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A piece of wood of uniform cross section...

A piece of wood of uniform cross section and height 15 cm floats vertically with its height 10 cm in water and 12 cm in spirit. Find the density of (i) wood and (ii) spirit

Text Solution

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Cross section of the piece of wood= `"A cm"^2`
Volume of the wood = `"15 A cm"^(3)`
Density of wood = `rho_"WOOD" "g cm"^(-3)`
Mass of wood = `15 A rho_"WOOD"g`
Weight of wood = `15A rho_"WOOD" gf`
Volume of wood under water = `10 "A cm"^3`
Density of water = `rho_"WATER" "g cm"^(-3)`
Mass of displaced water = 10 A `rho_"WATER"g` (where `rho_"WATER"` = Density of water)
Weight of displaced water = `10A rho_"WATER"gf`
So, upthrust = `10A rho_"WATER"gf`
According to the law of flotation,
Weight of the wood = Upthrust
`therefore 15A rho_"WOOD" = 10 A rho_"WATER"`
Or, `15A rho_"WOOD"= 10 A` (putting , `rho_"WATER"= 1 "g cm"^(-3)`)
Or, `15 rho_"WOOD"= 10`
Or, `rho_"WOOD" ` = 10/15
`therefore rho_"WOOD" = 0.667 "g cm"^(-3)`
Volume of wood under spirit = `12 A "cm"^3`
Mass of displaced spirit = `12A rho_"WATER"g` (where `rho_"SPIRIT"` = Density of spirit)
Weight of displaced spirit = `12A rho_"SPIRIT" gf`
So, upthrust = `12A rho_"SPIRIT" gf`
According to the law of flotation,
Weight of the wood = Upthrust
`therefore 15 Arho_"WOOD" = 12 Arho_"SPIRIT"`
Or, 15 x 0.667 = `12 rho_"SPIRIT"` (Putting , `rho_"WOOD"` = 0.667 g `"cm"^(-3)`)
Or , `rho_"SPIRIT"` = 15 x 0.667/12
`therefore rho_"SPIRIT"= 0.83 "g cm"^(-3)`
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