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.

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 h = 1000 m and ρ = 1.03 x 10 ³ kg m^-3
Gauge pressure is P −P_a = ρgh = P_g
P _g = 1.03 x 10³ kg m^–3 x 10 ms² x 1000 m
= 10³ x 10⁵ Pa
≈ 103 atm

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

(a) Capillary action.
(b) Capillary rise will be halved as \(h\propto\frac{1}{r}\)
(c) Paraffin wax.

Volume of 1000 small drops = volume of a large drop          
 \(1000\times \frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi { R }^{ 3 }\)     
 r = \(\frac { R }{ 10 } \) 
Surface area of large drop =\(\frac { 4 }{ 3 } \pi { R }^{ 2 }\)
Surface area of 1000 drop = \(4\pi\times1000r^{2}\) = 40\(\pi R^{2}\)
Increase in surface area = (40-4)\(\pi R^{2}\) = 36\(\pi R^{2}\)
The increase in surface energy 
= Surface tension \(\times\)increase in surface area
= 36\(\pi R^{2}\)\(\times\)0.07 = 36\(\times\)3.14(4\(\times\)10^-3)²\(\times\)0.07
= 1.26\(\times\)10^-4 J

Given, surface tension of water, S = 7.28\(\times\)10^-2 N/m 
Vapour pressure, p = 2.33\(\times\)10³ Pa
The drop will evaporate if the water pressure is greater than the vapour pressure
Let a water droplet of radius R can be formed without evaporating.
\(\therefore\)Vapour pressure = Excess pressure in drop
\(\therefore\)    \(p=\frac { 2S }{ R } \)
or   \(R=\frac { 2S }{ p } \frac { 2\times 7.28\times { 10 }^{ -2 } }{ 2.33\times { 10 }^{ 3 } } \approx 6.25\times { 10 }^{ -5 }m\) 

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

Given r₁= 4 cm, r₂ = 6 cm, S = 30dyne cm^-1
So, change in surface area =2 × 4π(6²-4²)
[Soap solution has two surface]
= 8 π 20  = 160π cm²
Work done = S × change in surface area
= 30 × 160 × 31.42
= 15081.6 cm

When some is dissolved in water, the surface tension of water decreases. Thus, the less energy required of spray water.

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.

greater

Here, depth of hole below the water level h = 2.4 m, density of sea water p = 1.03 x 10³ kg m^-3 and surface area of hole A = 4 cm² = 4 x 10^-4 m².
\(\therefore\) Minimum force required to hold on a patch covering the hole from inside the ship
F = Pressure at height h of sea water column x Surface area of hole
= hpg A = 2.4 x 1.03 x 10³ x 9.8 x 4 x 10^-4
= 9.69 N.

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., \(v_c\propto\frac{1}{D}\)

Velocity of efflux through the hole, v = \( \sqrt {2gh}\)
\(\because\) Distance moved by water in one second v =  \(\sqrt {2gh}\)
\(\therefore\) rate of the momentum = ( \(\rho \)A\( \sqrt{2gh}\) )( \(\sqrt{2gh}\) )
= 2ghA\(\rho\)
According to Newton's second law of motion,
Force due to the velocity of efflux = 2 gh A\(\rho\)
Now, according to newtons third law of motion ,
Force on the vessel = Rate of the momentum 
Force on the vessel = 2 gh A\(\rho\)
The vessel will move, if force on the vessel = force of friction
or 2gh A\(\rho\) = \(\mu\)Mg
A = \(\frac{\mu M}{2h\rho} \)
=  \(\frac{0.4 \times100}{2\times2\times1000}\)
=\( \frac{1}{100}\)
Since, the hole is circular.
A = \(\pi\) r²
= \(\frac{\pi D^2}{4}\)
D = \(\sqrt{\frac{4A}{\pi}}\)
= \(\sqrt{\frac{4\times1}{100\times3.14}}\)
= 0.113 m 
So, diameter of a hole D = 0.113 m

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.