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This is one of our many quizzes on Bernoulli's law. Click on this to access more such competitive quizzes on Fluid dynamics.

Bernoulli's law Quiz 2

physics

 If Alex puts a bucket in a lake and there is a hole inside the bucket from where the water is entering with a rate of $150\,\text{litres}\,{{\text{s}}^{\text{-1}}}$ , then what will be the area of cross-section of the hole if the height  of the bucket is $20\,\text{m}$? Take $g=10\,\text{m}{{\text{s}}^{\text{-2}}}$.

 First, calculate the velocity of efflux and then use the concept of the rate at which water enters to evaluate the value of the area of cross-section. 

  $\text{Velocity of the efflux,} \\ v=\sqrt{2gh} \\ \Rightarrow h=20\,\text{m} \\ \Rightarrow v=\sqrt{2\times 10\times 20} \\ \Rightarrow v=20\,\text{m}{{\text{s}}^{\text{-1}}} \\ \text{Also} \\ \text{Rate at which water enters,} \\ \text{Q=av} \\ \Rightarrow \text{a=}\dfrac{Q}{v} \\ \Rightarrow Q=150\,\text{litres}\,{{\text{s}}^{\text{-1}}} \\ \Rightarrow a=\dfrac{150}{20} \\ \therefore a=7.5\,{{\text{m}}^{\text{2}}}$

 There is a container in Ravi's house that is filled with water and also a hole is present in that container which is covered by tape so that the water will not emerge from that hole. The gauge pressure at that time is $4\times {{10}^{4}}\,\text{N}{{\text{m}}^{\text{-2}}}$. After that, the tape was removed by Ravi accidentally and the water present in that container starts coming out from that hole. If the density of water is $1000\,\text{kg}{{\text{m}}^{\text{-3}}}$ then find the velocity of the flowing water from the hole of the container. $($ Take $g=10\,\text{m}{{\text{s}}^{\text{-2}}})$ 

 First, calculate the height then calculate the value of the velocity of efflux. 

  $\text{Pressure is given by,} \\ P=h\rho g \\ \Rightarrow h=\dfrac{P}{\rho g} \\ \text{here P=4}\times \text{1}{{\text{0}}^{4}}\,\text{N}{{\text{m}}^{\text{-2}}} \\ \text{Density of the water}\left( \rho  \right)\text{=1000 kg}{{\text{m}}^{-3}} \\ \Rightarrow h=\dfrac{4\times \text{1}{{\text{0}}^{4}}}{1000\times 10} \\ \Rightarrow h=4\,\text{m} \\ \text{Now velocity of the efflux,} \\ v=\sqrt{2gh} \\ \Rightarrow v=\sqrt{2\times 10\times 4} \\ \Rightarrow v=\sqrt{80} \\ \therefore v=8.94\,\text{m}{{\text{s}}^{-1}}$

 If the area of the cross-section of the narrow neck is $\dfrac{a}{3}$ and the area of the cross-section of the wider part of the venturi meter is $a$ and the height of the manometer is  changed to $2h$ then what will be the value of the new volume of the liquid flowing out per second through the flow meter? 

 Recall and apply the concept of the volume of a liquid flowing out per second through a venturi meter. 

 $\text{Volume of a liquid flowing out per second,} \\ Q={{a}_{1}}{{a}_{2}}\sqrt{\dfrac{2gh}{a_{1}^{2}-a_{2}^{2}}} \\ {{a}_{1}}=a,\,{{a}_{2}}=\,\dfrac{a}{3},\,h=2h \\ \Rightarrow Q= a\times \dfrac{a}{3}\sqrt{\dfrac{2g\times 2h}{{{a}^{2}}-\dfrac{{{a}^{2}}}{9}}} \\ \Rightarrow Q= \dfrac{{{a}^{2}}}{3}\sqrt{\dfrac{36gh}{8{{a}^{2}}}} \\ \therefore Q= a\sqrt{\dfrac{gh}{2}}$

 If Archi noted the value of pressure in a closed pipe with the help of a pressure meter is $11\times {{10}^{4}}\,\text{N}{{\text{m}}^{\text{-2}}}$, then find the value of the pressure when the pipe is opened  and the speed of the water flowing in the pipe is $12\,\text{m}{{\text{s}}^{\text{-1}}}$. 

 Apply the formula of Bernoulli’s equation to find the value of the pressure in the case of flowing water. 

 $\text{Before opening the pipe:} \\ \text{Pressure(}{{\text{P}}_{1}}\text{) =11}\times \text{1}{{\text{0}}^{4}}\,\text{N}{{\text{m}}^{\text{-2}}} \\ \text{velocity(}{{\text{v}}_{1}})=0 \\ \text{After opening the pipe:} \\ \text{velocity(}{{\text{v}}_{2}})=12\,\text{m}{{\text{s}}^{\text{-1}}} \\ \text{In case of horizontal flow,} \\  \text{P.E. remains constant}\text{.} \\ \text{Apply Bernoulli }\!\!'\!\!\text{ s Equation:} \\ {{P}_{2}}+\dfrac{1}{2}\rho v_{2}^{2}={{P}_{1}}+\dfrac{1}{2}\rho v_{1}^{2} \\ \Rightarrow {{P}_{2}}+\dfrac{1}{2}\times 1000\times 144=\text{11}\times \text{1}{{\text{0}}^{4}} \\ \Rightarrow {{P}_{2}}=11\times \text{1}{{\text{0}}^{4}}-7.2\times \text{1}{{\text{0}}^{4}} \\ \therefore {{P}_{2}}=3.8\times \text{1}{{\text{0}}^{4}}\,\text{N}{{\text{m}}^{\text{-2}}}$

 If Raman joined two different pipes with different radius and the water is flowing in that joined pipe with two different velocities. If the velocity of water in case of bigger radius pipe is $5\,\text{m}{{\text{s}}^{\text{-1}}}$ and the pressure exerted in bigger radius pipe is $20.75\times {{10}^{4}}\,\text{N}{{\text{m}}^{\text{-2}}}$ then what will be the pressure in case of a smaller radius pipe? Assume that the radii of the two pipes are $1.5\,\text{m}$and $3\,\text{m}$ respectively.

 First, find the velocity of water in case of small diameter using the equation of continuity and then apply Bernoulli’s principle to find pressure in case of small radius.

 $\text{According to the equation of continuity,} \\ {{a}_{1}}{{v}_{1}}={{a}_{2}}{{v}_{2}} \\ \Rightarrow \pi r_{1}^{2}{{v}_{1}}=\pi r_{2}^{2}{{v}_{2}} \\ \Rightarrow {{v}_{2}}={{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}{{v}_{1}} \\ \Rightarrow {{v}_{1}}=5\,\text{m}{{\text{s}}^{\text{-1}}},\,{{r}_{1}}=3\,\text{m, }{{r}_{2}}=1.5\,\text{m} \\ \Rightarrow {{v}_{2}}={{\left( \dfrac{3}{1.5} \right)}^{2}}\times 5 \\ \Rightarrow {{v}_{2}}=20\,\text{m}{{\text{s}}^{\text{-1}}} \\ \text{Apply bernoulli }\!\!'\!\!\text{ s equation for horizontal flow:} \\ {{P}_{2}}+\dfrac{1}{2}\rho v_{2}^{2}={{P}_{1}}+\dfrac{1}{2}\rho v_{1}^{2} \\ \Rightarrow {{P}_{2}}={{P}_{1}}+\dfrac{1}{2}\rho \left( v_{1}^{2}-v_{2}^{2} \right) \\  here,\,{{P}_{1}}=20.75\times {{10}^{4}}\,\text{N}{{\text{m}}^{\text{-2}}} \\ \Rightarrow {{P}_{2}}=20.75\times {{10}^{4}}+\dfrac{1}{2}\times 1000\times \left( 25-400 \right) \\ \Rightarrow {{P}_{2}}=20.75\times {{10}^{4}}-18.75\times {{10}^{4}} \\ \therefore {{P}_{2}}=2\times {{10}^{4}}\,\text{N}{{\text{m}}^{\text{-2}}}$