An iron casting containing a number of cavities weight `6000N` in air and `4000N` in water. What is the volume of the cavities in the casting? Density of iron is `7.87g//cm^(3)`. Take `g = 9.8 m//s^(2)` and density of water `= 10^(3) kg//m^(3)`.

Options

A. `0.16 m^(3)`
B. `0.2 m^(3)`
C. `0.12 m^(3)`
D. `0.14 m^(3)`

Answer

C

Explanation

Let v be the volume of cavities and V the volume of solid iron.
Then `V=("mass")/("density")=((6000//9.8)/(7.87xx10^(3)))=0.078 m^(3)`
Further, decrease in weight = upthrust
`therefore (6000-4000)=(V+v)rho_(w)g`
or `2000=(0.078+v)xx10^(3)xx9.8`
or `0.078+v cong 0.2`
`therefore v=0.12 m^(3)`