A solid weight 45N and 15N respectively in air and water. Determine the relative density of the solid
- 0.33
- 0.50
-
1.50
- 3.00
The correct answer is: C
Explanation
R.d = \(\frac{\text{weight in air}}{\text{apparent loss of weight in water}} = \frac{45}{45 - 15}\)= \(\frac{45}{30} = 1.50\)