A glass ball weighs $0.188 k g f$ in air, $0.116 k g f$ in water and $0.125 k g f$ in turpentine. Calculate the density of glass and relative density of turpentine.

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Solution

Step 1: Given data,
The weight of the ball in the air = $0.188 k g f$
The weight of the ball in water = $0.116 k g f$
The weight of the ball in turpentine = $0.125 k g f$
Step 2: Finding the relative density of turpentine,
$r e l a t i v e d e n s i t y o f t u r p e n t i n e (R D) = \frac{w e i g h t o f t h e b o d y i n a i r - w e i g h t o f t h e b o d y i n t u r p e n t i n e}{w e i g h t o f t h e b o d y i n a i r - w e i g h t o f t h e b o d y i n w a t e r} S o, R D = \frac{0.188 k g f - 0.125 k g f}{0.188 k g f - 0.116 k g f} = 0.875$
Hence, the relative density of turpentine is $0.875$ .
Step 3: Finding the density of the glass,
$r e l a t i v e d e n s i t y o f t h e b a l l = \frac{w e i g h t o f t h e b a l l i n a i r}{w e i g h t o f t h e b a l l i n a i r - w e i g h t o f t h e b a l l i n w a t e r} = \frac{0.188 k g f}{0.188 k g f - 0.116 k g f} = 2.61 A l s o, r e l a t i v e d e n s i t y o f t h e b a l l = \frac{d e n s i t y o f t h e b a l l}{d e n s i t y o f w a t e r} S o, d e n s i t y o f t h e b a l l = r e l a t i v e d e n s i t y o f t h e b a l l \times d e n s i t y o f w a t e r = 2.61 \times 1000 k g / m^{3} = 2610 k g / m^{3}$
Hence, the density of the glass is $2610 k g / m^{3}$ .

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