JAMB - Physics (1981 - No. 43)
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
12
3
2
\(\frac{3}{2}\)
\(\frac{5}{4}\)
Explanation
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
\(\frac{4.8}{4.8 -3.2}\) = \(\frac{4.8}{2}\) = 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
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