Question

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

• A. $6\times 5\times 5\times 5gf$
• B. $4\times 4\times 4\times 7gf$
• C. $5\times 5\times 5\times 5gf$
• D. $4\times 4\times 4\times 5gf$

Answer 1

The correct option is A $6\times 5\times 5\times 5gf$

Apparent weight = actual weight - buoyant force

= V do g - V dw g

= Vg (do - dw)

Here do and dw are densities of object and water, V is the volume of object.

Relative density = do / dw

do/dw = 7

subtract 1 from both sideso - dw ) / dw = 6

do - dw = 6 dw

since density of water in cgs is 1

do - dw = 6

Apparent weight = Vg(do - dw)= 6Vg

= 6 x 5 x 5 x 5 g

= 750g

g is the acceleration due to gravity.