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A cube of 1 kg is floating at the inte...

A cube of 1 kg is floating at the interface of two liquids of density ` rho_(1) = 3//4 gm//cm^(3) and rho_(2) = 2.5 gm//cm^(3)` The cube has `(3//5)^(th)` part in `rho_(2)` and `(2//5)^(th)` part in `rho_(1) ` .Find the density of the cube .

Text Solution

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The weight of the cube = (force acting by the liquid of density `rho_(1)`) + (force acting by the liquid of density of `rho_(2)`)
Let mass of the block = m
` mg = v_(1) rho_(1)g +v_(2)rho_(2)g`
`rArr (ApL) = { A (2//5)L}rho_(1)g +v_(2) rho_(2)g`
` rArr (Arho L)= { A(2//5)L} rho_(1) + { A(3//5)L} rho_(2)`
`rArr = (2rho_(1) +3rho_(2))/6`
`rArr rho = (2(3//4) +3 (2.5))/5 = (1.5 + 7.5)/5 = 1.8 gm/cm^(3)`
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