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A Solid of Density 5000 Kg M-3 Weighs 0.5 Kgf in Air. It is Completely Immersed in Water of Density 1000 Kg M-3. Calculate the Apparent Weight of the Solid in Water. - Physics

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प्रश्न

A solid of density 5000 kg m-3 weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg m-3. Calculate the apparent weight of the solid in water. 

योग
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उत्तर

Density of solid = 5000 kg m-3

Weight of solid = 0.5 kgf

Density of water = 1000 kg m-3

Here , Upthrust = Volume of the solid × density of water × g

= `(0.5/"g")/5000 xx 1000 xx "g"`

= `0.5/(5000 xx "g") xx 1000 xx "g" = 0.1` kgf

Apparent weight = True weight - Upthrsut 

= 0.5 - 0.1 = 0.4 kgf

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Determination of Relative Density of a Solid Substance by Archimedes’ Principle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Upthrust and Archimedes’ Principle - Subjective Questions

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