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
12
B
3
C
2
D
\(\frac{3}{2}\)
E
\(\frac{5}{4}\)
Verified Correct Answer: B.
Physical Principles & Numerical Solution
Relative density of solid = \(\frac{\text{weight of solid}}{\text{weight of an equal volume of water}}\)
\(\frac{4.8}{4.8 -3.2}\) = \(\frac{4.8}{2}\) = 2.4
Ratio of 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
= \(\frac{\text{density of solid}}{\text{density of kerosine}}\)
= \(\frac{2.4}{0.8}\) = 3
Source: Myschool Past Questions Archive (Record #29044)