Name: Density of Substances
Uploaded: 2019-07-19
Duration: 6 min 34 s
Description: What is the density of a substance? Density is a very important thing to understand but is a bit tricky. Let’s fully grasp the idea of density from this video.

Question

State and prove Bernoulli&rsquo;s theorem.

Accepted Answer

Bernoulli’s theorem states that, for the streamline flow of an ideal fluid, sum of the pressure energy, potential energy and kinetic energy remains constant.

Proof: Consider the streamline flow of an ideal fluid through the tube of variable cross-section.

Let $P_{1}, P_{2}; A_{1}, A_{2}; h_{1}, h_{2}; v_{1}, v_{2}$ be the pressures, cross-sectional areas, heights and velocities at A and B respectively. Further let $ρ$ be the density of liquid.

The mass of liquid crossing per second through any section of the tube is:
$A_{1} v_{1} ρ = A_{2} v_{2} ρ = m$ (According to equation of continuity)

$∴ A_{1} v_{1} = A_{2} v_{2}$

Since $A_{1} > A_{2} ∴ v_{2} > v_{1}$

Force at A, $F_{1} = P_{1} A_{1}$

Force at B, $F_{2} = P_{2} A_{2}$
Work done on per second on liquid at $A = P_{1} A_{1} \times v_{1} = P_{1} (A_{1} v_{1}) = P_{1} V$

Work done on per second on liquid at $B = P_{2} A_{2} \times v_{1} = P_{2} (A_{2} v_{2}) = P_{2} V$

Here, $V$ represents volume flow rate.

Net work done by pressure energy to move liquid from $A$ to $B$ $= P_{1} V - P_{2} V$

Increase of potential energy of liquid from $A$ to $B$ $= m g h_{2} - m g h_{1}$

Increase of kinetic energy of liquid from $A$ to $B$ $= \frac{1}{2} m {v_{2}}^{2} - \frac{1}{2} m {v_{1}}^{2}$

Now according to work energy theorem:

Work done by pressure energy $=$ increase of potential energy $+$ increase of kinetic energy

. $P_{1} V - P_{2} V = (m g h_{2} - m g h_{1}) + (\frac{1}{2} m {v_{2}}^{2} - \frac{1}{2} m {v_{1}}^{2})$

$P_{1} V + m g h_{1} + \frac{1}{2} m {v_{1}}^{2} = P_{2} V + m g h_{2} + \frac{1}{2} m {v_{2}}^{2}$

On dividing both side by $m$ , we have
$\frac{P_{1} V}{m} + g h_{1} + \frac{1}{2} {v_{1}}^{2} = \frac{P_{2} V}{m} + g h_{2} + \frac{1}{2} {v_{2}}^{2}$

$\frac{P_{1}}{ρ} + g h_{1} + \frac{1}{2} {v_{1}}^{2} = \frac{P_{2}}{ρ} + g h_{2} + \frac{1}{2} {v_{2}}^{2}$

$\frac{P}{ρ} + g h + \frac{1}{2} v^{2}$ $=$ constant

State and prove Bernoulli’s theorem.

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