Class 11th

The area of the wing of the plane, A = 2 $*25m^2$ = 50 $m^2$

Speed of air over the lower wing, $V_1$ = 180 km/h = 50 m/s

Speed of air over the upper wing, $V_2$ = 234 km/h = 65 m/s

Density of air, $\rho$ = 1 kg/ $m^3$

Let us assume, pressure over the lower wing = $P_1$ and pressure over the upper wing = $P_2$

The upward force on the plane can be obtained using Bernoulli's equation:

On lower wing= $P_1$ + $\frac{1}{2}\rho V_1^2$ , On upper wing = $P_2$ + $\frac{1}{2}\rho V_2^2$

From equilibrium of momentum

$P_1$ + $\frac{1}{2}\rho V_1^2$ = $P_2$ + $\frac{1}{2}\rho V_2^2$

( $P_1$ - $P_2)=\frac{1}{2}\rho (V_2^2-V_1^2$ )

The net upward force = ( $P_1$ - $P_2)A$ = $\frac{1}{2}\rho (V_2^2-V_1^2$ ) $*A$

= $\frac{1}{2}1({65}^2-{50}^2$ ) $*50$ = 43125 N

Using the relation F = mg, we get m = F/g

= (43125/9.8) Kg = 4400.51 kg

Hence the mass of the plane is about 4400 kg.

Atmospheric pressure,  $P_0$ = 76 cm of Hg

For figure (a)

Difference between the levels of mercury in the two limbs gives gauge pressure

Hence, gauge pressure = 20 cm of Hg

Absolute pressure = Atmospheric pressure + Gauge pressure = 76 + 20 = 96 cm of Hg

For figure (b),

Difference between the levels of mercury in the two limbs gives gauge pressure

Hence, gauge pressure = - 18 cm of Hg

Absolute pressure = Atmospheric pressure + Gauge pressure = 76 - 18 = 58 cm of Hg

When 13.6 cm of water is poured into the right limb of figure (b)

Relative density of mercury = 13.6

Hence, a column of 13.6 cm of water is equivalent to 1 cm of Mercury.

Let h be the difference between the levels of mercury in the two limbs

The pressure in the right limb is given as:

$P_g$ = Atmospheric pressure + 1 cm of Hg = 76 + 1 = 77 cm of Hg

The mercury column will rise in the left limb. Hence, pressure in the left limb

$P_l$ = 58 + h

Since $P_g$ = $P_l$ , we get h = 77 – 58 = 19 cm

11.22

The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporization curves. It departs from ideal gas behavior as pressure increases.

(a) At 1 atm pressure and at – 60 °C, it lies left of -56.6 $?$ (triple point O). Hence, it lies in the region of vapour and solid phases. Thus,  CO₂  condenses into solid state directly, without going through liquid phases.

(b) At 4 atm pressure,  CO₂ lies below 5.11 atm (triple point O). Hence it lies in the region of vaporous and solid phases. Thus, CO₂ condenses into solid state directly, without going through liquid state.

(c) When the temperature of a mass of solid  CO₂ ( at 10 atm pressure and at -65 $?$ is increased, it changes to the liquid phase and then to the vaporous phase. It forms a line parallel to the temperature axis at 10 atm. The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporization curves.

(d) If CO2 is heated to 70 $?)$ compressed isothermally, then it will not exhibit any transition to the liquid state. This because 70 $?\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{}$ is higher than the critical temperature of  CO₂ . It will remain in the vapour state, but will depart from its ideal behavior as pressure increases.

Base area of the given tank, A = 1.0 $m^2$

Area of the hinged door, a = 20 ${cm}^2$ = 20 $*{10}^{-4}$ $m^2$

Density of water, ${\rho }_1$ = ${10}^3$ kg/ $m^3$

Density of acid, ${\rho }_2$ = 1.7 $*{10}^3$ kg/ $m^3$

Height of the water column, $h_1$ = 4 m

Height of the acid column, $h_2$ = 4 m

Acceleration due to gravity, g = 9.8 m/ $s^2$

The pressure due to Water column, $P_1$ = ${\rho }_{1}g{h}_1$ = ${10}^3*9.8*4$ = 3.92 $*{10}^4$ Pa

The pressure due to Acid column, $P_2$ = ${\rho }_{2}g{h}_2$ = 1.7 $*{10}^3*9.8*4$ = 6.64 $*{10}^4$ Pa

Pressure difference between two columns ΔP = $P_2$ - $P_1$ = 2.774 $*{10}^4$ Pa

Force exerted on the small hinged door = ΔP $*a$ = 2.774 $*{10}^4*$ 20 $*{10}^{-4}$ N = 55.48 N

Soap bubble radius, r = 5.0 mm = 5 $*{10}^{-3}$ m

Surface tension of the soap bubble, S = 2.50 $*{10}^{-2}$ N/m

Relative density of soap solution = 1.20, hence density of soap solution,  $?$ = 1.2 $*{10}^3$ Kg/ $m^3$

Air bubble formed at a depth, h = 40 cm = 0.4 m

1 atmospheric pressure = 1.01 $*{10}^5$ Pa

Acceleration due to gravity, g = 9.8 m/ $s^2$

We know, the excess pressure inside the soap bubble is given by the relation:

P = $\frac{4S}{r}$ = $\frac{4*2.50*{10}^{-2}}{5*{10}^{-3}}$ Pa = 20 Pa

Hence, the excess pressure inside the air bubble is given by the relation, P' = $\frac{2S}{r}$ = 10 Pa

At a depth of h, the total pressure inside the air bubble = Atmospheric pressure + h $?$ g +P'

= 1.01 $*{10}^5$ + (0.4 $*1.2*{10}^3*9.8)$ + 10 Pa = 105714 Pa = 1.06 $*{10}^5$ Pa

11.21 

(a) The P-T phase diagram for CO₂ is shown here. O is the triple point of the CO₂ phase diagram. This means that at the temperature and pressure corresponding to this point, the solid, liquid and vaporous phases of CO₂ exists in equilibrium.

(b) The fusion and the boiling points of CO₂ decreases with a decrease in pressure.

(c) The critical temperature and critical pressure of CO₂ are 31 $?$ 72.9 atm respectively. Even if it is compressed to a pressure greater than 72.9 atm, CO₂ will not liquefy above the critical temperature.

(d) It can be concluded from the P-T phase diagram of CO2 that:

CO₂ is gaseous at -70 $?$, under 1 atm pressure.

CO₂ is solid at -60 $?$ , under 10 atm pressure.

CO2 is liquid at 15 $?$ , under 56 atm pressure.