JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Fluid Mechanics

A wooden block floating in a bucket of water has

\frac{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:-

(A)0.5
(B)0.7
(C)0.6✓ Correct
(D)0.8

Answer: (C)

Solution

Floating means the weight of displaced fluid matches the block's own weight. At first, with four-fifths submerged in water,

ρ_{b} V g = \frac{4}{5} V ρ_{w} g

, so

ρ_{b} = \frac{4}{5} ρ_{w}

. After the oil is added and the block sits with half its volume in water and half in oil, the two buoyant contributions together support it:

ρ_{b} V g = \frac{1}{2} V ρ_{o} g + \frac{1}{2} V ρ_{w} g

, i.e.

2 ρ_{b} = ρ_{o} + ρ_{w}

. Substituting

ρ_{b} = \frac{4}{5} ρ_{w}

gives

\frac{8}{5} ρ_{w} = ρ_{o} + ρ_{w}

, so

ρ_{o} = \frac{3}{5} ρ_{w}

and the oil's density relative to water is

0.6

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Fluid Mechanics

Solution

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