A piece of metal weighs of $0.05 k g f$ in air and $0.045 k g f$ when immersed in an oil of specific gravity $0.8$ . Find the relative density of the metal.

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Solution

Step 1: Given data,
The weight of the metal in the air = $0.05 k g f$
The weight of the metal in oil = $0.045 k g f$
The specific gravity or relative density of the oil = $0.8$
We know that the density of water is = $1000 k g / m^{3}$
Step 2: Finding the relative density of the metal,
$S o, w e i g h t o f m e t a l i n w a t e r = w e i g h t o f t h e m e t a l i n a i r - \frac{w e i g h t o f t h e m e t a l i n a i r - w e i g h t o f t h e m e t a l i n o i l}{r e l a t i v e d e n s i t y o f o i l} w e i g h t o f m e t a l i n w a t e r = 0.05 k g f - \frac{0.05 k g f - 0.045 k g f}{0.8} = 0.04375 k g f r e l a t i v e d e n s i t y o f m e t a l = \frac{w e i g h t o f b o d y i n a i r}{w e i g h t o f b o d y i n a i r - w e i g h t o f b o d y i n w a t e r} = \frac{0.05}{0.05 - 0.04375} = 8$
Hence, the relative density of the metal is $8$ .

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