A solid weighs $32 g f$ in air and $28.8 g f$ in water. Find the volume and relative density of the solid. Find how much will it weigh in a liquid of density $0.9 g / c m^{3}$ .

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Solution

Step 1: Given data,
Weight of solid in air = $32 g f$
Weight of solid in water = $28.8 g f$
Density of liquid = $0.9 g / c m^{3}$
Step 2: Finding the relative density of the solid,
$R e l a t i v e d e n s i t y = \frac{W e i g h t o f t h e s o l i d}{L o s s o f w e i g h t o f t h e s o l i d i n w a t e r} = \frac{32 g f}{32 g f - 28.8 g f} = 10$
Hence, the relative density of solid is $10$
Step 3: Finding the density of the solid,
$R e l a t i v e d e n s i t y = \frac{d e n s i t y o f s o l i d}{d e n s i t y o f w a t e r a t 4^{\circ} C} S o, d e n s i t y o f s o l i d = R e l a t i v e d e n s i t y \times d e n s i t y o f w a t e r a t 4^{\circ} C = 10 \times 1 g / c m^{3} = 10 g / c m^{3}$
Step 4: Finding the volume of the solid,
$D e n s i t y = \frac{m a s s}{v o l u m e} S o, v o l u m e = \frac{m a s s}{d e n s i t y} = \frac{32 g}{10 g / c m^{3}} = 3.2 c m^{3}$
Hence, the volume of solid is $3.2 c m^{3}$
Step 5: Finding the weight of the solid in an unknown liquid,
$R e l a t i v e d e n s i t y = \frac{d e n s i t y o f s o l i d}{d e n s i t y o f l i q u i d} = \frac{10 g / c m^{3}}{0.9 g / c m^{3}} = 11.11$
$R e l a t i v e d e n s i t y o f l i q u i d i s a l s o = \frac{w e i g h t o f t h e s o l i d i n a i r}{l o s s o f w e i g h t o f t h e s o i l d i n l i q u i d} = \frac{32 g f}{32 g f - a p p a r e n t w e i g h t} = 11.11$
By cross-multiply we get, $a p p a r e n t w e i g h t = 32 g f - \frac{32 g f}{11.11} = 29.12 g f$
Hence, the apparent weight of solid in an unknown liquid is $29.12 g f$

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