A cylindrical solid of area of cross-section 0.0004m2 and length 0.30 m is
completely immersed in water.Calculate (0) Weight of solid in SI system (ii) Upthrust
acting on solid in SI system (iii) Apparent weight of solid in (a) Water (b) Alcohol
[Take g =10m/s2; Density of water =1000 kg/m3; Density of Alcohol =780 kg/m3;
Answers
Given info : A cylindrical solid of area of cross-section 0.0004m² and length 0.30 m is completely immersed in water.
To find : (i) weight of solid in SI system
(ii) upthrust acting on the solid
(iii) apparent weight of solid in (a) water and (b) alcohol.
Solution : volume of solid = cross sectional area × length = 0.0004m² × 0.3 m = 12 × 10¯⁵ m³
so, mass of solid = density of solid × volume of solid
= 1500 kg/m³ × 12 × 10¯⁵ m³
= 18000 × 10¯⁵ kg
= 0.18 kg
so, weight of solid = 0.18 kg × 9.8 m/s² = 1.764 N
(b) upthrust in water = volume of solid × density of liquid (water) × g
= 12 × 10¯⁵ × 1000 × 9.8
= 117.6 × 10¯²
= 1.176 N
Upthrust in alcohol = volume of solid × density of liquid (alcohol) × g
= 12 × 10¯⁵ × 780 × 9.8
= 0.91728 N
(c) apparent weight in water = weight - upthrust
= 1.764 - 1.176 = 0.588 N
Apparent weight in alcohol = weight - upthrust
= 1.764 - 0.91728 = 0.84672 N
Answer:
Explanation:
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