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A solid body floating in water has `(1)/(5^(th))` of its volume immersed in it. What fraction of its volume will be immersed, if it floats in a liquid of specific gravity 1.2 ? |
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Answer» Let volume of the solid body = `V_(s)` Density of the solid body = `rho_(s)` Density of water = `rho_(w)` Given that the body floats in water with one-fifth of its volume immersed i.e., immersed volume = `(V_(s))/(5)` `because (rho_(s))/(rho_(w)) = ("Immersed volume of the body")/("Total volume of the body")` `rArr (rho_(s))/(rho_(w))=(V_(s)//5)/(V_(s))` `rArr (rho_(s))/(rho_(w))=(1)/(5)" "...(i)` Let `(1)/(n^(th))` volume of the solid body is immersed when it floats in a liquid of relative density 1.2. Density of the liquid `rho_(l)=(1.2) rho_(w)` So, `(rho_(s))/(rho_(l))=(V_(s)//n)/(V_(s))` `rArr (rho_(s))/(1.2 rho_(w))=(1)/(eta)` `rArr (1)/(1.2 xx 5)=(1)/(eta)" "("from equation (i)")` `rArr (1)/(eta)=(1)/(6)` Thus, one-sixth of the volume remains immersed. |
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