Let the atmospheric pressure be P and area of cross section of the tube be a.
Case (i): Length of air column = 30 cm
\(\therefore\) Volume of air, V₁ = (20 x a) c.c,
Length of mercury column = 100 - (30 + 15) = 55 cm.
Pressure P₁ of air in tube = (P - 55) cm
Case (ii): Length of air column = 20 cm.
\(\therefore\) Volume of air, V₂ = (20 x a) c.c.
Length of mercury column = 100 - (20 + 35) = 45 cm.
Pressure P₂ of air in tube = (P - 45) cm.
Applying Boyle's law,
P₁V₁ = P₂V₂
(P - 55) x 30 a =  (P - 45) x 20 a
Solving, we get, P = 75 cm.

Here  D = 10^-3 m
r = \({D\over 2}\) = 0.5 x 10^-3 m
g = 9.8 ms^-2, \(\theta\) = 0, T = 72 x 10^-3 Nm^-1
cos\(\theta\) = 1
Density of water at 4°C = 103 kg m^-3
Using the relation
 \(h=\frac{2T\cos\theta}{r\rho g}=\frac{2\times 72\times 10^{-3}\times 1}{5\times 10^{-4}\times 10^3\times 9.8}\) 
= 2.939 x 10^-2 m.

When a body floats in a liquid with a part submerged in the liquid, the weight of the liquid displaced by the submerged part is always equal to the weight of the body.
Let   V = volume of the body
 \(\sigma\) = density of its material
 \(\rho\) = density of the liquid in which the body floats such that its
volume V' is outside the liquid.
Then volume of the body inside the liquid = V - V'
Weight of the displaced liquid = (V - V') Pg
Also weight of the body = V \(\sigma\) g
For the body to float,
weight of the liquid displaced by the submerged part = weight of the body
i.e., (V - V') \(\rho g\) = V \(\sigma\) g
or      \(V'=\frac{(\rho-\sigma)V}{\rho}\) 

Surface energy of first bubble
= Surface area x surface tension
= 2 x 4\(\pi r_1^2\)T = 8\(\pi r_1^2\)T
Surface energy of second bubble = 8n\(\pi r_2^2\)T
Let r be the radius of the coalseced bubble.
\(\therefore\) Surface energy of new bubble = 8\(\pi\)r²T
According to the law of conservation of energy,
8\(\pi\)r²T = 8\(\pi r_1^2\)T + 8\(\pi r_2^2\)T
= \(8\pi(r_1^2 + r_2^2)T\)
\(\therefore\)   r² = \(r_1^2+r_2^2\) = 3² + 4² = 9 + 16 = 25
\(\therefore\)   r = 5 cm.

Increase in surface area = 2[4\(\pi\)(3r)² - 4\(\pi\)r²]
Increase in surface energy = \(\sigma\) x 2 x 4\(\pi\) x 8r² = 8W
Additional work done = 8W.

The special design of the wings increases velocity at the upper surface and decreases velocity at the lower surface. So, according to Bernoulli's theorem, the pressure on the upper side is less than the pressure on the lower side. This difference of pressure provides lift.

Suppose the mass of the metal of specific gravity 11.4 be m. Now the mass of the second metal of specific gravity 7.4 will be (96 - m).
Volume of first metal = \(\frac{m}{11.4}cm^3\)
Volume of second metal = \(\frac{96-m}{7.4}cm^3\)
Total volume = \(\frac{m}{11.4}+\frac{96-m}{7.4}\)
Apparent loss of wt. in water = \((\frac{m}{11.4}+\frac{96-m}{7.4})\) gm wt.
Apparent wt. in water = \(96-[(\frac{m}{11})+\frac{(96-m)}{7.4}]\)
According to the given problem,
\(96-[(\frac{m}{11})+\frac{(96-m)}{7.4}]=86\)
or \(\frac{m}{11.4}+\frac{(96-m)}{7.4}=10\)
Solving we get,   m = 62.7 gm.
\(\therefore\) Mass of second metal = 96 - 62.7 = 33.3 gm.

Given     \(\frac{4T}{r_1}=\frac{3\times 4T}{r_2}\) or r₂ = 3r₁
\(\frac{V_1}{V_2}=\frac{(\frac{4}{3})\pi r_1^3}{(\frac{4}{3})\pi r_2^3}=(\frac{r_1}{r_2})^3=(\frac{1}{3})^3=\frac{1}{27}\)

Height of water column, h₁ = 10.0 cm
Density of water, \({ \rho }_{ 1 }\) = 1g/ cm 3
Height of spirit column, h₂ = 12.S cm
Density of spirit,\({ \rho }_{ 2 }\) =?
The mercury column in both arms of the U-tube are at same level, therefore, pressure in both arms will be same.
Pressure exerted by water column = pressure exerted by spirit column
p₁= p₂
\({ h }_{ 1 }{ \rho }_{ 1 }g={ h }_{ 2 }{ \rho }_{ 2 }g\ or\ { \rho }_{ 2 }=\frac { { h }_{ 1 }{ \rho }_{ 1 } }{ { h }_{ 2 } } =\frac { 10\times 1 }{ 12.5 } =0.80g/{ cm }^{ 3 }\) 
Specific gravity of spirit=\(\frac { Density\ of\ spirit }{ Density\ of\ water } =\frac { 0.80 }{ 1 } =0.80\) 

Let the speed of the flow be v.
Given, diameter of tap = d = 1.25 cm
Volume of water flowing out per second.
Q = v \( \times\) \(\frac{\pi d^2}{4} \) \(\Rightarrow\) v = \(\frac{4Q}{d^2\pi}\)
Estimate Reynold's number, Re = \(\frac{4 \rho Q}{\pi d \eta}\)
Q = 0.48 L / min 
= 8 \(\times\) 10^-3 L/s
= 8 \(\times\)10^-6 m³/s
Re =  \(\frac{4\times10^3\times8\times10^{-6}}{3.14\times1.25\times10^{-2}\times10^{-3}} \)
Re = 815                        [ i.e below 1000, the flow is steady ]
After some time, when
Q = 3L/min 
= 5\(\times\)10^-5m³/s,
Re =  \(\frac{4\times10^3\times5\times10^{-5}}{3.14\times1.25\times10^{-2}\times10{-3}} \)
= 5095
ஃ The flow will be turbulent.

Volume of a bigger drop = n \(\times\) Volume of a smaller droplet
or\( \frac{4}{3} \) \(\pi\) R₃ = n\(\times\) \(\frac{4}{3}\) \(\pi \)r³  or R³ = nr³
or R = n^1/3 r
Terminal velocity of a small droplet is given by
v_s = \(\frac{2}{9}\)\(\frac{R^2}{\eta}\)( \(\rho - rho ^{'}\) ) g                 .........(i)
Terminal velocity of a bigger drop is given by
v_b = \(\frac{2}{9}\)\(\frac{r^2}{\eta}\)(\(\rho - rho^{'}\))g                 .............(ii)
Dividing Eq. (ii) by Eq. (i), we get
\(\frac{v_b}{v_s}\) = \(\frac{R^2}{r^2}\)
But   R = n^1/3 r
and  v_s = 10 cm/s
vb = v_s \( \times\) ( \(\frac{R^2}{r^2}\))
= 10 \( \times\) \( \frac{n^{2/3}r^2}{r^2}\)
v_b = 10 n^2/3 cm/s

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

(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 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 \(v \propto{\text { (radius })^2}^2\)
\( \therefore  \frac{v_2}{v_1}=\frac{R^2}{r^2}=\frac{(3 r)^2}{r^2}=9 \)
\( \therefore  \frac{v_2}{0.15}=9 \ \therefore \ v_2=9 \times 0.15=1.35 \mathrm{~m} / \mathrm{s}\)