Question

3. A solid of area of cross-section 0.004 mº and length 0.60 m is completely immersed in water of
density 1000 kgm.
Calculate : (i) Wt. of solid in SI system
(ii) Upthrust acting on the solid in SI system
(iii) Apparent weight of solid in water,
(iv) Apparent weight of solid in brine solution of density 1050 kgm
[Take g = 10 N/kg; Density of solid = 7200 kgm 31
[(i) 172.8 N (ii) 24 N (iii) 148.8 N (iv) 147.6 N​

Answer 1

Explanation:

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PHYSICS

A solid of density 5000 Kg m

−3

weight 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.

Density ρ = 5000 kg/m

3

Weight = 0.5 kg f in air

mass M = 0.5 kg

Volume V = M / ρ

= 0.5 / 5000 = 10

−4

m

3

mass of water displaced = V

∗

density of water

= 10

−4

∗1000 = 0.1 kg

Apparent weight = wt in air - wt of water displaced

= 0.5 kg f - 0.1 kg f

= 0.4 kg f