Class 11th

Speed of the wind on the upper surface of the wing, $V_1$ = 70 m/s

Speed of the wind on the lower surface of the wing, $V_2$ = 63 m/s

Area of the wing, A = 2.5 $m^2$

Density of air, $\rho$ = 1.3 kg/ $m^3$

According to Bernoulli's theorem, we have the relation:

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

( $P_2$ - $P_1$ )= $\frac{1}{2}\rho ({V_1}^2$ - ${V_2}^2)$ , where $P_1$ = pressure on the upper surface of the wing and $P_2$ - pressure on the lower surface of the wing

The pressure difference provides lift to the aeroplane

Lift on the wing = ( $P_2$ - $P_1$ )A = $\frac{1}{2}\rho ({V_1}^2$ - ${V_2}^2)$ A = $\frac{1}{2}*1.3*({70}^2$ - ${63}^2)*2.5$ N = 1512.8 N = 1.5 $*{10}^3$ N

11.19 (a) A body with a large reflectivity is a bad absorber. A bad absorber will in turn be a poor emitter of radiations.

(b) Brass is a good conductor of heat, when one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a colder value and one feels cold.

On the other hand, wood is a poor conductor of heat. Very little heat is conducted from the body to the wooden tray. Resulting in negligible drop in body temperature.

Thus a brass tumbler feels colder than a wooden tray on a chilly day.

(c) Black body radiation equation is given by:

E = $\sigma$ ( $T^4$ - $T_o^4$ ), where

E = energy radiation

$\sigma$ = constant

$T=$ temperature of the optical pyrometer

$T_o$ = Temperature of open space

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}$

When the same piece of iron placed in a furnace, the radiation energy is E = $\sigma T^4$

(d) In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from Earth's surface.

(e) A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat ( 540 cal/g)

Length of the horizontal tube, l = 1.5 m

Radius of the tube, r = 1 cm = 0.01 m

Diameter of the tube, d = 2r = 0.02m

Glycerine mass flow rate, M = 4 $*{10}^{-3}$ kg/s

Density of glycerine, $\rho$ = 1.3 $*{10}^3$ kg

Viscosity of glycerine, $\eta$ = 0.83 Pa-s

Now, volume of glycerine flowing per sec V = $\frac{M}{\rho }$ = $\frac{4*{10}^{-3}}{1.3\mathrm{}*{10}^3}$ $m^3$ /s = 3.08 $*{10}^{-6}$ $m^3$ /s

According to Poiseville's formula, we know the flow rate

V = $\frac{\pi *p*r^4}{8*\eta *l}$ where p is the pressure difference between two ends of the tube

p = $\frac{V*8*\eta *l}{\pi *r^4}$ = $\frac{3.08\mathrm{}*{10}^{-6}*8*0.83*1.5}{\pi *{\left(0.01\right)}^4}$ = 976.47 Pa

Reynolds's number is given by the relation

Re = $\frac{4\rho V}{\pi d\eta }$ = $\frac{4*1.3\mathrm{}*{10}^3*3.08\mathrm{}*{10}^{-6}}{3.1416*0.02*0.83}$ = 0.3

Since the Reynolds's number is 0.3, the flow is laminar

11.18 Base area of the boiler, A = 0.15 m2

Thickness of the boiler, l = 1.0 cm = 0.01 m

Boiling rate of water, R = 6.0 kg/min

Let us assume the mass of the boiling water, m = 6 kg and the time to boil, t = 1 min = 60 s

Thermal conductivity of brass, K = 109 J s–1 m–1 K–1

The amount of heat flowing into water through the brass base of the boiler is given by:

$\theta$ = $\frac{KA\left(T_1-T_2\right)t}{l}$ , where

$T_1$ = Flame temperature in contact with the boiler

$T_2$ = Boiling point of water = 100 $?$

Heat required for boiling water $\theta$ = mL, where L = heat of vaporization of water = 2256 * 103 J kg–1

By equating for $\theta$ we get

6 $*2256*{10}^3$ = 109 $*0.15*\left(T_1-T_2\right)*60/0.01$

$\left(T_1-T_2\right)$ = 137.98 $?$

$T_1$ = 237.98 $?\left(\mathrm{F}\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{b}\mathrm{o}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{r}\right)$

11.17 Size of the sides of cubical ice box, s = 30 cm =0.3 m

Thickness of the icebox, l = 5 cm = 0.05 m

Mass of ice kept in the box, m = 4 kg

Time gap, t = 6 h = 6 $*60*60$ s

Outside temperature, T = 45 °C

Coefficient of thermal conductivity of thermocole, K = 0.01 J $s^{-1}{m}^{-1}{K}^{-1}$

Heat of fusion of water, L = 335 $*{10}^3$ J ${kg}^{-1}$

Let m' be the mass of the ice melts in 6 h

The amount of heat lost by the food: $\theta$ = $\frac{KA\left(T-0\right)t}{l}$ , where

A = Surface area of the box = 6 $s^2$ = 6 $*{0.3}^2{m}^2$ = 0.54 $m^2$

$\theta$ = $\frac{0.01*0.54\left(45-0\right)*6\mathrm{}*60*60}{0.05}$ = 104976 J

We also know $\theta =m\text{'}L$ so m' = 104976/ (335 $*{10}^3)$ = 0.313 kg

Hence the amount of ice remains after 6 h = 4 – 0.313 = 3.687 kg