Question 27: Properties of Matter

Question

A 400 g solid cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water?(Given: density of water = 1000 kg m-3)

Accepted Answer

First, letÃ¢ÂÂs find the weight of the cube and set it equal to the upward buoyant force that keeps the cube floating.
Step 1: Set up the equation for floating
The weight of the cube (mass ÃÂ gravity) is balanced by the buoyant force (density of water ÃÂ volume of cube under water ÃÂ gravity):
$$\mathrm{Mg} = \mathrm{F}_{\mathrm{B}} \Rightarrow (400 	imes 10^{-3}) = 10^3 	imes \mathrm{V}_{\mathrm{d}}$$
Step 2: Solve for the volume under water
We find the volume of the cube under water ($$\mathrm{V}_\mathrm{d}$$):
$$\mathrm{V}_{\mathrm{d}} = 400 	imes 10^{-6}~\mathrm{m}^3$$
Step 3: Find the total volume of the cube
Each edge of the cube is 10 cm, so total volume is:
$$ (10 	imes 10^{-2})^3 $$
Step 4: Find the volume outside water
Subtract the volume under water from the total volume to get the volume outside water:
$$(	ext {Vol.})_{	ext{outside}} = (10 	imes 10^{-2})^3 - 400 	imes 10^{-6}$$
Calculate the answer:
$$= 600 	imes 10^{-6}~\mathrm{m}^3 = 600~\mathrm{cm}^3$$

Answer

400 cm3

Answer

600 cm3

Answer

1400 cm3

Answer

4000 cm3

Properties of Matter

A 400 g solid cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water?(Given: density of water = 1000 kg m-3)

Choose the correct answer: