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

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Solution

Step 1 Given,
Weight of solid body in air = $2.10 N$
Relative Density( RD) of liquid = $1.2$
R.D of solid = $8.4$
Density of water( $ρ_{w a t e r}$ ) = $1000 k g m^{- 3}$
Step 2 Formula used,
$R . D o f s o l i d = \frac{D e n s i t y o f s o l i d (ρ_{s o l i d})}{D e n s i t y o f w a t e r} R . D o f s o l i d = \frac{w e i g h t o f s o l i d i n a i r}{w e i g h t o f w a t e r d i s p l a c e d b y b o d y}$
Step 3 Solution,
$8.4 = \frac{ρ_{s o l i d}}{ρ_{w a t e r}} ρ_{s o l i d} = 8.4 \times ρ_{w a t e r} ρ_{s o l i d} = 8.4 \times ρ_{w a t e r} = 8.4 \times 1000 = 8400 k g / m^{3}$
$R . D o f s o l i d = \frac{w e i g h t o f s o l i d i n a i r}{w e i g h t o f w a t e r d i s p l a c e d b y b o d y} 8.4 = \frac{2.10}{w e i g h t o f w a t e r d i s p l a c e d b y b o d y} w e i g h t o f w a t e r d i s p l a c e d b y b o d y = \frac{2.10}{8.4} = 0.25 N$
$U p t h r u s t d u e t o w a t e r = w e i g h t o f w a t e r d i s p l a c e d b y t h e b o d y$ = $0.25 N$
$U p t h r u s t d u e t o l i q u i d$ $U p t h r u s t d u e t o w a t e r \times R . D o f l i q u i d = 0.25 \times 1.2 = 0.30 N$
$W e i g h t o f b o d y i n l i q u i d = w e i g h t o f t h e b o d y i n t h e a i r - u p t h r u s t d u e t o l i q u i d$
$2.10 - 0.30 = 1.80 N$
The weight of the body in liquid is $1.80 N$

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