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Q.

A cubical block of wood of side a and density $ρ$ floats in water of density 2 $ρ$ . The lower surface of the cube just touches the free end of a massless spring of force constant K fixed at the bottom of the vessel. The weight W put over the block so that it is completely immersed in water without wetting the weight is

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a

$a (a^{2} ρg + K)$

b

$a (aρg + 2 K)$

c

$a (\frac{aρg}{2} + 2 K)$

d

$a (a^{2} ρg + \frac{K}{2})$

answer is D.

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Detailed Solution

Since density of block = ½ (density of water). 50% of its volume is immersed in
water. In the second case, half of the volume of the block is further immersed in
water
Question Image
Therefore. W= Extra upthrust + Spring force
$\begin{array}{l} = a \times a \times \frac{a}{2} \times 2 ρ \times g + K (\frac{a}{2}) \\ = a (a^{2} ρg + \frac{K}{2}) \end{array}$

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