Question

The specific gravity of a fluid with a weight density of 31.2 lb/ft is a. 2.00 b. 0.969 c. 0.500 d. 1.03

Answer 1

Answer is c 0.500

Step-by-step explanation:

`SpecificGravity=(\rho _(fluid)*g)/(\rho _(water)*g)`

We know that
`\rho_(water)=62.42lb/ft^(3)`

Applying values we get

`SpecificGravity=(31.2)/(62.4)=0.5`