A piece of sealing wax weighs $0.27 k g f$ in air and $0.12 k g f$ when immersed in water. Calculate its:
relative density, and
apparent weight in a liquid of density $800 k g / m^{3}$

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Solution

Step 1: Given data,
Weight of sealing wax in air = $0.27 k g f$ (Mass will be = $0.27 k g$ )
Weight of sealing wax in water = $0.12 k g f$
Density of an unknown liquid = $800 k g / m^{3}$
Also, we know that the density of water at $4^{\circ} C$ is = $1000 k g / m^{3}$
Step 2: Finding the relative density of sealing wax,
$T h e l o s s i n w e i g h t o f s e a l i n g w a x i n w a t e r = W e i g h t o f s e a l i n g w a x i n a i r - W e i g h t o f s e a l i n g w a x i n w a t e r = 0.27 k g f - 0.12 k g f = 0.15 k g f$
$R e l a t i v e D e n s i t y = \frac{W e i g h t o f t h e b o d y}{L o s s o f w e i g h t o f t h e b o d y i n w a t e r} = \frac{0.27 k g f}{0.15 k g f} = 1.8$
Hence, the relative density of the given sealing wax is $1.8$
Step 3: Finding the weight of sealing wax in the unknown liquid,
As we know, weight of water displaced = loss of weight of sealing wax in water = $0.15 k g f$
So, the mass of water displaced will be = $0.15 k g$
Volume of sealing wax = Volume of water displaced = $\frac{m a s s o f w a t e r d i s p l a c e d}{d e n s i t y o f w a t e r} = \frac{0.15 k g}{1000 k g / m^{3}} = 0.00015 m^{3}$
Density of sealing wax = $\frac{m a s s}{v o l u m e} = \frac{0.27 k g}{0.00015 c m^{3}} = 1800 k g / m^{3}$
$R e l a t i v e D e n s i t y o f s e a l i n g w a x = \frac{D e n s i t y o f s e a l i n g w a x}{D e n s i t y o f t h e u n k n o w n l i q u i d} = \frac{1800 k g / m^{3}}{800 k g / m^{3}} = 2.3$
$R e l a t i v e D e n s i t y i s a l s o e q u a l t o = \frac{W e i g h t o f t h e b o d y}{L o s s o f w e i g h t o f t h e b o d y i n t h e u n k n o w n l i q u i d} = \frac{0.27 k g f}{0.27 k g f - a p p a r e n t w e i g h t i n t h e l i q u i d} = 2.3 (f r o m p r e v i o u s e q u a t i o n)$
So, by cross-multiply we get, $A p p a r e n t w e i g h t i n t h e l i q u i d = 0.27 k g f - \frac{0.27 k g f}{2.3} = 0.15 k g f$
Hence, the apparent weight of sealing wax = $0.15 k g f$

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A piece of sealing wax weighs 0.27kgf in air and 0.12kgf when immersed in water. Calculate its:relative density, andapparent weight in a liquid of density 800kg/m3

A piece of sealing wax weighs $0.27 k g f$ in air and $0.12 k g f$ when immersed in water. Calculate its:
relative density, and
apparent weight in a liquid of density $800 k g / m^{3}$