Question

A SLENDER HOMOGENEOUS ROD OF LENGTH 2 L FLOATS PARTLY IMMERSED IN WATER BEING SUPPORTED BY A STRING FASTENED TO ONE OF ITS END . THE SPECIFIC GRAVITY OF THE ROD IS 0.75 .THE LENGTH OF THE THAT EXERTS OUT OF WATER IS 1. L 2. L/2 3. L/4 4. L/3

Answer 1

Rod length = 2 * L 
Let the length of the rod immersed in water = x.
Let the cross section of the rod = A.
Let density of water = d.

Given that the specific gravity of the rod = 0.75
The density of rod = 0.75 d

So mass of the rod = 2 L A * 0.75 d.
Volume of water displaced = x * d

Apply Archimedes principle.

   2 L A 0.75 d = x A * d
   x = 1.5 L 

So the length of the rod that is outside water = 2 L - 1.5 L = 0.5 L.

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Formula :
       Length of object immersed = relative density * length of object.

Answer 2

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