A solid of density 5000kg-m3 weighs 0-5 kg in air -It is completely immersed in a liquid of density 8000 kg-m3-Calculate the apparent weight of the solid in the liquid-

Step 1:Given dataweight of solid =0.5 kg.Density of solid = 5000 kg/m^3.Density of liquid = 8000 kg/m^3.Step 2: Formula used:Density =Mass/volume.Volume =Mass ...

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Standard IX

Physics

Fundamental and Derived Units

A solid of de...

Question

A solid of density 5000 $\frac{k g}{m^{3}}$ weighs 0.5 kg in air .It is completely immersed in a liquid of density 8000 $\frac{k g}{m^{3}}$ .Calculate the apparent weight of the solid in the liquid.

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Solution

Step 1:Given data
weight of solid =0.5 kg.
Density of solid = 5000 $\frac{k g}{m^{3}}$ .
Density of liquid = 8000 $\frac{k g}{m^{3}}$ .
Step 2: Formula used:
Density = $\frac{M a s s}{v o l u m e}$ .
Volume = $\frac{M a s s}{D e n s i t y}$ .
Mass of water displaced = Volume $\times$ density of water.
Apparent weight =Weight in air - weight of water displaced .
Step 3: calculation
Volume = $\frac{M a s s}{D e n s i t y}$ .
Volume = $\frac{0.5 k g}{5000 \frac{k g}{m^{3}}} = \frac{1}{10^{4}} m^{3} o r 10^{- 4} m^{3}$ .
Mass of water displaced = Volume $\times$ density of water.
Mass of water displaced = $10^{- 4} \times 8000 = 0.8 k g .$
Apparent weight =Weight in air - weight of water displaced .
Apparent weight $= 0.5 k g - 0.8 k g$
Apparent weight $= - 0.3 k g = - 0.3 k g w t .$
Hence, Apparent weight $- 0.3 k g$

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A solid of density 5000kgm3 weighs 0.5 kg in air .It is completely immersed in a liquid of density 8000 kgm3.Calculate the apparent weight of the solid in the liquid.

A solid of density 5000 $\frac{k g}{m^{3}}$ weighs 0.5 kg in air .It is completely immersed in a liquid of density 8000 $\frac{k g}{m^{3}}$ .Calculate the apparent weight of the solid in the liquid.