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This is one of our many quizzes on Fluids. Click on this to access more such competitive quizzes on Mechanics.

Fluids Quiz 2

physics

If pressure head of a fluid of density $20\,kg/{{m}^{3}}$ is given as $25$ units. Then, what is the pressure which the fluid exerts of the pipe? $\text{(Take }g=10\,m/{{s}^{2}})$ 

The pressure head can be found by taking the ratio of the pressure to the product of density and acceleration due to gravity. 

 The pressure head has been given as, pressure head $ = \dfrac{P}{\rho g}$ $ \\ $ Substituting the values in it will give, $ \\ \text{25=}\dfrac{P}{20\times 10} \\  \therefore P=5000\text{ }N $ 

 If Suman says that the Reynold’s number of a fluid flowing in a dam is approximately $4000$. What does it represent about the nature of the flow of fluid? 

Reynold’s number can be defined as the ratio of the inertial forces to the viscous forces. This number helps us to get some knowledge about the flow of liquid. 

 The Reynold’s number which is greater than $2000$ will be showing turbulent flow. Hence we can conclude that the flow will be turbulent itself.  

 In her imagination, Priyanka saw that she had landed in Antarctica, where she saw an iceberg floating on the cold water. She noticed that the iceberg was approximately $20\%$ immersed in the cold water. If a native of the island tells her that density of the cold water is about $1000\,kg/{{m}^{3}}$, then the density of the piece of iceberg which she saw is approximately, 

 Here, according to the Archimedes principle, the weight of the iceberg will be equivalent to that of the water displaced. 

 As per the Archimedes principle, we can write that the weight of the iceberg will be equal to the weight of water displaced. $ \\  V\times {{\left( d \right)}_{ice}}=v\times 1000\text{ }kg{{m}^{-3}} \\  \Rightarrow {{\left( d \right)}_{ice}}=\dfrac{v}{V}\times 1000 \\ $ We have to find the exposed volume. $ \\  \Rightarrow \dfrac{v}{V}=\dfrac{80}{100} \\  \therefore \dfrac{v}{V}=\dfrac{4}{5} \\ $ Substituting the values in it, $ \\ {{d}_{ice}}=\dfrac{4}{5}\times 1000 \\  \therefore {{d}_{ice}}=800\text{ }kg{{m}^{-3}}\approx 900\text{ }kg{{m}^{-3}}$

If suppose ${{d}_{s}}$ is the density of a football and ${{d}_{L}}$ is the density of the fluid. Now, suppose a child throws the ball in the fluid. Based on the understanding of the laws of flotation, match the column A and the Column B $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b341a5b2-023b-4f9b-a278-e4ef804c7b912711822542788733458.png"/>

When the density of the body matches the density of fluid, the body will float in the fluid. 

This can be done as per the laws of the floatation, In the first case, we can say that, as $ {{d}_{S}} >  {{d}_{L}}$, the solid will sink in the liquid.$ \\ $ In the second case, As ${{d}_{S}} <  {{d}_{L}}$ the solid will be floating in the liquid. $ \\ $ In the third case, ${As} {{d}_{S}}={{d}_{L}}$ the solid will sink but will keep floating.

 Suppose in Disneyland, Hong Kong there is a large cylindrical container $120cm$ tall, which contains a chocolate fluid inside it. A child hits the bottom of the tank with a gemstone. As a result, the gemstone strikes the bottom of the container and makes a small orifice over there. The fluid now starts to flow out of the container. If $60cm$ of the column of the fluid is emptied in $20$ minutes approximately, then another $60cm$ will flow out in, 

The volume will be the product of the area of the cross section and the height of the body. Where the height can be found by taking the product of velocity and time.  

The volume of liquid coming out of the hole in a time dt,$ \\  dV=av\times dt\\ $ Substituting the equation of $v$ in it, $ \\  dV=a\sqrt{2gh}\cdot dt\\ $ The volume of liquid coming out of the hole must be equal, $ \\  dV=A\left( -dH \right)\\  \Rightarrow\int\limits_{0}^{t}{dt}=\dfrac{A}{a\sqrt{2}g}\int\limits_{H}^{0}{{{y}^{-\frac{1}{2}}}}dy............\text{ }\left( 1 \right)\\ $ Substituting the proper limits in equation $(1)$, $ \\  \Rightarrow \int\limits_{0}^{{{t}_{1}}}{dt}=\dfrac{A}{a\sqrt{2}g}\int\limits_{H}^{\frac{H}{2}}{{{y}^{-\frac{1}{2}}}}dy\\  \Rightarrow{{t}_{1}}=\dfrac{A}{a}\sqrt{\dfrac{H}{g}}\left( \sqrt{2}-1. \right)..........\text{ }\left( 2 \right) \\  \Rightarrow {{t}_{2}}=\dfrac{A}{a}\sqrt{\dfrac{H}{g}}...........\text{ }\left( 3 \right) \\ $ The ratio of $ \\ {{t}_{1}} \text{and} {{t}_{2}} \text{will be,}\\\dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{2}-1=1.414-1 \\  \Rightarrow\dfrac{{{t}_{1}}}{{{t}_{2}}}=0.414 \\  \Rightarrow {{t}_{1}}=0.414\times {{t}_{2}} \\  \therefore {{t}_{2}} <  {{t}_{1}}$ 