A piece of iron of density 7.8³ km^-3 and volume 100 cm³ is completely immersed in water (p = 1000 kg m^-3). Calculate: (i) the weight of iron piece in air. (ii) the upthrust, and (iii) its apparent weight in water. (g = 10 m s^-2)
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Given :
- Volume of iron piece = 100 cm³
= 100 x 10^-6 m³
= 10^-4 m³
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To Find :
- (i) the weight of iron piece in air.
- (ii) the upthrust.
- (iii) its apparent weight in water.
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Solution :
(i) Weight of iron piece in air
= Volume x density of iron x g
= 10^-4 × (7.8 × 10³) × 10
= 7.8 N
(ii) Upthrust = (Volume of Water displaced) x density of water × g
But volume of water displaced = volume of iron piece when it is completely immersed = 10^-4 m³
Upthrust = 10^-4 × 1000 × 10
= 1 N
(iii) Apparent weight = True weight - Upthrust
= 7.8 - 1
= 6.8 N.
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