physics ncert solutions class 11th

The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact ( $\theta )$ , as shown in the diagram

$S_{la}$ = Interfacial tension between liquid-air interface

$S_{sl}$ = Interfacial tension between solid -liquid interface

$S_{sa}$ = Interfacial tension between solid-air interface

At the line of contact of contact, the surface forces between the three media must be in equilibrium. Hence

$\cos ?\theta$ = $\frac{S_{sa}-S_{la}}{S_{la}}$

The angle of contact $\theta$ is obtuse, if $S_{sa}<S_{la}$ , as in the case of mercury on glass

This angle is acute if $S_{sl}<S_{la}$ , as in the case of water on glass

Mercury molecules (which makes an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction towards solids. Hence, they tend to form drops

Water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction towards solids. Hence, they tend to spread out

Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is independent of the area of the liquid surface.

Water with detergent dissolved in it has a small angle of contact( $\theta ).$ This is because for a small $\theta ,$ there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportion to the cosine of the angle $\theta .$ If $\theta$ is small, then cos $\theta$ will be large and then the rise of detergent in the cloth will be faster.

A liquid tend to acquire the minimum surface area because of the presence of surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape.

The pressure of a liquid is given by the following relation:

P = h $\rho$ g, where P = Pressure, h = height of the liquid column,  $\rho$ = is the density of the liquid and g= acceleration due to gravity

From the above relation, because of the h factor (height of the human body), the pressure is more at the feet and less at the brain

The said phenomenon is due to the $\rho$ factor. Density of air is maximum at the sea level. At height, density decreases and pressure also decreases. At 6 km height, the density of air is nearly half of that of a t sea level

When pressure is applied on the liquid, the pressure is transmitted in all directions in the liquid, Hence in absence of fixed direction, it is a scalar quantity

11.5 (a) For Thermometer A

Triple point of water, T = 273.16 K

At this temperature, the pressure in thermometer A , $P_A$ = 1.250 $*{10}^5$ Pa

Let $T_1$ be the temperature for the normal melting point of sulphur and $P_1$ be the corresponding pressure. It is given, $P_1$ = 1.797 $*{10}^5$ Pa

From Charles' law, we get $\frac{P_A}{T}$ = $\frac{P_1}{T_1}$ , $T_1$ = $\frac{P_1*\mathrm{T}}{P_A}$ = $\frac{1.797\mathrm{}*{10}^5*273.16}{1.250\mathrm{}*{10}^5}$ = 392.69 K

For Thermometer B

Triple point of water, T = 273.16 K

At this temperature, the pressure in thermometer B , $P_B$ = 0.2 $*{10}^5$ Pa

Let $T_1$ be the temperature for the normal melting point of sulphur and $P_1$ be the corresponding pressure. It is given, $P_1$ = 0.287 $*{10}^5$ Pa

From Charles' law, we get $\frac{P_B}{T}$ = $\frac{P_1}{T_1}$ , $T_1$ = $\frac{P_1*\mathrm{T}}{P_B}$ = $\frac{0.287\mathrm{}*{10}^5*273.16}{0.2\mathrm{}*{10}^5}$ = 391.98 K

(b) The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gas, hence there is a slight difference in readings.

(c) To reduce the difference between two readings, the measurement should be carried out at low pressure so that gases behave like an ideal gas.

11.4 (a) The triple point of water has a unique value of 273.16 K, irrespective of pressure and volume. Whereas, melting point of ice and boiling point of water, the temperature value depends on pressure and volume.

(b) The other fixed point on Kelvin scale is 0 K.

(c) The temperature 273.16 K is the triple point of water, it is not the melting point of ice. The melting point of ice is specified in Celsius scale as 0 $°C.$ Hence the absolute temperature in Kelvin scale, $T_k$ is related to temperature $T_c$ in Celsius scale as

$T_c=T_k-273.15$

(d) Let $T_f$ and $T_k$ be the temperature in Fahrenheit and absolute scale. From the co-relation of the scale we know

$\frac{T_f-32}{9}$ = $\frac{T_k-273.15}{5}$ ……(i)

Let $T_{f1}$ and $T_{k1}$ be the temperature in Fahrenheit and absolute scale. From the co-relation of the scale we know

$\frac{T_{f1}-32}{9}$ = $\frac{T_{k1}-273.15}{5}$ ……(ii)

It is given $T_{k1}-T_k$ = 1 K

Now subtracting (i) from (ii), we get

$\frac{T_{f1}-32}{9}$ - $\frac{T_f-32}{9}$ = $\frac{T_{k1}-273.15}{5}-\frac{T_k-273.15}{5}$

$\frac{T_{f1}-T_f}{9}$ = $\frac{T_{k1}-T_k}{5}$

$T_{f1}-T_f$ = $\frac{9}{5}$

Triple point of water = 273.16 K

Hence triple point of water in the absolute scale = 273.16 $*$ $\frac{9}{5}$ = 491.69