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प्रश्न
A solid of R.D. 4.2 is found to weigh 0.200 kgf in air. Find its apparent weight in water.
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उत्तर
Relative density of solid = R.D. = 4.2
Weight of solid in air = W = 0.200 kgf
R.D. = `"Weight of solid in air"/("wt. of solid in air" - "wt. of solid in water")`
Also, wt. of solid in air - Wt. of solid in water = Upthrust
R.D. = `"Wt. of solid in air"/"Upthrust"`
`4.2 = 0.200/"Upthrust"`
Upthrust = `0.200/4.2 = 0.0476` kgf
So, apparent wt. of solid in water = wt. of solid in air - Upthrust
= 0.200 - 0.0476 = 0.1524 ≅ 0.15 kgf
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संबंधित प्रश्न
A body of volume V and density ρ is kept completely immersed in a liquid of density ρL. If g is the acceleration due to gravity, then write expressions for the following:
(i) The weight of the body, (ii) The upthrust on the body,
(iii) The apparent weight of the body in liquid, (iv) The loss in weight of the body.
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[Hint: Both have equal volume inside the water].
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A piece of metal weighs 44.5 gf in air, 39.5 gf in water. What is the R.D. of the metal?
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