Class 11th

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.

11.14 Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal,  $T_1$ = 150 °C, Final temperature of the metal,  $T_2$ = 40 °C

The water equivalent mass of the calorimeter, m' = 0.025 kg = 25 g

Volume of water, V = 150 cm3

Mass of water, M at T = 27 °C, = 150 $*1=150g$

Fall in metal temperature,  $?$ T = $T_1-T_2$ = 150 – 40 = 110 °C

Specific heat of water,  $C_w$ = 4.186 J/g/ $°$ K

Let the specific heat of metal = C

Then, heat loss by the metal,  $\theta$ = mC $?$ T ……. (i)

Rise in the water of the calorimeter system $?$ T' = 40 – 27 = 13°C

Heat gained by the water and calorimeter system; $\theta \text{'}$ = (M + m') $C_w?$ T' …… (ii)

Now heat lost by the metal = heat gained by the water and calorimeter system

mC $?$ T = (M + m') $C_w?$ T'

C = $\frac{(\mathrm{M}\mathrm{}+\mathrm{}\mathrm{m}\mathrm{\text{'}})C_w?\mathrm{T}\mathrm{\text{'}}}{m?\mathrm{T}}$ = $\frac{(150+25)*4.186*13}{200*110}$ = 0.433 J/g/ $°$ K

11.12 Power of the drilling machine, P = 10 kW= 10 $*{10}^3$ W

Mass of the Aluminium block, m = 8.0 kg = 8 $*{10}^3$ gm

Time for which the machine is used, t = 2.5 minute = 2.5 $*60$ s = 150 s

Specific heat of Aluminium, c = 0.91 J/gK

Let the rise of temperature in the block after drilling be $\delta$ T

Total energy consumed by the drilling machine= P $*t$ = 10 $*{10}^3*150$ J = 1.5 $*{10}^6$ J

It is given 50% of energy is useful.

So useful energy,  $?Q$ = 50% of Pt= 0.5 $*$ 1.5 $*{10}^6=$ 7.5 $*{10}^5$ J

We know,  $?Q$ = mc $?$ T or T = $\frac{?Q}{mc}$ = $\frac{7.5\mathrm{}*{10}^5}{8\mathrm{}*{10}^3*0.91}$ = 103 $?$

Therefore 2.5 minute drilling will raise the block temperature by 103 $?$

When air is blown under a paper, the velocity of air is more than the upper portion of the paper. As per Bernoulli's principle, atmospheric pressure reduces under the paper and makes it fall. To keep the paper horizontal, the air needs to be blown on the upper surface of the paper.

For a smaller opening, the flow of fluid is more than when it is bigger. When we try to close the tap with our fingers, water gushes through the small openings. Area and velocity are inversely proportional to each other.

Small opening of a syringe needle controls the velocity of the blood oozing out. At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure.

When a fluid flows out from a small hole in a vessel, the vessel receives backward thrust, when fluid flows out from a small hole, it attains high velocity. Due to the law of conservation of momentum, the vessel attains backward velocity.

A spinning cricket ball in air has two simultaneous motions – rotary and linear. These two types of motion, opposes each other. The velocity below the ball decreases, the pressure on upper side becomes lesser than the lower side of the ball. An upward force acts on the ball and it takes a curved path, does not follow parabolic path.