Question

A block of steel of size $ 5\,cm\times 5\,cm\times 5\,cm $ is weighed in water. If the relative density of steel is $7$, its apparent weight is

• A. $ 4\times 4\times 4\times 6g $
• B. $ 5\times 5\times 5\times 9g $
• C. $ 4\times 4\times 4\times 7\text{ }g $
• D. $ 6\times 5\times 5\times 5\text{ }g $

Answer 1

Answer: (D)

From Archimedes principle, when a solid body is immersed in a liquid, then there is some apparent loss in its weight. This apparent loss in weight of the body is due to upward force applied by the liquid on the body. This force is called the buoyant force or upthrust.
$ \therefore $ Apparent weight = weight in air $ - $ force of buoyancy
$ =(V{{d}_{steel}}-V{{d}_{water}})g $
$ =(5\times 5\times 5\times 7)g-(5\times 5\times 5\times l)g $
$ =(5\times 5\times 5\times 6)g $