A solid weighs 4.8g in air, 2.8g in water and 3.2g in Kerosine. The ratio of density of the solid to that of the kerosine is
A
2
B
\(\frac{3}{2}\)
C
\(\frac{2}{3}\)
D
3
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Relative density of solid
\(\frac{\text{weight of solid}}{\text{weight of an equal volume of water}}\)
= \(\frac{4.8}{4.8 - 2.8}\)
= \(\frac{4.8}{2}\) = 2.4
Relative density of liquid
= \(\frac{\text{Weight of liquid kerosine}}{\text{Weight of an equal volume of water}}\)
= \(\frac{4.8 - 3.2}{4.8 - 2.8}\) = \(\frac{1.6}{2}\) = 0.8
= \(\frac{\text{Ratio of relative density of solid}}{\text{Relative density of Kerosine}}\)
= \(\frac{\text{density of solid}}{\text{density of Kerosine}}\)
= \(\frac{2.4}{0.8}\)
= 3
Source: Myschool Past Questions Archive (Record #56906)