Tag: option b: engineering physics

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.