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Question

A block of silver of mass 4kg hanging from a string is immersed in a liquid of a relative density of 0.72. If the relative density of silver is 10, then tension in the string will be


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Solution

Step1: Given data

A block of silver of mass 4kg hanging from a string is immersed in a liquid of a relative density of 0.72.

Relative density of silver is 10.

Step2: Relative density

  1. The relative density of the substance is defined as the ratio of the density of the substance to that of the density of water.
  2. The string is assumed to be massless. Therefore the tension in the string will be due to the net force acting on the silver block downwards.

Step3: Formula used

FB=ρLgV[FB=buoyantforce,ρL=densityofliquid,g=accelerationduetogravity,v=volumeoftheobjectinsidewater.]FG=mg[FG=gravitationalforce,m=mass]Step4: Calculating net force on the block

  1. Let us say there is uniform tension T in the string.
  2. From the below diagram, we can see that the buoyant force (FB) acts in the upward direction
  3. While the gravitational force (FG) acts in the downward direction.
  4. The net force on the block is zero i.e. there is no relative motion of the block. Hence from the above free body diagram of the silver block, the equation for net force on the body is given by, FG-FB-T=0

Step5: Calculating tension

Let us say a body of volume V is completely immersed in a liquid of density ρL .

Therefore the buoyant force on the body is given by, FB=ρLgV

Hence the equation of motion for the silver block is,

mg-ρLgV-T=0T=mg-ρLgV.............(i)

Step6: Calculating the relative density

Let us say the density of silver be ρs, density of the liquid be ρLand density of water be ρw.

Therefore the relative density of silver is given by, ρs/w=ρsρw.........(ii)

And the relative density of the liquid is given by, ρL/w=ρLρw.........(iii)

The relative density of silver and the liquid is given to us. Hence dividing equation 3 by 2 we get,

ρL/wρs/w=ρLρwρsρw0.7210=ρLρsρLρs=0.072

Step7: Calculating tension in the string

Using equation (i)

T=mg-ρLgVT=mg-ρLgmvT=4kg10-ρL(10)ρsAs,mv=ρsT=4[10-0.072(10)]As,ρLρs=0.072T=4[9.28)]T=37.12N

Hence, the tension in the string will be 37.12 N.


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