A solid of density $ρ = 5000 k g m^{- 3}$ weighs $= 0.5 k g f i n a i r .$ . It is completely immersed in a liquid of density $1000 k g m^{- 3}$ . Calculate the apparent weight of the solid in the liquid

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Solution

Step 1: Given,
Density of solid, $ρ = 5000 k g m^{- 3}$ .
Weight $= 0.5 k g f i n a i r .$
Mass, $M$ $= 0.5 k g .$
Step 2: Formula used,
$V o l u m e = \frac{M a s s}{D e n s i t y} .$
$M a s s o f l i q u i d d i s p l a c e d = V o l u m e \times D e n s i t y o f l i q u i d .$
$A p p a r e n t w e i g h t = W e i g h t i n a i r - W e i g h t o f w a t e r d i s p l a c e d .$
Step 3: Calculate the volume,
$V = \frac{M}{ρ}$ $= \frac{0.5}{5000} = 10^{- 4} m^{3} .$
Step 4: Calculate the mass of water displaced,
weight of liquid displaced $=$ Volume $\times$ Density of liquid.
$= 10^{- 4} \times 1000$
$= 0.1 k g f$
Step 5: Calculate the apparent weight,
Apparent weight $=$ Weight in air $-$ Weight of water displaced.
$= 0.5 k g f - 0.1 k g f = 0.4 k g f .$
Hence, the apparent weight of the solid in the given liquid is, $0.4 k g f .$

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