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A body of volume 100 cm3 weighs 5 kgf in air. It is completely immersed in a liquid of density 1.8 × 103 kg m-3. Find: The upthrust due to liquid and The weight of the body in liquid.

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Question

A body of volume 100 cm3 weighs 5 kgf in air. It is completely immersed in a liquid of density 1.8 × 103 kg m-3. Find:

  1. The upthrust due to liquid and
  2. The weight of the body in liquid. 
Sum
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Solution

Volume of body = V = 100 cm3

= 100 × 10-6

= 10-4m3

Weight in air W = 5 kgf

Density of liquid d = 1.8 × 103 kg/m3

(i) Upthrust due to liquid = Volume of the solid submerged × density 0

= 10-4 × 1.8 × 103 × g

= 0.18 gN

= 0.18 kgf

(ii) Weight of body in liquid = Weight of body in air - upthrust

= 5 kgf - 0.18 kgf

= 4.82 kgf

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Relation Between Volume of Submerged Part of a Floating Body, the Densities of Liquid and the Body
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Chapter 5: Upthrust in Fluids, Archimedes’ Principle and Floatation - Exercise 5 (A) [Page 110]

APPEARS IN

Selina Concise Physics [English] Class 9 ICSE
Chapter 5 Upthrust in Fluids, Archimedes’ Principle and Floatation
Exercise 5 (A) | Q 1 | Page 110

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