NCERT Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids

Ans : (a) The pressure Of a liquid is given by the relation:  
p = hpg 
where, 
p = Pressure 
h = Height of the liquid column 
p = Density of the liquid  
g = Acceleration due to the gravity  
It can be inferred that pressure is directly proportional to height. Hence, the blood 
pressure in human vessels depends on the height Of the blood column in the body. The 
height Of the blood column is more at the feet than it is at the brain. Hence, the blood 
pressure at the feet is more than it is at the brain.  
(b) Density of air is the maximum near the sea level. Density of air decreases with 
increase in height from the surface. At a height of about 6 km, density decreases to 
nearly half of its value at the sea level. Atmospheric pressure is proportional to density. 
Hence, at a height of 6 km from the surface, it decreases to nearly half of its value at  the sea level.
(c) When force is applied on a liquid, the pressure In the liquid is transmitted in all 
directions. Hence, hydrostatic pressure does not have a fixed direction and it is a scalar 
physical quantity. 

Ans : (a) 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 (B), as shown in the given figure. 

\(\begin{array}{l}{S_{\text { lat }} S_{\text { sat }} \text { and } S_{\text { sl }} \text { are the respective interfacial tensions between the liquid-air, solid-air, }} \\ {\text { and solid-liquid interfaces. At the line of contact, the surface forces between the three }} \\ {\text { media must be in equilibrium, i.e., }}\end{array}\)
\(\cos \theta=\frac{S_{\mathrm{a}}-S_{\mathrm{sl}}}{S_{\mathrm{la}}}\)
\(\begin{array}{l}{\text { The angle of contact } \theta, \text { is obtuse if } S_{\mathrm{sa}} < S_{1 a}(\text { as in the case of mercury on glass). This }} \\ {\text { angle is acute if } S_{\mathrm{sl}} < S_{\mathrm{la}}(\text { as in the case of water on glass). }}\end{array}\)
(b) Mercury molecules (which make an obtuse angle with glass) have a strong force Of 
attraction between themselves and a weak force of attraction toward solids. Hence, they 
tend to form drops. 
On the other hand, water molecules make acute angles with glass. They have a weak 
force of attraction between themselves and a strong force of attraction toward solids. 
Hence, they tend to spread out. 
(c) 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 also independent Of the area Of the liquid surface. 
(d) Water with detergent dissolved in it has small angles of contact (9). This is because 
for a small 9, there is a fast capillary rise of the detergent in the cloth. The capillary rise 
of a liquid is directly proportional to the cosine of the angle of contact (B). If \( \theta\) is small, 
then \(\cos \theta\) will be large and the rise of the detergent water in the cloth will be fast. 
(e) A liquid tends 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. 

Ans : (a) decreases 
The surface tension of a liquid is inversely proportional to temperature. 
(b) increases; decreases 
Most fluids Offer resistance to their motion. This is like internal mechanical friction, 
known as viscosity. Viscosity Of gases increases with temperature, while viscosity Of 
liquids decreases with temperature. 
(c) Shear strain; Rate of shear strain 
With reference to the elastic modulus of rigidity for solids, the shearing force is 
proportional to the shear strain. With reference to the elastic modulus of rigidity for 
fluids, the shearing force is proportional to the rate of shear strain. 

Ans : (a) When air is blown under a paper, the veloc ty Of air is greater under the paper than it is above it. As per Bernoulli's principle, atmospheric pressure reduces under the paper. This makes the paper fall. To keep a piece of paper horizontal, one should blow over it, This increases the velocity of air above the paper. As per Bernoulli's principle, atmospheric pressure reduces above the paper and the paper remains horizontal. 
(b) According to the equation of continuity: Area x Velocity = Constant 
For a smaller opening, the velocity Of flow Of a fluid is greater than it is when the 
opening is bigger. When we try to close a tap Of water with our fingers, fast jets Of water gush through the openings between our fingers, This is because very small openings are left for the water to flow out of the pipe. Hence, area and velocity are inversely proportional to each other. 
(c) The small opening Of a syringe needle controls the velocity Of the blood flowing Out. 
This is because Of the equation Of continuity. At the constriction point Of the syringe 
system, the flow rate suddenly increases to a high value for a constant thumb pressure 
applied. 
(d) When a fluid flows out from a small hole in a vessel, the vessel receives a backward 
thrust. A fluid flowing out from a small hole has a large velocity according to the 
equation of continuity : Area x Velocity Constant According to the law of conservation of momentum, the vessel attains a backward velocity because there are no external forces acting on the system.
(e) A spinning cricket ball has two simultaneous motions — rotary and linear. These 
two types of motion oppose the effect of each other. This decreases the velocity of air 
flowing below the ball. Hence, the pressure on the upper side of the ball becomes lesser 
than that on the lower side. An upward force acts upon the ball. Therefore, the ball takes 
a curved path. It does not follow a parabolic path. 

Ans : \(\begin{array}{l}{\text { Mass of the girl, } m=50 \mathrm{kg}} \\ {\text { Diameter of the heel, } d=1 \mathrm{cm}=0.01 \mathrm{m}} \\ {\text { Radius of the heel, } r=\frac{d}{2}=0.005 \mathrm{m}}\end{array}\)
\(\begin{array}{l}{\text { Area of the heel }=\pi r^{2}} \\ {=\pi(0.005)^{2}} \\ {=7.85 \times 10^{-5} \mathrm{m}^{2}} \\ {\text { Force exerted by the heel on the floor: }} \\ {F=m g}\end{array}\)
\(\begin{array}{l}{=50 \times 9.8} \\ {=490 \mathrm{N}} \\ {\text { Pressure exerted by the heel on the floor: }} \\ {P=\frac{\text { Force }}{\text { Area }}}\end{array}\)
\(\begin{array}{l}{=\frac{490}{7.85 \times 10^{-5}}} \\ {=6.24 \times 10^{6} \mathrm{Nm}^{-2}} \\ {\text { Therefore, the pressure exerted by the heel on the horizontal floor is }} \\ {6.24 \times 10^{6} \mathrm{Nm}^{-2}}\end{array}\)