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

Buoyancy Quiz 2

physics

 A body is dropped in a water tank. If the body moves down by a distance of $x$ then calculate the work done by the body. Assume the density of the body to be $d$ and that of water to be $\rho $. 

 Work done by a force is equal to the product of force and the displacement that occurred due to it.Here a net downwards force is acting on the body due to which it shows some displacement. 

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e725463c-217f-469b-a000-225d35a8e5463256954879986812363.png"/> The density of the body is given by; $d=\dfrac{m}{V} \\ \Rightarrow m=Vd $ From the free body diagram; $F=\rho Vg-mg \\ \Rightarrow F=\rho Vg-Vdg \\ \Rightarrow F=\left( \rho -d \right)Vg $ The work done by the body is given by; $W=Fx \\ \therefore W=\left( \rho -d \right)Vgx$

Refer to the following diagram and evaluate the mass of the cube. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7689f32d-249e-4305-a280-8d4bec07d8ee1304441157767010770.png"/>

 There is some downward force along with the self-weight of the cube that is acting on the top of the cube and balanced by the buoyant force. 

The pressure acting on the top surface of the block is given by; $P={{P}_{0}}+{{\rho }_{o}}g{{h}_{0}}$ The force on the top surface will be; $F=PA \\ A={{a}^{2}} \\ \Rightarrow F=\left[ {{P}_{0}}+{{\rho }_{o}}g{{h}_{0}} \right]{{a}^{2}}$ The cube is floating, so the forces acting on it will be given by; $F=mg-{{\rho }_{w}}Vg \\ V={{a}^{3}} \\ \Rightarrow F=mg-{{\rho }_{w}}g{{a}^{3}}$ Substituting the values; $\left[ {{P}_{0}}+{{\rho }_{o}}g{{h}_{0}} \right]{{a}^{2}}=mg-{{\rho }_{w}}g{{a}^{3}} \\ \Rightarrow {{P}_{0}}{{a}^{2}}+{{\rho }_{o}}g{{h}_{0}}{{a}^{2}}+{{\rho }_{w}}g{{a}^{3}}=mg \\ \therefore m=\dfrac{{{P}_{0}}{{a}^{2}}}{g}+{{\rho }_{o}}{{h}_{0}}{{a}^{2}}+{{\rho }_{w}}{{a}^{3}}$

 A block of mass $m$ and dimensions $2\times 4\times 6\text{ m}$ is floating with some part of it under water and some above. If the density of water is $1.028\times {{10}^{3}}\text{ kg}{{\text{m}}^{-3}}$ , calculate the percentage fraction of the block that will be underwater. 

 The density of the body is inversely proportional to the volume of the body. Inverse ratio of density gives the fraction of volume that is underwater. 

The density of the block is given by; $d=\dfrac{m}{V} \\ \Rightarrow V=2\times 4\times 6=48\text{ }{{\text{m}}^{3}} \\ \Rightarrow d=\dfrac{m}{48} $ But the block is floating; $\rho vg=dVg \\ \Rightarrow \dfrac{v}{V}=\dfrac{d}{\rho } \\ \Rightarrow \dfrac{v}{V}=\dfrac{m}{48\times 1.028\times {{10}^{3}}} \\ \Rightarrow \dfrac{v}{V}=\dfrac{m\times {{10}^{-3}}}{19.344}$ The percentage fraction is given by; $\dfrac{v}{V}\times 100=\dfrac{m\times {{10}^{-3}}\times 100}{19.344}=\dfrac{m}{193.44}\%$

 A wooden plank of density $d$ and thickness $t$ is floating such that only the top surface area is above the water. Determine the maximum number of wooden logs that the plank can hold without sinking if a single wooden log has mass $m$. The surface area of the top of the plank is $A$. 

 The total weight of the logs along with the self-weight of the plank should be equal to the upward buoyant force.

The buoyant force $B$ is given by; $B=\rho Vg-mg \\ \Rightarrow V=At \\ \Rightarrow m=dV=dAt \\ \Rightarrow B=\rho Atg-dAtg $ Let the maximum number of logs that the plank can hold be $n$ $nmg\le B \\ \Rightarrow nmg\le \rho Atg-dAtg \\ \therefore n\le \dfrac{\left( \rho -d \right)At}{m}$

 A ball of mass $m$ and density $d$ is inside a vertical cylinder of length $l$. The cylinder has some fluid of density $\rho $ and the cylinder is moving up with an acceleration of $a$. Calculate the time taken by the ball to reach the bottom of the cylinder. 

 The ball has a net downward acceleration and it is proportional to the buoyant force and self-weight of the ball. 

The buoyant force for vertically accelerated fluid is given by; $B=\rho V{{g}_{eff}} \\ \Rightarrow {{g}_{eff}}=g-a \\ V=\dfrac{m}{d} \\ \Rightarrow B=\rho \left( \dfrac{m}{d} \right)\left( g-a \right)$ But the net acceleration of the ball is given by; ${{a}_{net}}=\dfrac{B-mg}{m} \\ \Rightarrow {{a}_{net}}=\dfrac{\left( \dfrac{\rho m\left( g-a \right)}{d} \right)-mg}{m} \\ \Rightarrow {{a}_{net}}=\dfrac{\rho \left( g-a \right)-dg}{d} $ The displacement $l$ is given by; $l=\dfrac{1}{2}{{a}_{net}}{{t}^{2}} \\ \Rightarrow 2l=\left( \dfrac{\rho \left( g-a \right)-dg}{d} \right){{t}^{2}} \\ \Rightarrow {{t}^{2}}=\dfrac{2dl}{\rho \left( g-a \right)-dg} \\ \therefore t=\sqrt{\dfrac{2dl}{\rho \left( g-a \right)-dg}}$