    (H - h) = \(\frac{1}{2}gt^2\) ..............(2)
The distance R is given by
R = v_A x t ........................(3)
From equation (2) t = \(\sqrt{\frac{2(H-h)}{g}}\) ...............(4)
Substituting the value of v_A from equation (1) and the value of t from equation (4) in equation (3), we get
R = \(\sqrt{(2gh)}\times \sqrt{\left\{2(H-h)/g \right\} }\)
= \(2\sqrt{\left\{ h(H-h)\right\} }\)
The range R will be maximum when
dR/dh = 0.
\(\therefore 22.\frac{1}{2}h^{-1/2}-2h^{1/2}.{1\over 2}(H-h)^{1/2}=0\)
Saving, we get h = H/2.

Three forms of energy possessed by a flowing fluid are as follows:
1. Pressure energy:
Let an ideal fluid of density P be contained in a rectangular vessel, provided with a small side tube at a depth h_o below the free surface of fluid in the vessel. At the level of side tube, pressure of fluid along the axis of side tube
p = h₀pg
If we want to introduce more fluid into the vessel at this very pressure, we can force it through the side tube by doing work on the piston, If 'A' be the cross-section area of the piston, then force acting on the piston F = PA.
\(\therefore\) Work done in moving the piston through a small distance \(\Delta r\) will be
\(\Delta W=F\Delta x=PA\Delta x=P\Delta V\)
As a result of motion of piston, the mass of the fluid forced in the vessel
\(\Delta m=\rho A\Delta x=\rho\Delta V\)
The work done is stored up in the liquid in the form of its pressure energy.
\(\therefore\) Pressure energy of liquid per unit mass = Work done per unit mass
 \(=\frac{\Delta W}{\Delta m}=\frac{P\Delta V}{\rho\Delta V}=\frac{p}{\rho}\)
and pressure energy per unit volume = p.
2. Gravitational potential energy:
Let at any stage of its flow a fluid element of mass "m' be situated at a height 'h' from the reference line (generally taken to be earth's surface), then its gravitational potential energy in that position is mgh.
\(\therefore\) Gravitational potential energy per unit mass = \(\frac{mgh}{m}=gh\)
and gravitational potential energy per unit volume = pgh.
3. Kinetic energy:
Let at any stage of its flow, a fluid element of mass 'm' be moving with a speed 'v', then the kinetic energy of this fluid element is \(\frac{1}{2}v^2\)
\(\therefore\) Kinetic energy per unit mass = \(\frac{\frac{1}{2}mv^2}{m}=\frac{1}{2}v^2\)
and kinetic energy per unit volume = \(\frac{1}{2}\rho v^2\)
These three forms of energy possessed by a flowing fluid are mutually convertible from one form to another.

For law of floatation, see the chapter in the NCERT Textbook.
The weight of the boy W_b and the weight of the preserver W_p acting downward are just balanced by the upward buoyant force of the preserver B_p and the buoyant force of the boy B_b. Therefore,
\(E_b+W_p=B_b+B_p\)         ........... (1)
But   \(B_p=V_b\rho_wg,W_b=V_b\rho_bg=60\times g\)
and  \(B_p=V_b\rho_wg,W_p=V_p\rho_p g\)
where g is the acceleration due to gravity. V and \(\rho\) denote volume and density respectively.
\(\rho_b\) = 0.9 x 1000 = 900 kg m^-3
\(\rho_p\) = 0.20 x 1000 = 200 kg m^-3
and   \(V_b=\frac{W_b}{g\rho_b}\)
From Eq. (1), we have
\(\frac{3}{4}V_b\rho_wg+V_p\rho_wg=W_b+V_p\rho_pg\)
or   \(\frac{3}{4}\frac{W_b}{g\rho_b}\rho_wg+V_p\rho_wg=60\times g+V_p\rho_pg\)
or   \(\frac{3}{4}\times \frac{60\times g}{\rho_b}p_w+V_p\rho_wg=60g+V_p\rho_pg\)
or   \(45\frac{\rho_w}{\rho_b}+V_p\rho_wg=60+V_p\rho_p\)
or   \(V_p(\rho_w-\rho_p)=60-45\frac{\rho_w}{\rho_b}\)
 \(=60-\frac{45\times 1000}{900}=10\)
\(V_p=\frac{10}{\rho_w-\rho_p}=\frac{10}{800}\) = 1.25 x 10^-2 m³
Volume of the life preserver = 0.0125 m³.

(a) Inside a liquid drop
Let r = radius of a spherieal liquid drop of centre O. T = surface tension of the liquid. Let P_i and P_o be the values of pressure inside and outside the drop.
\(\therefore\) Excess pressure inside the liquid drop = P_i - P_o
Let \(\Delta\)r be the increase in its radius due to excess pressure. It has one free surface outside.
\(\therefore\)  increase in surface area of the liquid drop
= \(4\pi(r+\Delta r)^2-4\pi r^2\)
= \(4\pi[r^2+(\Delta r)^2+2r\Delta r-r^2]\)
= \(8\pi r\quad \Delta r\)  .................(i)
\(\therefore\) increase in surface energy of the drop is
W = Surface tension x increase in area 
= \(T\times 8\pi r\quad \Delta r\)  .............. (ii)
Also W = Force due to excess of pressure x displacement
= Excess pressure x Area of drop x increase in radius
= \((p_i-p_0)4\pi r^2\Delta r\) ............(iii)
\(\therefore\) From eqns (ii) and (iii), we get
\((p_i-p_0)\times 4\pi r^2\quad \Delta r=T\times 8\pi r\quad \Delta r\)
\(\Rightarrow p_i-p_0=\frac{2T}{r}\)
(b) Inside a liquid bubbles:
A liquid bubble has air both inside and outside it and therefore it has two free surfaces.
Thus increase in its surfaces area
= \(2[4\pi (r+\Delta r)^2-4\pi r^2]\)
= \(2\times 8\pi r\Delta r=16\pi r\quad\Delta r\)
\(\therefore\)   W = \(T\times 16\pi r\Delta r\)     ...............(i)
Also  W = \((p_i-p_0)4\pi r^2\times \Delta r\)  ........(ii)
From eqnd (i) and (ii), we get
\((p_i-p_0)\times 4\pi r^2\times \Delta r=T.16\pi r\quad \Delta r\)
or \((p_i-p_0)=\frac{4T}{r}\)
(c) Inside an air bubble:
Air bubble is formed inside liquid, thus air bubble has one free surface inside it and liquid is outside.
If   r = radius of air bubble
\(\Delta\)r = increase in its radius due to excess of pressure
(P_i - P₀) inside it.
T = surface tension of the liquid in which bubble is formed.
\(\therefore\) increase in surface area = \(8\pi r \Delta r\).
\(\therefore\)    W = T x \(8\pi r \Delta r\)
Also W = \((p_i-p_0)\times 4\pi r^2\Delta r\)
\(\therefore (p_i-p_0)\times 4\pi r^2\Delta r=T\times 8\pi r^2\Delta r\)
or  \(p_i-p_0=\frac{2T}{r}\)

Given, diameter , D = 1.30 cm = 1.3\(\times \)10^-2 m.
Coefficient of viscosity of water, \(\eta \) = 10^-3 Pa-s
The volume of the water flowing out per second is
V = vA =v \(\times \)\(\pi { r }^{ 2 }=v\pi \frac { { D }^{ 2 } }{ 4 } \)
\(\therefore \) Speed of flow, v = \(\frac { 4v }{ \pi { D }^{ 2 } } \)
Reynold's number, R_e = \(\frac { \rho vD }{ \eta } =\frac { 4\rho v }{ \eta \pi D } \)
Case I when V = 0.58 L /min = \(\frac { 0.58\times 10^{ -3 }{ m }^{ 3 } }{ 1\times 60\quad s } \)
\(\\ =\quad 9.67\times { 10 }^{ -6 }{ m }^{ 3 }{ s }^{ -1 }\)   
\(\\ { R }_{ e }=\frac { 4\times { 10 }^{ 3 }\times 9.67\times 10^{ -6 } }{ { 10 }^{ -3 }\times 3.14\times 1.3\times { 10 }^{ -2 } } =948\)
\(\\ \because \ { R }_{ e }<1000,\ so\ the\ flow\ is\ steady\ or\ streamlime\)
Case II when V = 4 L/min​​​
\(=\frac { 4\times { 10 }^{ 3 } }{ 60 } { m }^{ 3 }{ s }^{ -1 }=6.67\times { 10 }^{ -5 }{ m }^{ 3 }{ s }^{ -1 }\)
\(\\ { R }_{ e }=\frac { 4\times { 10 }^{ 3 }\times 6.67\times { 10 }^{ -5 } }{ { 10 }^{ -3 }\times 3.14\times 1.3\times { 10 }^{ -2 } } =6536\)
\(\\ \because \ { R }_{ e }>3000,\ so\ the\ flow\ will\ be\ turbulent\)

Let the radius of the small drop is r and that of big drop is R. The volume of the big drop = 27\(\times \) volume of each small drop
\(\frac { 4 }{ 3 } \pi { R }^{ 3 }=27\times \frac { 4 }{ 3 } \pi { r }^{ 3 }\Rightarrow R\quad =3r\)
Let the terminal velocities of small and big drop are v₁ and v₂, respectively. Then,
\(v=\frac { 2{ r }^{ 2 }(\rho -\sigma )g }{ 9\eta } \Rightarrow v\infty { r }^{ 2 }\)
\(\\ Hence,\frac { { v }_{ 2 } }{ { v }_{ 1 } } =\frac { { R }^{ 2 } }{ { r }^{ 2 } } \Rightarrow { v }_{ 2 }={ v }_{ 1 }\times \frac { { R }^{ 2 } }{ { r }^{ 2 } } =0.2\left( \frac { 3r }{ r } \right) ^{ 2 }=0.2\times 9\)
\( { v }_{ 2 }=1.8m/s\)

Given, v = 160km/h = \(160\times \frac { 5 }{ 18 } m/s=44.44m/s\)
l = 6 m and \(\eta \) = 10^-2 poise = 10^-3 Pa-s
Shearing stress = \(\frac { F }{ A } =\eta \frac { v }{ 1 } ={ 10 }^{ 3 }\times \frac { 44.44 }{ 6 } \)
Shearing stress = 7.407 \(\times \)10^-3 Nm^-2

(b) Consider v₁ and v₂ are the speeds of air on lower and upper surfaces of the wings and  \(\rho _{ 1 }and{ \rho }_{ 2 }\) are the corresponding pressures.
From Bernoulli's principle
\({ \rho }_{ 1 }+\frac { 1 }{ 2 } \rho v^{ 2 }_{ 1 }=\rho _{ 2 }+\frac { 1 }{ 2 } { \rho v }^{ 2 }_{ 2 }\)
\(\\ \therefore { \rho }_{ 1 }-{ \rho }_{ 2 }=\Delta \rho =\frac { \rho }{ 2 } \left( { v }^{ 2 }_{ 2 }-{ v }^{ 2 }_{ 1 } \right) \)
\(\\ \left( { v }_{ 2 }-{ v }_{ 1 } \right) =\frac { 2\Delta \rho }{ \rho ({ v }_{ 2 }+{ v }_{ 1 }) } \)
Average speed, \({ v }_{ av }=\frac { { v }_{ 2 }+{ v }_{ 1 } }{ 2 } \) = 960 km/h = 267 m/s
So,  \(\frac { { v }_{ 2 }+{ v }_{ 1 } }{ 2 } =\frac { \Delta \rho }{ \rho v^{ 2 }_{ av } } =\frac { 6.5\times { 10 }^{ 3 } }{ 1.2\times (267{ ) }^{ 2 } } \approx 0.08=8%x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Then, the speed above the wing needs to be only 8% higher than that below.

The Bernoulli's equation for the horizontal flow is
\(\rho _{ 1 }+\frac { 1 }{ 2 } { \rho v }^{ 2 }_{ 1 }=\rho _{ 2 }+\frac { 1 }{ 2 } { \rho v }^{ 2 }_{ 2 }\)
By equation of continunity,
\({ a }_{ 1 }{ v }_{ 1 }={ a }_{ 2 }{ v }_{ 2 }\quad or\quad { v }_{ 2 }=\frac { { a }_{ 1 }{ v }_{ 1 } }{ { a }_{ 2 } } \)
\(\\ \therefore \quad { \rho }_{ 1 }-{ \rho }_{ 2 }=\frac { 1 }{ 2 } \frac { \rho { a }^{ 2 }_{ 1 }{ v }^{ 2 }_{ 1 } }{ { a }^{ 2 }_{ 2 } } -\frac { 1 }{ 2 } \rho v^{ 2 }_{ 1 }\left[ \left( \frac { { a }^{ 2 }_{ 1 } }{ { a }^{ 2 }_{ 2 } } \right) -1 \right] \)
\(\\ Here,{ \rho }_{ 1 }-{ \rho }_{ 2 }=24Pa\)
\(\\ \rho (blood)=1.06\times 10^{ 3 }kgm^{ -3 },{ a }_{ 1 }/{ a }_{ 2 }=8/4=2\)
\(\\ \therefore { v }_{ 1 }=\sqrt { \frac { 2({ p }_{ 1 }-{ p }_{ 2 }) }{ \rho \left( \frac { { a }^{ 2 }_{ 1 } }{ { a }^{ 2 }_{ 2 } } -1 \right) } } =\sqrt { \frac { 2\times 24 }{ 1.06\times { 10 }^{ 3 }\times ({ 2 }^{ 2 }-1) } } =0.1228\quad m/s\)

неге \(D_1=2.0 \mathrm{~cm}=0.02 \mathrm{~m}\)
\(v_1=1.0 \mathrm{~ms}^{-1}, D_2=?, v_2=4.0 \mathrm{~ms}^{-1}\)
Fromequation of continuity \(a_1 v_1=a_2 v_2\)
\( \text { or } \frac{\pi D_1^2}{4} \times v_1=\frac{\pi D_2^2}{4} \times v_2 \)
\( \text { or } D_2^2=\frac{v_1}{v_2}=D_1^2=\frac{1}{4} \times(0.2)^2=(0.01)^2 \)
\( \text { or } D_2=0.01 \mathrm{~m}=1.0 \mathrm{~cm} \)

0.02m