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If a block of iron (density 5gcm⁻³) is size 5 cm xx 5 cm xx 5 cm was weight while completely submerged in water, what would be the apperent weight ?

A. 5xx5xx5xx5 gf
B. 4xx4xx4xx5 gf
C. 3xx5xx5xx5 gf
D. 4xx5xx5xx5 gf

User Kian Cross
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1 Answer

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Final answer:

The apparent weight of a block of iron when completely submerged in water is found by subtracting the weight of the water displaced from the actual weight of the iron block. In this case, the apparent weight of the iron block is 500 g, which corresponds to option D, 4 x 5 x 5 x 5 g.

Step-by-step explanation:

To find the apparent weight of a block of iron when completely submerged in water, we need to consider the concept of buoyancy. Buoyancy is the upward force exerted by a fluid that opposes the weight of an object immersed in it. According to Archimedes' principle, the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

The density of the block is given as 5 g/cm³, and its volume is the product of its dimensions, which is 5 cm x 5 cm x 5 cm. So, the volume is 125 cm³. The weight of the water displaced is equal to the volume of the iron block times the density of water (1 g/cm³), which equals 125 g.

The actual weight of the iron block is its volume times its density, which is 125 cm³ x 5 g/cm³ = 625 g. To find the apparent weight of the block, we subtract the weight of the water displaced from the actual weight of the block: 625 g - 125 g = 500 g. Therefore, the block would apparently weigh 500 g when completely submerged in water, which can be represented as 4 x 5 x 5 x 5 g or option D.