Fluid MechanicsHard
Question
A 10 cm side cube weighing 5N is immersed in a liquid of relative density 0.8 contained in a rectangular tank of cross section a area 15cm x 15cm. If the tank contained liquid to a height of 8 cm before the immersion, the level of the liquid surface is :
Options
A.
cm
cmB.
cm
cmC.10 cm
D.11 cm
Solution
Given: ρc × (10)3 × g = 5 × 105 dynes ⇒ ρc = 0.5; g = 1000 cm/s2, Buoyant force should balance weight ⇒ ρl Vl g = 5 × 105 ⇒ Vl = 625 cm3, ∴ depth upto which cube is dipped is ⇒ 625 = 10 × 10 × d ⇒ d = 6.25 cm. Now, 15 × 15 × h = 15 × 15 × 8 + 625 ⇒ h = 97/9 [ ρc (density of cube), ρl (density of liquid), Vl (volume of liquid displacement)]
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