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A cube of wood of mass 0.5kg and density $800 k g m^{- 3}$ is fastened to the free end of vertical spring of spring constant $k = 50 N m^{- 1}$ fixed at the bottom. Now, the entire system is completely submerged in water. The elongation or compression of the spring in equilibrium is $\frac{β}{2} c m$ . Find the value of $β$ (Given $g = 10 m s^{- 2}$ )

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answer is 5.

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Detailed Solution

$p_{s} < p_{w} s o, b l o c k$ tends to move up and thus the spring has elongation

$\begin{array}{l} B_{F} = m g + k x \\ V_{P w} g = V_{P s} g + k x \\ \frac{0.5}{800} \times 1000 \times 10 = (0.5) (10) + 50 x \\ x = 0.025 m = 2.5 c m = \frac{5}{2} c m \\ β = 5 \end{array}$

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