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Question

A piece of iron of density 7.8×103 kg m3 and volume 100 cm3 is totally immersed in water (ρ = 1000 kg m3). Calculate its apparent weight in water.


A

7 N

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B

6 N

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C

1 N

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D

6.8 N

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Solution

The correct option is D

6.8 N


Given,

Volume of iron piece = 100 cm3 =100×106m3=104m3
Weight of iron piece in air = Volume × density of iron × g
Weight of iron piece in air =104×(7.8×103)×10
Weight of iron piece in air = 7.8 N
Upthrust = (Volume of water displaced) × density of water ×g
But volume of water displaced = Volume of iron piece when it is completely immersed = 104m3
Upthrust = 104×1000×10=1N
Apparent Weight = True weight – Upthrust = 7.81 = 6.8 N.


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