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Question

A solid weighs $$1.5 \ kgf$$ in air and $$0.9 \ kgf$$ in a liquid of density $$1.2\times 10^3  \ kg  m^{-3}$$. Calculate relative density of solid.


A
13
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B
24
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C
3
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D
14
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Solution

The correct option is C $$3$$
The solid apparently weighs less in fluid than in air because of the upward buoyant force acting on it in the fluid. 
According to Archimedes principle, the upward buoyant force is equal to the weight of the volume of fluid displaced by the object.
Weight of fluid displaced = Apparent decrease in weight = $$1.5kgf - 0.9kgf = 0.6kgf$$
Mass of fluid displaced = 0.6kg
So, Volume of Solid = Volume of fluid displaced= $$\dfrac{Mass}{Density} = \dfrac{0.6kg}{1.2 \times {10}^{3}kg/{m}^{3}} = 0.5 \times {10}^{-3}{m}^{3}$$
Actual weight of solid = weight of solid in air =1.5kgf (since buoyant force due to air is negligible)
Mass of solid = 1.5 kg
Density of solid = $$\dfrac{Mass}{Volume}= \dfrac{1.5}{0.5 \times {10}^{-3}} = 3 \times {10}^{3} kg/{m}^{3}$$
Relative density of solid = $$\dfrac{{Density}_{solid}}{{Density}_{water}} = \dfrac{3 \times {10}^{3} kg/{m}^{3}}{1 \times {10}^{3} kg/{m}^{3}} = 3$$

Physics

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