A solid body weighs $2.10 N$ in air. Its relative density is $8.4$ . How much will the body weigh if placed in:
Water
A liquid of relative density 1.2

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Solution

Step 1: Given data,
Weight of body in air = $2.10 N$ (mass will be $0.21 k g$ taking acceleration due to gravity as $10 m / s^{2}$ )
Relative density of the body = $8.4$
We know that the density of water at $4 C$ = $1000 k g / m^{3}$
Step 2: Finding the weight of the body in water,
$R e l a t i v e d e n s i t y = \frac{w e i g h t o f t h e b o d y i n a i r}{w e i g h t o f t h e b o d y i n a i r - w e i g h t o f t h e b o d y i n w a t e r} S o, w e i g h t o f t h e b o d y i n w a t e r = w e i g h t o f t h e b o d y i n a i r - \frac{w e i g h t o f t h e b o d y i n a i r}{r e l a t i v e d e n s i t y} = 2.10 N - \frac{2.10 N}{8.4} = 1.85 N$
Hence, the weight of the body in water is $1.85 N$
Step 3: Finding the density of liquid with RD 1.2,
$r e l a t i v e d e n s i t y o f t h e l i q u i d = \frac{d e n s i t y o f l i q u i d}{d e n s i t y o f w a t e r} S o, d e n s i t y o f l i q u i d = r e l a t i v e d e n s i t y \times d e n s i t y o f w a t e r = 1.2 \times 1000 k g / m^{3} = 1200 k g / m^{3}$
Step 4: Finding the density of the body,
$r e l a t i v e d e n s i t y o f t h e b o d y = \frac{d e n s i t y o f t h e b o d y}{d e n s i t y o f w a t e r} S o, d e n s i t y o f t h e b o d y = d e n s i t y o f t h e b o d y \times d e n s i t y o f w a t e r = 8.4 \times 1000 k g / m^{3} = 8400 k g / m^{3}$
Step 5: Finding the weight of the body in a liquid with RD 1.2,
$r e l a t i v e d e n s i t y o f t h e b o d y = \frac{d e n s i t y o f t h e b o d y}{d e n s i t y o f l i q u i d} = \frac{8400 k g / m^{3}}{1200 k g / m^{3}} = 7 A l s o r e l a t i v e d e n s i t y o f t h e b o d y = \frac{w e i g h t o f t h e b o d y i n a i r}{w e i g h t o f t h e b o d y i n a i r - w e i g h t o f t h e b o d y i n l i q u i d} S o, w e i g h t o f t h e b o d y i n l i q u i d = w e i g h t o f t h e b o d y i n a i r - \frac{w e i g h t o f t h e b o d y i n a i r}{r e l a t i v e d e n s i t y} = 2.10 N - \frac{2.10 N}{7} = 1.8 N$
Hence, the weight of the body in the liquid is $1.8 N$ .

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