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# When a tap is closed, the manometer attached to the pipe reads $3.5 \times {10^5}\;{\rm{N}}\;{{\rm{m}}^{ - 2}}$. When the tap is opened. The reading of the manometer falls to$3 \times {10^5}\;{\rm{N}}\;{{\rm{m}}^{ - 2}}$. The velocity of water in the pipe is:A. 0.1 $m{s^{ - 1}}$B. 5 $m{s^{ - 1}}$C. 10 $m{s^{ - 1}}$D. 1 $m{s^{ - 1}}$

Last updated date: 24th Jul 2024
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Hint: The study of an external force by a fluid on a surface is known as pressure calculation. The most common unit of measurement for pressure is force per unit of surface area. The calculation of friction and vacuum has spawned a slew of new techniques.

The theory of conservation of energy can be used to derive Bernoulli's principle. The number of all sources of energy in a fluid along a streamline is the same at all points on the streamline in a steady flow. The sum of kinetic energy, potential energy, and internal energy must remain stable for this to happen.

As a result, an increase in the fluid's motion – meaning an increase in its kinetic energy – is accompanied by a reduction in the fluid's potential energy and internal energy. Since the energy per unit volume in a reservoir is the same everywhere, the amount of all sources of energy on all streamlines is the same if the fluid is flowing out of it.

According to Bernoulli's theorem,
${P_1} + \dfrac{1}{2}\rho v_1^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho v_2^2 + \rho g{h_2}$
The total mechanical energy of the flowing fluid, which includes the gravitational potential energy of elevation, fluid pressure energy, and fluid motion kinetic energy, remains constant.
So according to Bernoulli's theorem, liquid at same height
${P_c} + \dfrac{1}{2}\rho v_c^2 = {P_0} + \dfrac{1}{2}\rho v_o^2$
Lets say
${P_0}$ = pressure in the pipe when tap is closed
$\Rightarrow {P_c}$= pressure in the pipe when tap is opened
$\Rightarrow v_o^{}$= velocity of water when tap is opened
$\Rightarrow v_c^{}$= velocity of water when tap is opened =0
$\Rightarrow {{\rm{P}}_{\rm{c}}} - {{\rm{P}}_{\rm{o}}} = \dfrac{1}{2}\rho {\rm{v}}_{\rm{o}}^2$
$\Rightarrow {{\rm{v}}_{\rm{o}}} = \sqrt {\dfrac{{2\left( {{{\rm{P}}_{\rm{c}}} - {{\rm{P}}_{\rm{o}}}} \right)}}{\rho }}$
$\Rightarrow {{\rm{v}}_{\rm{o}}} = \sqrt {\dfrac{{2\left( {3.5 \times {{10}^5} - 3 \times {{10}^5}} \right)}}{{1000}}}$
$\Rightarrow {{\rm{v}}_{\rm{o}}}= \sqrt {\dfrac{{2 \times 0.5 \times {{10}^5}}}{{1000}}} \\ \Rightarrow {{\rm{v}}_{\rm{o}}}= \sqrt {{{10}^2}} \\ \therefore {{\rm{v}}_{\rm{o}}}= 10\;{\rm{m}}/{\rm{s}}$

Hence option C is correct.

Note: Bernoulli's theorem can be generalised to a variety of fluid flows, yielding a number of different Bernoulli's equations. For incompressible flows, Bernoulli's equation in its simplest form is valid (e.g. most liquid flows and gases moving at low Mach number). At higher Mach numbers, more sophisticated forms can be extended to compressible flows (see the derivations of the Bernoulli equation).