How Much of the Wooden Block Emerges When Floating in Water Compared to Alcohol?

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SUMMARY

A wooden block floats in alcohol with 3/8 of its length above the surface, indicating that 5/8 of the block is submerged. Given the density of alcohol at 0.8 g/cc, the buoyant force can be calculated using the formula F = (5/8)V(0.8 g/cc)g. When the block is placed in water, which has a density of 1.0 g/cc, the submerged fraction can be determined to be 0.5, meaning 1/2 of the block's length will be above the water surface.

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Homework Statement


A wooden block floats in alcohol with 3/8th of it's length above alcohol. If it is made to float in water , what fraction of it's length is above water? [Density of alcohol = 0.8 gm/cc]

Homework Equations


density = mass/Volume

The Attempt at a Solution



5/8 of the block is under the alcohol. So the volume of the displaced fluid is 5/8V where V is volume of the block. density = mass/volume so: 0.8g/cc = m/[(5/8)V] m = (5/8) V (0.8g/cc) Then I got the buoyant force -: F = (5/8) V (0.8g/cc) g . After this i create such equation for water with water's density and fraction of block is submerged but how can i relate them :?
 
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Well the buoyant force will be the same (weight of block doesn't change) but the 5/8 figure will be some thing different, say K.

(5/8)*V*0.8 = K*V*1.0

so cancel out the V's and we get

K = (5/8)*0.8 = 0.5
 
ah yes , duh @ me :|
Thanks a lot Carid :)
 

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