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This is the second of our many quizzes on Archimedes' principle. Click on this to access more such competitive quizzes on Mechanics.

Archimedes' principle Quiz 2

physics

Calculate the density of a body, if it is $70\%$ submerged in water.$[\text{density of water}\,{{\rho }_{w}}=1000kg/{{m}^{3}}]$ 

If a body is submerged partially in a liquid, then the density of the body will be a fraction of the body submerged in a liquid multiplied by the density of the liquid. 

  Weight of the body, $W = {{\rho }_{b}}{{V}_{b}}g$ $ \\ $ Buoyant force acting on the body, $ \\{{F}_{A}}={{\rho }_{w}}{{V}_{b1}}g$  $ \\ $ Volume of the body submerged in water is,$ \\ {{V}_{b1}}=0.70\times {{V}_{b}} \text{(since, it is 70 % submerged in water)}$ $ \\ $ Body is in equilibrium, so by comparing buoyant force and weight of body we have, $ \\ {{F}_{A}}=W\\  \Rightarrow {{\rho }_{w}}{{V}_{b1}}g={{\rho }_{b}}{{V}_{b}}g\\  \Rightarrow {{\rho }_{b}}=\dfrac{{{V}_{b1}}}{{{V}_{B}}}{{\rho }_{w}}\\  \Rightarrow {{\rho }_{b}}=\dfrac{0.70V}{V}\times 1000\\  \therefore {{\rho }_{b}}=700Kg/{{m}^{3}}$

 A raft has a mass of 8000 kg with one-third of its volume submerged in water. If 4 more people get into the raft, with each of them having a mass of 80 kg. Find the additional volume of water displaced due to them. 

This problem can be solved easily without calculating buoyant force acting on a raft because when a body is submerged in liquid the weight of the liquid displaced equals the weight of extra weight added to the system. 

  The weight of the 4 people will be equal to the weight of the displaced water. $ \\ $ Mass of displaced water$ = 4\times 80=320Kg$ $ \\ $ Density of water $ {{\rho }_{w}}=1000Kg/{{m}^{3}} $ $ \\ $ Now, volume of water displaced will be given by $ \\ {{V}_{w}}=\dfrac{{{m}_{w}}}{{{\rho }_{w}}}\\  \Rightarrow {{V}_{w}} =\dfrac{320}{1000}\\  \therefore {{V}_{w}}=0.32{{m}^{3}}$

 A ball of density ${{\rho }_{b}}=300Kg/{{m}^{3}}$ and radius 3 m is tied to a weightless rope, connected to a bottom of container filled with water as shown in the figure. Find the net tension on the rope.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/287bd34b-3bac-4f85-be6d-164c9cd52e307014672818413134833.png"/> 

 The density of the ball is lesser than that of liquid, hence upward buoyant force will be greater than its weight, so the ball will try to go in the upward direction and produce tension in the rope. 

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a4503972-49b9-43b5-9d39-e65493e9781c8381198128535436005.png"/> $\\\text{Radius of ball r =3 m.}\\\text{Volume of ball,}\\  {{V}_{B}}=\dfrac{4}{3}\pi {{r}^{3}}\\  \Rightarrow {{V}_{B}}=\dfrac{4}{3}\pi {{(3)}^{3}}\\\Rightarrow {{V}_{B}}=36\pi\\  \text{Now, weight of ball,}\\  {{W}_{b}}={{\rho }_{b}}{{V}_{B}}g\\  \Rightarrow {{W}_{b}} =300\times 36\pi \times 10\\  \Rightarrow {{W}_{b}} =108000\pi\\  \text{Upward buoyant force will be,}\\  {{F}_{A}}={{\rho }_{W}}{{V}_{B}}g\\  \Rightarrow {{F}_{A}} =1000\times 36\pi \times 10\\  \Rightarrow {{F}_{A}} =360000\pi\\  \text{Now, tension T in rope will be} = \text{buoyant force} – \text{weight of ball}\\  T=360000\pi -108000\pi\\  \Rightarrow T =252000\pi\\ \Rightarrow T=252000\pi\\  \Rightarrow T =791,681\\  \therefore T =7.91\times {{10}^{5}}N$

A cube of side X is 50 % submerged in water in equilibrium, and an external force is applied on it which is to be completely submerged in water. When force is removed, find the net acceleration of cube.$[\text{assume density of water}\,\rho]$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/16d68479-e4bb-4e98-a68d-8ad8d4cf163f3639113808963678785.png"/> 

The upward buoyant force is responsible for the net acceleration of the cube, as it is greater than the weight of the cube. 

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/871427c7-4658-4cdc-9476-bfc16a11109a7117031886413478517.png"/>$ \\ $ Density of the cube is half of the density of water, as it is $50\%$ submerged in water. $ \\ $ Mass of cube $ =\sigma V=\dfrac{\rho }{2}{{X}^{3}}$ $\\ $ When force is removed, the net force in upward direction will be $ \\  \rho {{X}^{3}}g-mg=ma\\  \Rightarrow \rho {{X}^{3}}g-\dfrac{\rho }{2}{{X}^{3}}g=\dfrac{\rho }{2}{{X}^{3}}a\\  \therefore a=g=10m/{{\operatorname{s}}^{2}}$

 A liquid of density $\rho $ is filled inside a U-tube of uniform cross-sectional area A and the length of the liquid column is L. If the liquid from one side is displaced by a distance of $x$, it starts oscillating along with its mean position at time period T. If the density of the liquid is doubled, the cross-sectional area is doubled and length of the liquid column is reduced to $\dfrac{1}{4}^{th}$ what will be the new time period? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4cf77b0d-d3d6-43e1-b513-6581989032dc4680834840751021019.png"/> 

We just need to observe net force acting on the system and it is following the property of SHM or not. 

Total mass of liquid $(m) = \rho AL$ $ \\ $ Total force buoyant force exerted due to liquid will be,$ \\ $ As liquid is displaced from the two positions i.e. right and left so, $ \\  F=-2\rho Axg\\ $ By comparing with standard equation of SHM $\text{(simple harmonic motion)}$ $ \\  F=-kx\\  \Rightarrow k=2\rho Ag\\ $ Now, time period T is given by, $ \\  T=2\pi \sqrt{\dfrac{m}{k}} \\ $ By putting values from the above we have, $ \\  T=2\pi \sqrt{\dfrac{\rho AL}{2\rho Ag}}\\  \Rightarrow T=\sqrt{\dfrac{L}{2g}}\\ $ Now, we can notice that time period doesn’t depends on area or density, so the only factor is length therefore, $ \\  T=\sqrt{\dfrac{L}{2g}}\\  \Rightarrow T=\sqrt{\dfrac{{}^{L}/{}_{4}}{2g}}\\  \therefore T =\dfrac{1}{2}\sqrt{\dfrac{L}{2g}}$