Evaporation causes cooling and the rate of evaporation increases with an increase in the surface area. Since, saucer has a large surface area, hence tea/  coffee taken in a saucer cools quickly.

The cohesive forces in mercury are much stronger than the force of adhesion between glass and mercury. Therefore, mercury-glass contact angle is greater than 90^o . As a result, the vertical component of the surface tension forces acts vertically downward, thereby lowering the level of mercury column in the capillary tube.

A liquid boils when its vapour pressure becomes equal to the atmospheric pressure. An increase in pressure on liquid, therefore, causes a rise in the boiling temperature of the liquids.

Radius of the balloon = 10 m
\(\therefore\)Volume of the balloon = \(\frac{4}{3}\pi\)r² = \(\frac{4}{3}\times\frac{22}{7}\times\)(10m)³ = 4190.5 m³
Volume of He filled at 1.66 bar and 27°C = 4190.5 m3
Calculation of mass of He
PV = nRT = \(\frac { w }{ M } \)RT
or \(w\) = \(\frac{MPV}{RT}\) = \(\frac { (4\times { 10 }^{ -3 }kgmol^{ -1 })(1.66bar)(4190.5\times 103dm^{ 3 }) }{ (0.083bardm^{ 3 }K^{ -1 }mol^{ -1 })(300K) } \)
= 1117.5 kg
Total mass of the balloon along with He = 100 + 1117.5 = 1217.5 kg
Maximum mass of the air that can be displaced by balloon to go up = Volume x Density
= 4190.5 m³ x 1.2 kg m^-3 = 5028.6 kg
\(\therefore\) Pay load = 5028.6 - 1217.5 kg = 3811.1 kg

It states that, the pressure of a fixed mass of a gas is inversely proportional to its volume if temperature is kept constant.
P\(\alpha\)\(\frac{1}{V}\)
PV = constant (n and T are constant)
P₁V₁ = P₂V₂.
Graphical representation:
        
Fig. Graph of pressure, P vs. Volume,       
V of a gas at different temperatures.

 Fig. Graph of pressure of a gas, P VS. 1N

Weight of LPG originally present = 29.0 - 14.8 = 14.2 Kg
Pressure = 2.5 atm
Weight of LPG present after use = 23.2 - 14.8 
= 8.4 Kg
Since volume of the cylinder is constant, applying
pV = nRT    \(\cfrac { { p }_{ 1 } }{ { p }_{ 2 } } =\cfrac { { n }_{ 1 } }{ { n }_{ 2 } } =\cfrac { { w }_{ 1 }/m }{ { w }_{ 2 }/m } =\cfrac { { w }_{ 1 } }{ { w }_{ 2 } } \)
\(\cfrac { 2.5 }{ { p }_{ 2 } } =\cfrac { 14.2 }{ 8.4 } \)  or p₂ = \(\cfrac { 2.5\times 8.4 }{ 14.2 } \)  = 1.48 atm
Weight of used gas  = 14.2 - 8.4 = 5.8 Kg
Moles of gas = \(\cfrac { 5.8\times { 10 }^{ 3 } }{ 58 } \)= 100mol
Normal Conditions p = 1 atm;
t = 273 + 27 = 300K
Volume of 100m L of LPG at 1 atm and 300 K
V = \(\cfrac { nRT }{ p } =\cfrac { 100\times 0.082\times 300 }{ 1 } \)
= 2460 L = 2.460 m³

The volume of a sphere = \(\frac { 4 }{ 3 } \pi { r }^{ 2 }\) where r is radius of the sphere. For N₂ molecule,
\(r=0.2nm=0.2\times { 10 }^{ -9 }m=2\times { 10 }^{ -8 }cm\)
Volume of a molecule of
\({ N }_{ 2 }=\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times (2\times { 10 }^{ -8 })^{ 3 } \ cm^{ 3 }=3.35\times { 10 }^{ -23 }{ cm }^{ 2 }\)

Total K.E = 74826 J Average K.E, K.E = 6.1 x 10^-21 J mole^-1 J

^pN₂ = 8.71 bar; ^pO₂ = 0.778

Density at STP = 3 g/dm³