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JEE Main 2019 (9 Apr, Shift 2) · Archimedes' Principle & Floatation

Archimedes' Principle & Floatation question from JEE Main 2019 (9 Apr, Shift 2)

Q19JEE Main 2019 · 9 APR · SHIFT 2Medium

Archimedes' Principle & Floatation

A wooden block floating in a bucket of water has 4/5 of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is :-

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Answer: (c)

  1. In water alone: ρb=45ρw=0.8ρw\rho_b = \dfrac{4}{5}\rho_w = 0.8\rho_w.
  2. With oil: ρbV=V2ρw+V2ρoil\rho_b V = \dfrac{V}{2}\rho_w + \dfrac{V}{2}\rho_{oil}
  3. 0.8=0.5+0.5 ρoilρw⇒ρoilρw=0.60.8 = 0.5 + 0.5\,\dfrac{\rho_{oil}}{\rho_w} \Rightarrow \dfrac{\rho_{oil}}{\rho_w} = 0.6
  4. So the answer is (c) 0.6.
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