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Q.

A block of wood is floating in water such that 1/2 of it is submerged in water. When the same block is floated in alcohol, l/3^rd of it's volume is submerged. Now a mixture of water and alcohol is made taking equal volume of both and block is floated in it. What is the % of it's volume that is now submerged?

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a

60

b

50

c

40

d

30

answer is B.

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Detailed Solution

$\begin{array}{l} mg = ρ_{1} \times 0 .5 V_{0} g \Rightarrow ρ_{1} = \frac{2 m}{V_{0}} \\ mg = ρ_{2} \times \frac{V_{0}}{3} g \Rightarrow ρ_{2} = \frac{3 m}{V_{0}} \\ ρ = \frac{ρ_{1} V + ρ_{2} V}{2 V} = \frac{ρ_{1} + ρ_{2}}{2} ∣ \\ mg = ρVg = \frac{(ρ_{1} + ρ_{2}) Vg}{2} \end{array} \begin{array}{l} \Rightarrow m = \frac{1}{2} (\frac{2 m}{V_{0}} + \frac{3 m}{V_{0}}) V \\ m = \frac{5 mV}{2 V_{0}} \\ V = 0 .4 V_{0} \Rightarrow 40 % \end{array}$

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