physics ncert solutions class 11th

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

11.16 Initial body temp of the child,  $T_1$ = 101°F

Final body temp of the child,  $T_2$ = 98°F

Change in temperature,  $?$ T = $T_1-T_2=(101°\mathrm{F}-\mathrm{}$ 98°F) = 3 °F = $\frac{5}{9}$  (3-32) $°$ C = 1.666 $?$

Time taken t achieve this temperature, t = 20 min

Specific heat of human body = Specific heat of water, c = 1000 cal/kg/ $?$

Latent heat of evaporation of water, L = 580 cal/g

Mass of the child, m = 30 kg

The heat lost by the child is given as $?\theta =mc?\mathrm{T}$ = 30 $*1000*1.666$ = 49980 cal

Let $m_1$ be the mass of water evaporated from the child's body in 20 mins.

Loss of heat $?\theta$ = $m_1*L$ = 580 $m_1$ = $49980cal$

$m_1=$ 86.2 gm

Rate of evaporation = $\frac{m_1}{t}$ = 4.31 g/min

11.15 The gases listed above are diatomic. Besides the translational degree of freedom, they have other degrees of freedom. Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the modes of motion. Hence the molar specific heat of diatomic gases is more than that of monatomic gases.

If only rotational mode of motion considered, then the molar specific heat of a diatomic gas

= $\frac{5}{2}$ R = $\frac{5}{2}*1.98$ = 4.95 cal mo1^–1 K^–1

With the exception of Chlorine, all the observations given above agrees with ( $\frac{5}{2}$ R). This is because at room temperature, chlorine also has vibrational modes besides rotational and translational modes of motion.