A metal cube of side 5 cm and density 7.9 g cm⁻³ is suspended by a thread and is immersed completely in a liquid of density 1.1 g cm⁻³.
Find :
(a) the weight of cube,
(b) the upthrust on cube and
(c) the tension in thread.
Answers
Side of the metal cube = 5cm (Given)
Thus, Volume of metal cube = a³
(5)³ = 125 x 10−6 m3.
Density of metal = 7.9 g/cm3 = 7900 kg/m3.
Mass of the metal cube m = volume x density
= 125 x 10−6 × 7900
= 0.987 kg
Weight of the metal cube = mg = 0.987 kg x 10 m/s2 = 9.87 N
Volume of liquid displaced = Volume of metal cube = 125 x 10−6 m3.
Mass of liquid displaced = Volume x density of liquid
= 125 x 10−6 m3 × 1100 kg/m3.
= 0.13 kg
Weight of liquid displaced = 0.13 kg x 10 m/s2
= 1.3 N = upthrust
Weight of metal cube in liquid = Weight of metal cube in air - the upthrust
= 9.87 − 1.3
= 9.75 N
Therefore tension in the thread = 8.57 N.
Explanation:
Volume of metal cube = (5 cm)3 = 125 x 10−6 m3.
Density of metal = 9 g/cm3 = 9000 kg/m3.
Mass of the metal cube m = volume x density
= 125 x 10−6 m3.9000 kg/m3.
= 1.125 kg
Weight of the metal cube in air = mg = 1.125 kg x 10 m/s2 = 11.25 N
Volume of liquid displaced = Volume of metal cube = 125 x 10−6 m3.
Mass of liquid displaced = Volume x density of liquid
= 125 x 10−6 m3.1200 kg/m3.
= 0.15 kg
Weight of liquid displaced = 0.15 kg x 10 m/s2 = 1.5 N = upthrust
Weight of metal cube in liquid = Weight of the metal cube in air - upthrust
= 11.25 − 1.5
= 9.75 N
Therefore tension in the string = 9.75 N.
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