Mechanical Properties of Fluids

Surface tension is the force acting on the surface of the liquid.

Bernoulli's principle states that in a steady flow, the sum of pressure, kinetic energy per unit volume, and potential energy per unit volume remains constant.

Yes, the Mechanical properties of fluids class 11th physics is important in NEET. On average, 1-2 questions would be asked from this chapter, which you can cover from the Class 11th Mechanical Properties of Fluids notes.

The main mechanical properties of fluids are exerting pressure, resisting flow or viscosity, forming surface tension, following Bernoulli's principle, and moving in a streamline.

Since velocity does not change, so acceleration will be zero.

mg = FB + Fv $\Rightarrow \frac{4\pi }{3}{r}^3\rho g=\frac{4\pi }{3}{r}^3\sigma g+6\pi \eta rv$

$\Rightarrow v=\frac{2r^2\left(\rho -\sigma \right)g}{9\eta }=\frac{2*0.1*0.1*10^{-6}*\left(10^4-10^3\right)*10}{9*1.0*10^{-5}}$

$\Rightarrow h=\frac{400}{2g}=20m$

Kindly consider the following figure

m = 0.3g density,   ${\rho }_{\left(ball\right)}=8g/cc$                   $\therefore V=\frac{m}{\rho }=\frac{0.3}{8}cc$                     

${\rho }_{glycerene}=1.3g/cc=1.3*10^{3}kg/m^3$

F_v + F_B = mg.

Þ F_v = mg – F_B = .3 * 10^-3 * 10 – 1.3 *  $\frac{0.3}{8}*10^{-6}*10*10^3$

  $=3*10^{-3}-\frac{.39}{8}*10^{-2}=3*10^{-3}-.5*10^{-3}$

= 2.5 * 10^-3N

Tension in wire, $T=\frac{2*3*5*10}{3+5}=\frac{75}{2}N$

$\frac{T}{\pi r_{min}^2}=\frac{24}{\pi }*10^2$

$\Rightarrow r_{min}^2=\frac{75}{2*24}*10^{-2}=\frac{25}{16}*10^{-2}$
$\Rightarrow r_{min}=\frac{5}{4}*10^{-1}m=12.5cm$

Load of mass will be equally distributed among the four colours so force on each columns will be 125 * 10³ N.

Cross section area of the column = $\pi \left[{\left(1\right)}^2-{\left(0.5\right)}^2\right]=2.355m^2$

Using young's modulus : $\epsilon =\frac{\sigma }{Y}=\frac{F}{AY}=\frac{125*10^3}{2.355*2*10^{11}}=2.65*10^{-7}$  

g_h = g (1-2h/R). g_d = g (1-d/R). Given h=d.
g_h = g (R/ (R+h)² ≈ g (1-2h/R). This differs from the image.
The image has g/ (1+h/R)² = g (1-h/R). This leads to h² + hR - R²=0. h = R (√5-1)/2.