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A solid weighs 1.5 kgf in air and 0.9 kgf in a liquid of density 1.2 x 103 kg m-3. Calculate R.D. of solid.

Fluids Upthrust

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Answer

Given,

Weight of the solid in air (W1) = 1.5 kgf

Weight of the solid in liquid (W2) = 0.9 kgf

Density of the liquid = 1.2 x 103 kg m-3

R.D. of liquid=Density of liquid in kg m−31000 kg cm−3=1.2×1031000=1.2\text{R.D. of liquid} = \dfrac{\text {Density of liquid in kg m}^{-3}}{\text{1000 kg cm}^{-3}} \\[0.5em] = \dfrac{1.2 \times 10^3}{1000} \\[0.5em] = 1.2

R.D. of solid=(W1W1−W2)×R.D. of liquid=(1.51.5−0.9)×1.2=(1.50.6)×1.2=3.0\text{R.D. of solid} = \Big(\dfrac{W{1}}{W{1} − W_{2}}\Big) \times \text{R.D. of liquid} \\[0.75em] = \Big(\dfrac{1.5}{1.5 - 0.9}\Big) \times 1.2 \\[0.75em] = \Big(\dfrac{1.5}{0.6}\Big) \times 1.2 \\[0.75em] = 3.0

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