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This is the second of our many quizzes on Pascal's law. Click on this to access more such competitive quizzes on Mechanics.

Pascal's law Quiz 2

physics

 A cylindrical glass is spinning about its central axis. If the maximum height reached by the water level while the cylinder is spinning is $24cm$ and the minimum height attained by the water level is $16cm$ then what must be the height of water level at rest? $(\text{all heights are measured from the base of the glass})$ 

When the glass spins some water will rise and some will go down. The initial height is the average of the maximum and the minimum heights. 

  $\text{When a cylinder rotates the water level that is horizontal takes the shape of a 3D parabola. It is called a paraboloid.}\\\text{Just like a parabola there is a maxima and a minima for the paraboloid.}\\\text{The maxima is the maximum height attained or reached by the water at the corners.}\\\text{The minima is the minimum height attained by the water at the central portion.}\\\text{Thus, the initial height of the water level when the cylinder is at rest i.e.,}\text{not spinning is the average of the maximum and minimum heights attained by the water level.Thus,}\\{{y}_{0}}=\dfrac{{{y}_{\max }}+{{y}_{\min }}}{2} \\\Rightarrow {{y}_{0}}=\dfrac{24+16}{2} \\\therefore {{y}_{0}}=20\,cm$

 In a cylinder spinning with angular velocity $\omega $about its central axis. If the equation of the paraboloid formed on spinning the glass is given by $y=2{{x}^{2}}+4$ , what must be the angular velocity of the cylinder? (take $g=10m{{s}^{-1}}$ ) 

Remember the equation of the parabola for such a case, compare both the standard equation and the given equation. 

  $\text{When a cylinder rotates the water level that is horizontal takes the shape of a 3D parabola.}\\\text{It is called a paraboloid.}\\\text{The equation of this paraboloid in 2D is given by;}\\y=\dfrac{{{x}^{2}}{{\omega }^{2}}}{2g}+{{y}_{\min }}\\\text{But we are given that the equation is}\\y=2{{x}^{2}}+4\\\text{Now comparing the two equations we get;}\\\dfrac{{{x}^{2}}{{\omega }^{2}}}{2g}=2{{x}^{2}} \\\Rightarrow {{\omega }^{2}}=4g \\\Rightarrow {{\omega }^{2}}=4\times 10 \\\therefore \omega =2\sqrt{10}$ 

In a glass filled with a liquid, a cube is dipped such that it reaches the base of the glass and throughout its motion the base and top surfaces are horizontal. What should be the value of density of the liquid for the cube to reach the base of the glass? 

For the cube to be continuously moved down there should be a net acceleration of the cube that is in the downward direction. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4812d9fc-3afe-401d-b1b8-630d7b3d32927179063984851400536.png"/>$\\ \text{Now when the cube is dropped in the liquid we all know for the cube to sink,}\text{the density of the cube should be more than that of the liquid.}\\\text{But here we need to find the maximum value of the density of the liquid}\text{for which the cube will reach the base.}\\\text{From the above FBD we can write the equation for maximum density as;}\\{{P}_{b}}A-{{P}_{t}}A-mg=ma \\\Rightarrow {{P}_{b}}A-{{P}_{t}}A-\rho Vg=\rho Va \\\Rightarrow \rho V\left( a+g \right)=\left( {{P}_{b}}-{{P}_{t}} \right)A \\\therefore {{\rho }_{\max }}=\dfrac{\left( {{P}_{b}}-{{P}_{t}} \right)A}{V\left( a+g \right)}$

For a cylindrical bucket of height $H$ its angular velocity about the central axis is $\omega $ . If the water is filled to a level of $h$ then what must be the maximum angular velocity for which the water will not spill out from the bucket? ( take radius of the bucket as $R$ ) 

For the water to not spill out of the bucket the maximum height of the water level should be less than the height of the bucket. However if it is equal to the height of the bucket then also the water will not spill. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/da7eaade-bb31-4e14-9998-28a58cadd2f63979945155719150950.png"/>$\\ \text{When the bucket will spin the horizontal water level will take the shape of a 3D parabola.}\\\text{Now for the water to not spill out the bucket}\,{{y}_{\max }}\le H\,\text{but we know that}\\{{y}_{\max }}={{y}_{0}}+\dfrac{{{\omega }^{2}}{{R}^{2}}}{4g}\\\text{Substituting all the values we can write;}\\H\ge h+\dfrac{{{\omega }^{2}}{{R}^{2}}}{4g} \\\Rightarrow {{\omega }^{2}}\le \dfrac{4g\left( H-h \right)}{{{R}^{2}}} \\\therefore \omega \le \dfrac{2}{R}\sqrt{g\left( H-h \right)}$

A tank is filled with water and is moving with a velocity of $v=10\sqrt{3}t$ towards right, where $t$ is time. At what height from the initial level the water must rise if $g=10m{{s}^{-1}}$ and base of the tank is $20m$ and the other side goes down by $10m$ 

Find the acceleration in the horizontal direction and then find the angle that will be formed when the tank accelerates. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d6d93b0b-c605-4369-844f-9824b16136242132954500220618345.png"/>$\\ \text{Now we are given the velocity of the tank that is changing with respect to time.}\\\text{So, there is some acceleration.Now acceleration is defined as;}\\a=\dfrac{dv}{dt} \\\Rightarrow a=10\sqrt{3} \\\text{This acceleration is towards the right.}\\\text{There is also a vertically downward acceleration due to gravity.}\\\text{Let the curve for angle}\,'\theta ' \text{touch}\,'{A}'{B}''.\,\text{Thus, we can say from the vector diagram that}\\\theta ={{\tan }^{-1}}\left( \dfrac{a}{g} \right) \\\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{10\sqrt{3}}{10} \right) \\\Rightarrow \theta ={{\tan }^{-1}}\left( \sqrt{3} \right) \\\therefore \theta ={{60}^{\circ }} \\\text{Thus, from the diagram we can say that;}\\\dfrac{x}{20}=\tan \left( \theta  \right)=\tan \left( {{60}^{\circ }} \right) \\\Rightarrow \dfrac{x}{20}=\sqrt{3} \\\Rightarrow x=20\times 1.73 \\\therefore x=34.6m \\\text{But we have to find}\,A{A}'\,\text{So,}\\A{A}'=34.6-B{B}' \\\therefore A{A}'=34.6-10=24.6\,m$