Viscosity - class-XI

Viscosity is the property by virtue of which a liquid offers resistance for the relative motion between liquid molecules. Friction is the force which opposes relative motion between any two objects. So, viscosity is most closely related to friction.

Viscosity is the property by virtue of which a liquid offers resistance for the relative motion between its layer.
Friction is the force between two solid surfaces which offers resistance for the relative motion between them. 
So, viscosity of the fluids is analogous to friction between the solid surfaces.

When the inter-molecular forces of attraction are strong within a liquid, there is a viscosity. Cohesive forces are attractive forces exist between molecules of the same substance. So, cohesive forces are the main cause of viscosity. Adhesive forces are attractive force between unlike molecules.

The arrow with  a pointed tip can move faster in air than that of with a blunt tip because blunt tip arrow has more area of cross section , so it requires more amount of force to create the same amount of pressure as that by the pointed tip .

Viscosity is the property by virtue of which a liquid offers resistance for the relative motion between it's layer because it's main cause is cohesive force which acts between like molecules and try to bind them. So, it offers resistance for relative motion between liquid molecules.

Due to strong cohesive forces between the molecules, any layer in a moving fluid tries to drag the adjacent layer to move with an equal speed and thus produces the effect of viscosity.

Viscosity is exhibited by liquids and gases only. In solids, the role of viscosity is played by friction. 

To reduce the fluid  friction body of an Aeroplane is streamlined .

Velocity gradient $=\dfrac{dv}{dx}$
so dimensions of velocity gradient is $\dfrac{[LT^{-1}]}{[L]}=[T^{-1}]$

The viscous drag force can be written as below:
$F =  \eta \ A \ (\cfrac{dV}{dX})$ where, F is viscous drag force (or tangential force)
$\eta$ is coefficient of viscosity
$A$ is surface area of layers in contact
$\dfrac{dV}{dX} $ is velocity gradient
$\eta$ is property of liquid.
So, we can see that drag is directly proportional to the velocity gradient, surface area and nature of liquid. Drag force acts in opposite direction of liquid flow. 

A fluid that has no resistance to shear stress between its layers is known as an ideal fluid.(Definition).Hence ideal fluid  has zero viscosity.

Using Bernoulli's equation, the pressure difference across the wing is given by delta P = (1/2) * rho * (v2^2 - v1^2). Multiplying this pressure difference by the total wing area gives the total lift force: F = 0.5 * 1.28 * (140^2 - 110^2) * 10 = 48000 N.

Viscosity depends of mass of body, temperature, velocity and density.
Where as friction depends only on normal reaction.

According to Stokes' Law, the viscous drag force F on a sphere of radius R moving at velocity v in a fluid is given by F = 6 * pi * eta * R * v. Thus, the force is directly proportional to the radius R.

Due to surface tension of oil capillary rise can be seen in cotton wick which makes lamp to burn.

The equation of motion for a falling sphere is m*a = mg - F_buoyant - F_viscous. At terminal velocity v_t, the net force is zero, so F_viscous = mg - F_buoyant. When v = v_t/3, the viscous force is F_viscous' = (1/3) * F_viscous. Substituting this, the acceleration becomes a' = (1 - 1/3) * g_effective = (2/3) * a_initial. The question asks for a/X, where a' = a/3, implying X=3.

If angle of contact is acute, liquid rises in a capillary tube whereas if angle of contact is obtuse, liquid is depressed in a capillary tube.

Terminal velocity v_t is proportional to r^2. The drag force F = 6 * pi * eta * r * v_t. Substituting v_t proportional to r^2, we get F proportional to r * r^2 = r^3. The ratio of forces for radii r and 2r is (r/2r)^3 = 1/8.

The rate at which a stirred liquid comes to rest depends on its viscosity. Glycerine has the highest viscosity among the given liquids, causing it to dissipate kinetic energy through internal friction most rapidly.

Compressibility K = -(1/V) * (dV/dP). Here, dV/V = 0.2 / 200 = 0.001. dP = 2 * 10^8 Pa. K = 0.001 / (2 * 10^8) = 0.5 * 10^-11 = 5 * 10^-12 m^2/N.

Given,

Area, $A=\dfrac{\pi {{d}^{2}}}{4}=\dfrac{\pi }{4}{{\left( 0.02 \right)}^{2}}$

Volume rate of flow, $\dot{V}=\dfrac{volume}{time}=\dfrac{20\,L}{60}=\dfrac{{{10}^{-3}}}{3}\,{{m}^{3}}{{s}^{-1}}$

Volume rate of flow = Cross-Section Area x Velocity of Flow

$ \dot{V}=Av $

$v=\dfrac{{\dot{V}}}{A}=\dfrac{{{10}^{-3}}}{3}\times \dfrac{4}{\pi {{\left( 0.02 \right)}^{2}}}=1.06\,m{{s}^{-1}}=106\,cm{{s}^{-1}}$

Hence, velocity of water leaves hose is$106\,cm{{s}^{-1}}$.

The force of buoyancy B is equal to the weight of  the displaced liquid. Thus

Poiseuille's law states Q = (pi * P * r^4) / (8 * eta * L). Solving for eta: eta = (pi * P * r^4) / (8 * Q * L). Plugging in P=20, r=1.510^-3, Q=10^-7, L=0.1, we get eta = (3.14 * 20 * (1.510^-3)^4) / (8 * 10^-7 * 0.1) = 4 * 10^-3 Pa-s.

By hydrostatic equilibrium, the pressure exerted by the oil column must balance the pressure exerted by the displaced water column. Using rho_oil * h_oil = rho_water * h_water with rho_oil = 0.8 g/cc and water rising by 25 cm (meaning the water level difference is 25 cm), we find the oil height is 31.25 cm, making it stand 12.5 cm higher than the final water level.

When sphere is in glycerine, three forces acts on it. which balances each other.

• weight $(W)$
  
• buoyant force $(F _B)$
• viscous force $(F _V)$

and $W=F _B+F _V$
$\Rightarrow Mg=V _g D _2g +F _V$
$\Rightarrow Mg=V _s D _2g +F _V           \because V _g=V _s$
$\Rightarrow Mg=\frac{M}{D _1} D _2g +F _V             \because V _s=\frac{M}{D _1}$
$\Rightarrow F _V=Mg \left [ 1- \frac {D _2}{D _1} \right ]$

The viscous drag on a spherical body is given as $F=6\pi \eta RV$. Here $\eta$ is the coefficient of viscosity, R is the radius of the sphere and V is its velocity. 

The potential energy gained by the water column is m * g * h_cm = (A * h * rho) * g * (h/2) = (A * rho * g * h^2) / 2. This energy is provided by the surface tension work, and the heat evolved is the difference between the work done by surface tension and the potential energy gained.

Using the relation $\tau=\mu\displaystyle\frac{du}{dy}$ we see that viscosity is velocity dependent. Also viscosity decreases with increasing temperature. Thus A and C are correct.

The Newton's viscous force acting between two liquid surfaces with relative velocity $dv$ and distance $dy$ between the layers is given as $-\eta A\dfrac{dv}{dy}$

Shearing stress  $=\eta \dfrac{dv}{dy}=1.5 \times 10^{-3} N /m^{2}$

If the bubble is moving up with uniform velocity, the frictional force acting downwards is equal to the buoyant force acting upwards.

$c. p \rightarrow (4); q \rightarrow (2); r \rightarrow (1); s \rightarrow (3)$

The profile of the velocity of fluid as shown in the figure The
velocity of the layer of fluid ,which in contact with metal plates
(fixed), is zero. As we move towards the centre from either plate the
velocity of the layer of fluid increases and it becomes maximum at the
location of moving plate. This maximum value is same as that of the
velocity of plate

The ball falls 19.6 m, reaching velocity v = sqrt(2gh) = sqrt(2 * 9.8 * 19.6) = 19.6 m/s. In water, the buoyant force exceeds gravity (density is half of water), so it decelerates. The time to stop and return is calculated using the effective acceleration in water.

If the velocity does not change upon entering water, the drag force must balance the net weight in water. F_drag = (m_ball - m_displaced) * g. Using Stokes' Law for F_drag and equating, we solve for the velocity v, then use v^2 = 2gh to find h.