Given, angle of contact (\(\theta\)) =140⁰ 
Radius of tube (r) = 1mm = 10^-3 m
Surface tension (S) = 0.465 N/m
Density of mercury (\(\rho\)) = 1.36\(\times\)10³ kg/m³
Height of liquid rise or fall due to surface tension (h)
\(\frac { 2S\quad cos\theta }{ r\rho g } =\frac { 2\times 0.465\times cos\quad { 140 }^{ 0 } }{ 1\times { 10 }^{ -3 }\times 13.6\times { 10 }^{ 3 }\times 9.8 } \)
\(\\ =\frac { 2\times 0.465\times (-0.7660) }{ { 10 }^{ -3 }\times 13.6\times { 10 }^{ 3 }\times 9. } =5.34\times { 10 }^{ -3 }m\)
\(\\ =-5.34\ mm\)
\(\\ Hence,\ the\ mercury\ level\ wil\ depressed\ by\ 5.34\ mm\)

Given, radius of drops(R) = 3.00 mm = 3\(\times\)10^-3m
Surface tension of mercury (S) = 4.65 \(\times\)10^-1N/m
Atmospheric pressure (p₀) = 1.01\(\times\)10⁵ Pa
Pressure inside the drop = Atmospheric pressure +Excess pressure in side the liquid drop = p₀ + \(\frac { 2S }{ R } \)
\(=1.01\times { 10 }^{ 5 }+\frac { 2\times 4.65\times { 10 }^{ -1 } }{ 3\times { 10 }^{ -3 } } \)
= 1.01\(\times\)10⁵ + 3.10\(\times\)10²
= 1.01\(\times\)10⁵+ 0.00310\(\times\)10⁵
= 1.01310\(\times\)10⁵Pa
Excess pressure inside the drop 
\((\Delta p)\frac { 2S }{ R } =\frac { 2\times 4.65\times { 10 }^{ -1 } }{ 3\times { 10 }^{ -3 } } \)
= 3.10\(\times\)10² =310 Pa

\(Here,\ p=4\times { 10 }^{ 5 }N/{ m }^{ 2 }\ and\ \rho =1000\ kg{ m }^{ -3 },\ g=10m/{ s }^{ 2 }\)
\(\\ Apply\ p=h\rho g\)
\(\\ \Rightarrow \ h\ =\frac { p }{ \rho g } =\frac { 4\times { 10 }^{ 5 } }{ 1000\times 10 } \)
\(\\ Velocity\ of\ efflux,\ v=\sqrt { 2gh } =\sqrt { \frac { 2\times 10\times 4{ \times 10 }^{ 5 } }{ 1000\times 10 } }\)
\( \\=\sqrt { 800 } =8.28m/s\)

Given , S = 0.0073 N/m
R = 0.005 cm 05\(\times\)10^-4m
\(\theta\) = 0⁰ , h = ?
\(From\ the\ formula,h=\frac { 2cos\theta }{ \rho gr } \)
\(\\ =\frac { 2\times 0.073\times cos{ 0 }^{ 0 } }{ { 10 }^{ 3 }\times 9.8\times 5{ \times 10 }^{ -4 } }\)
= 0.02979m

A fluid flowing out of small hole in vessel have a large velocity and therefore, a large momentum. As no external force is acting, therefore according to law of conservation of momentum equal momentum in attained by the vessel. Therefore, a backward thrust \(\left( F=\frac { dp }{ dt } \right) \) acts on the vessel.

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

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

Open tube manometer is used for measuring pressure difference.

If the volumes are same, then height of the liquid will be highest in which the cross-sectional area is least at the top. So, the vessel having least cross-sectional area at the top possess more pressure at the base(∵p=ρgh)

We have vt = 6.5 × 10^-2 ms^-1 , a = 2 × 10^-3 m, g = 9.8 ms^-2 , ρ = 8.9 × 10³ kg m^-3 , σ =1.5 ×10³ kg m^-3. From Eq
\(\eta =\frac{2}{9} \times \frac{\left(2 \times 10^{-3}\right)^2 \mathrm{~m}^2 \times 9.8 \mathrm{~m} \mathrm{~s}^{-2}}{6.5 \times 10^{-2} \mathrm{~m} \mathrm{~s}^{-1}} \times 7.4 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} \)
\(=9.9 \times 10^{-1} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1} \)

If the angle of the content between the water and the material of the raincoat is obtuse, then rainy water does not wet the rain coat. Thus the raincoat becomes waterproof.

increases, decreases

Here D = 0.5 cm = 0.005 m
\(\rho\) = 103 kg m-3
\(\eta \) = 0.00125 Nm^-2 s
For laminar flow, the Reynold number for water
NR = 2000
Let v be the maximum average velocity
\(\therefore\)     \(N_R=\frac{\rho v D}{\eta}\)
or       \(v=\frac{N_R.\eta}{\rho.D}\)
 \(=\frac{2000\times 0.00125}{1000\times 0.005}\)
= 0.5 m/s.

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

Given, v₂ = 120 m/s, v₁ = 90m/s, \(\rho\)  = 1.3 kg/m³
h₁ = 10 m , a₁ = 10\( \times\) 2 = 20 m²
According to Bernoulli's theorem,
\(\frac{p_1}{\rho} \) + gh₁ +\( \frac{1}{2}\) \(v_{1}^{2} \) = \( \frac{p_2}{\rho}\)  + gh₂ + \(\frac{1}{2 }\) \(v_{1}^{2}\)
For the horizontal flow, h₁ = h₂
\(\therefore \) \(\frac{p_1}{\rho} \)+ \( \frac{1}{2}\) \(v_{1}^{2} \) = \( \frac{p_2}{\rho}\) + \(\frac{1}{2}\) \( v_{2}^{2}\)
Given, V₁ =90 m / s , v₂ =120 m / s , P = 1.3 kg / m³
\(\frac{p_1 - p_2}{\rho}\) = \( \frac{1}{2}\)  \((v_{2}^{2}- v_{1}^{2})\)
 ( p₁ -p₂ ) = \(\frac{ \rho (v_{2}^{2}- v_{1}^{2})}{2} \)
= 1.3 \(\times \) \(\frac{(14400-8100)}{2} \)
=  \(\frac{1.3\times6300}{2}\)
p₁ - p₂ = 4.095 \(\times \)10³ N/m²
It is the pressure difference between the top and the bottom of the wing.
Gross lift of wing = ( p₁ - p₂ )\(\times \) Area of the wing 
= 4.095 \(\times \)10³ \(\times \)10 \(\times \)2
= 8.190\(\times \)10⁴ N

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