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Two non-mixing liquids of densities ρ and nρ (n>1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p<1) in the denser liquid. The density d is equal to:
$\eqalign{
  & A) \left\{ {1 + \left( {n - 1} \right)p} \right\}\rho \cr
  & B) \left\{ {1 + \left( {n + 1} \right)p} \right\}\rho \cr
  & C) \left\{ {2 + \left( {n + 1} \right)p} \right\}\rho \cr
  & D) \left\{ {2 + \left( {n - 1} \right)p} \right\}\rho \cr} $

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: According to the law of flotation, the cylinder floats in the liquid if the weight of the liquid displaced by it is equal to its weight.

Complete step by step solution:
Archimedes Principle states that a body is completely or partially immersed in a fluid (gas or liquid), the upward, or coercive, act is done by the force, the magnitude of which is equal to the weight of the fluid displaced by the body.
Using this principle we have
Weight of cylinder= $Up\,thrust_1$ + $Up\,thrust_2$
$\eqalign{
  & \Rightarrow LAdg = {\text{ }}(\rho L)A(n\rho )g{\text{ }} + {\text{ }}(1 - p)LA\rho g \cr
  & {\text{since volume}} = mass/density \cr
  & \Rightarrow d = (1 - p)\rho + (pn)\rho \cr
  & \therefore d = [1 + (n - 1)p)\rho \cr} $

Additional Information: Archimedes' theory is very useful for calculating the volume of an object that does not have a regular shape. An object of odd shape can be submerged, and the amount of fluid displaced is equal to the volume of the object. It can also be used to calculate the density or specific gravity of an object. When the object is submerged, it weighs less because of the buoyant force pushing upward. The object's specific gravity is then the object's weight in air divided by how much weight the object loses when placed in water. But most importantly, the principle describes the behaviour of any body in any fluid,
If the weight of an object is less than the displaced fluid, the object rises, such as in the case of a wooden block that is released below the surface of water or a helium-filled balloon that is loose in the air. An object heavier than the amount of the fluid it displaces, though it sinks when released, has an apparent weight loss equal to the weight of the fluid displaced. In fact, in some accurate weighings, a correction must be made in order to compensate for the buoyancy effect of the surrounding air.

Note: The buoyancy force, which always opposes gravity, is still due to gravity. The fluid pressure increases with depth due to the (gravity) load of the fluid above. This increased pressure applies a force to the submerged object that increases with depth. The result is buoyancy.