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A solid weighs 4.8g in air, 2.8g in water and 3.2g in kerosine. The ratio of den...

Question

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

Options

A) 12

B) 3

C) 2

D) $$\frac{3}{2}$$

E) $$\frac{5}{4}$$

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

Dicussion (1)

• 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