Sum

A block of mass 7 kg and volume 0.07 m^{3} floats in a liquid of density 140 kg/m^{3}. Calculate:

- Volume of block above the surface of liquid.
- Density of block.

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#### Solution

Mass of block = m = 7 kg

Volume of block = V = 0.07 m^{3}

Density of liquid =ρ_{l} = 140 kgm^{-3}

Let V’ = Volume of block immersed in the liquid

**By law of floatation:**

Weight of block = Weight of liquid displaced by the immersed pa

`"V"'rho"g" = "V"'rho_l "g"`

`"V"' = rho/rho_l` V

Density of block = `rho = "m"/"V" = 7/0.07 = 100` kg m^{-3}

From (1)

`"V"' = (100 "V")/140 = 5/7 "V" = 5/7 xx 0.7`

`"V"' = 1/20 = 0.05` m^{3}

Volume of block above the surface of liquid = V = V^{1}

= 0.07 - 0.05 = 0.02 m^{3}

Concept: Archimedes' Principle

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