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.

\(According\ to\ the\ equation\ of\ continutity,\ we\ get\)
\(\\ { a }_{ 1 }{ v }_{ 1 }={ a }_{ 2 }{ v }_{ 2 }\)
\(\\ \Rightarrow \ \pi { r }_{ 1 }{ v }_{ 1 }=\pi { r }_{ 2 }{ v }_{ 2 }\)
\(\\ \therefore { v }_{ 2 }={ \left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) }^{ 2 }\ { v }_{ 1 }\)
\(\\ Given,\ { r }_{ 1 }=\frac { 4 }{ 2 } =2cm=2\times { 10 }^{ -2 }m\)
\({ r }_{ 2 }=\frac { 6 }{ 3 } =3cm\ =3\times { 10 }^{ -2 }m\)
\( { v }_{ 1 }=5m/s\)
\({ v }_{ 2 }={ \left( \frac { 2 }{ 3 } \right) }^{ 2 }\times 5=2.22m/s\)
\(\\ Now,\ applying\ the\ Bernoulli's\ theorem\)
\(\\ { p }_{ 1 }+\frac { 1 }{ 2 } { \rho { v } }_{ 1 }^{ 2 }={ p }_{ 2 }\frac { 1 }{ 2 } { \rho v }_{ 2 }^{ 2 }\)
\({ p }_{ 2 }={ p }_{ 1 }+\frac { 1 }{ 2 } \rho ({ { v } }_{ 1 }^{ 2 }+{ v }_{ 2 }^{ 2 })\)
\(=3.0\times { 10 }^{ 4 }+\frac { 1 }{ 2 } \times { 10 }^{ 3 }({ 5 }^{ 2 }-{ 2.22 }^{ 2 })\)
\(=3\times { 10 }^{ 4 }+\frac { 1 }{ 2 } \times { 10 }^{ 4 }+500\times 20.08\)
\( { p }_{ 2 }=4\times { 10 }^{ 4 }N/{ m }^{ 2 }\) 

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

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 \(a_1 v_1=a_2 v_2\)
\( \text { or } \frac{\pi D_1^2}{4} \times v_1=\frac{\pi D_2^2}{4} \times v_2 \)
\( \text { or } D_2^2=\frac{v_1}{v_2}=D_1^2=\frac{1}{4} \times(0.2)^2=(0.01)^2 \)
\( \text { or } D_2=0.01 \mathrm{~m}=1.0 \mathrm{~cm} \)

D = 2 cm = 0.02 m
\(\rho\) = 10³ kg m^-3
\(\eta\) = 0.001 Nm^-2 s = 10^-3 Nm^-2s
Flow of water will be laminar if
N_R = 1000 where N_R is Reynold number
Let v = maximum average velocity
\(\therefore\) Using the relation
 \(N_R=\frac{\rho vD}{\eta}\)
or   \(v=\frac{N_R\eta}{\rho D}=\frac{1000\times 0.001}{1000\times 0.02}\) = 0.5 ms^-1.

Critical velocity (v_c) of a liquid is:
(i) directly proportional to the coefficient of viscosity of the liquid.
(ii) inversely proportional to the density of the liquid i.e., \(v_c\propto\frac{1}{\rho}\)
(iii) inversely proportional to the diameter of the tube through which it flows i.e., \(v_c\propto\frac{1}{D}\)

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 R1 and R2 be the radii of the two bubbles.
Then, R1/R2 =2/3
If S is the surface tension of soap solution, then excess of pressure inside the soap bubble of radius R1 is \(p_1=\frac{4 S}{R_1}\)
Excess of pressure inside the soap bubble of
\( \text { radius } R_2 \text { is } p_2=\frac{4 S}{R_2} \)
\(\therefore \frac{p_1}{p_2}=\frac{R_2}{R_1}=\frac{3}{2}\)
Work done in blowing up the soap bubble is
W=S xsurface area of bubble
\( \therefore W_1=S \times\left(2 \times 4 \pi R_1^2\right) \)
\( \text { and } W_2=S \times\left(2 \times 4 \pi R_2^2\right) \)
\( \frac{W_1}{W_2}=\frac{R_1^2}{R_2^2}=\left(\frac{2}{3}\right)^2=\frac{4}{9} .\)

Let R, and r be the radii of the big and small drops respectively.
The volume of the big drop =27 x volume of one small drops
\(\therefore \frac{4}{2} \pi R^3=27 \times \frac{4}{3} \pi r^3 \)
\( \therefore R^3=27 r^3 \ \therefore R=3 r\)
The terminal velocity of a drop \(v \propto{\text { (radius })^2}^2\)
\( \therefore  \frac{v_2}{v_1}=\frac{R^2}{r^2}=\frac{(3 r)^2}{r^2}=9 \)
\( \therefore  \frac{v_2}{0.15}=9 \ \therefore \ v_2=9 \times 0.15=1.35 \mathrm{~m} / \mathrm{s}\)

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

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.

This is done to reduce the backward drag of the atmosphere.

Due to change in viscosity with temperature.

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 =\(\quad \frac { thrust }{ area } =\frac { 294N }{ 0.02 } =14700N\)