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A uniform cylindrical body when placed in liquid a floats with one third of its length outside the liquid. When placed in liquid B, it floats with one third of its length immersed in the liquid. When the body is made to float vertically in a homogeneous mixture of equal volumes of the two liquids, 25 cm of its length is seen in air. Find the length of the body. |
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Answer» Solution :Let 'l' be the length of the cylindrical body. When the body floats in liquid A, `(1//3)` RD of its length is above the liquid. `:.` the fraction inside the liquid `=2//3` `:.(p)/(p_(A))=(2)/(3)rArrp_(A)=(3)/(2)p` where p and `p_(A)` are densities of the body and the liquid 'A', respectively. Similarly when the body is made to float in liquid B, one-third of it is inside the liquid. `:.(p)/(p_(B))=(1)/(3)rArrp_(B)=3p` where `p_(B)` is the density of the liquid 'B'. When equal volumes of the LIQUIDS are COMBINED to from a HOMOGENEOUS mixture, the density of the mixture `rArrp_(m)=(p_(A)+p_(B))/(2)=(1)/(2)(P_(A)+p_(B))` `=(1)/(2)((3)/(2)p+3p)=(9)/(4)p` `:.(p_(m))/(p)=(9)/(4)or(p)/(p_(m))=(4)/(9):.` Fraction of the body inside the mixture `=(4)/(9)` Fraction of the body outside the mixture `=1-(4)/(9)=(5)/(9)` Given length outside the mixture = 25 cm `:.(5l)/(9)=25cmrArr=45cm` `:.` the total length of the cylindrical body = 45 cm |
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