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)`.

Answer

A, B, 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)))=0.078m^(3)`
Further, decrease in weight =upthrust
`:. (6000-4000)=(V+v)rho_(w)g`
or,` 2000=(0.078+v)xx 10^(3)xx9.8`
or,` 0.078+v~~0.2`
`v=0.12m^(3)`.