Properties of Matter

Given, Height of small orifice from ground $(H–h) = 0.25$ m

Total height of water tank,

$(H)=1 \mathrm{~m}$

$\therefore$ Range of water stream,

$\begin{aligned} R & =2 \sqrt{h(H-h)}=2 \sqrt{(H-0.25)(0.25)} \\\\ & =2 \sqrt{(1-0.25) 0.25}=2 \sqrt{0.75 \times 0.25} \\\\ & =0.866 \mathrm{~m}=86.6 \mathrm{~cm} \end{aligned}$

Given :

Diameter at the first end, $ d_1 = 5 \, \text{cm} $

Velocity at the first end, $ v_1 = 4 \, \text{m/s} $

Diameter at the second end, $ d_2 = 2 \, \text{cm} $

$ \Delta P = ? $

Step 1 : Calculate the radii

$ r_1 = \frac{d_1}{2} = 2.5 \, \text{cm} = 2.5 \times 10^{-2} \, \text{m} $

$ r_2 = \frac{d_2}{2} = 1 \, \text{cm} = 10^{-2} \, \text{m} $

Step 2 : Apply the principle of continuity

$ A_1 v_1 = A_2 v_2 $

$ \pi r_1^2 v_1 = \pi r_2^2 v_2 $

$ \left(2.5 \times 10^{-2}\right)^2 \times 4 = \left(10^{-2}\right)^2 \cdot v_2 $

$ v_2 = 25 \, \text{m/s} $

Step 3 : Use Bernoulli's theorem

$ p_1 + \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2} \rho v_2^2 $

$ p_1 - p_2 = \frac{1}{2} \rho \left(v_2^2 - v_1^2\right) $

Given $ \rho = 10^3 \, \text{kg/m}^3 $:

$ p_1 - p_2 = \frac{1}{2} \times 10^3 \left(25^2 - 4^2\right) $

$ = 304500 \, \text{Pa} $

Given, surface tension of water, $T=0.1 \mathrm{~N} / \mathrm{m}$

Radius of small drops, $r=1 \mathrm{~mm}=10^{-3} \mathrm{~m}$

If $R$ be the radius of bigger drops, then volume remains conserved.

$\begin{aligned} \text{i.e.,} \quad V_1 & =V_2 \\ \frac{4}{3} \pi R^3 & =2 \times \frac{4}{3} \pi r^3 \\ R^3 & =2 r^3=2 \cdot\left(10^{-3}\right)^3 \\ \therefore \quad R & =2^{1 / 3} \times 10^{-3} \mathrm{~m} \end{aligned}$

$\therefore$ Change in energy during isothermal combination,

$\begin{aligned} \Delta E & =T \cdot \Delta A \\ & =T\left[A_2-A_1\right] \\ & =0.1\left[2 \times 4 \pi r^2-4 \pi R^2\right] \\ & \begin{aligned} & =0.1 \times 4 \pi\left[2 r^2-R^2\right] \\ & =0.4 \pi\left[2 \times 10^{-6}-2^{2 / 3} \times 10^{-6}\right] \\ & =0.4 \pi \times 10^{-6} \times 0.41 \\ & =0.52 \times 10^{-6} \mathrm{~J}=0.52 \mu \mathrm{J} \simeq 0.5 \mu \mathrm{J} \end{aligned} \end{aligned}$

Surface tension is responsible for the shape of tiny water drops when falling freely in the absence of other forces like gravity, shape of water drops is spherical because droplets of water tend to be pulled into a spherical shape by the cohesive forces of the surface layer.

In the presence of gravity, the shape of heavy water drops is not spherical when falling freely. It is oval shaped surface tension of methyl alcohol is lower than surface tension of water.

Hence, water has not minimum surface tension among all liquids.

Therefore, both Assertion and Reason are false.

Sometimes, insects can walk on the surface of water due to surface tension, when legs of insects are not being wet. In this situation, the gravitational force on insect is balanced by force due to surface tension. Hence, both assertion and reason are true and reason is correct explanation of assertion.

As excess pressure,

$p=\frac{2 s}{R} \Rightarrow p \propto \frac{1}{R}$

$\therefore$ Excess pressure inside smaller drop is large due to which smaller force resist deforming forces better.