# Answer to Question #72865 in Mechanics | Relativity for Abby

Question #72865

An object of density 20 grams per centimetre cube weighs 200 grams when suspended from a spring balance. It is then inserted into a liquid of density 0.6 grams per cm cube until 1/3 of its volume is immersed in the liquid. Calculate the tension in the spring in newton.

Expert's answer

Density of the object , D = 20 gm / cm cube

Mass , M = 200 gram

Therefore, Volume of the object is , V = M/D

= 200 / 20 cm cube

Volume immersed in the liquid is = V/3

= 10/3 cm cube

Therefore the buoyancy force exerted by the liquid = volume × density × g

=( 10/3)× 0.6 × 980

=1960 dyne

Tension force will be = weight - buoyancy

= Mg - 1960

=( 200× 980 )- 1960

=196000 - 1960

=194040 dyne

or 1.94 N

Mass , M = 200 gram

Therefore, Volume of the object is , V = M/D

= 200 / 20 cm cube

Volume immersed in the liquid is = V/3

= 10/3 cm cube

Therefore the buoyancy force exerted by the liquid = volume × density × g

=( 10/3)× 0.6 × 980

=1960 dyne

Tension force will be = weight - buoyancy

= Mg - 1960

=( 200× 980 )- 1960

=196000 - 1960

=194040 dyne

or 1.94 N

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