Question

# A uniform cylinder of length $L$ and mass $M$ having cross-sectional area $A$ is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density at the equilibrium position. The extension ${x}_{0}$ of the spring when it is in equilibrium is?

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Solution

## Extension of spring:Extension springs are used to absorb and store energy while also providing resistance to a pulling force.Step 1: Given:Length of cylinder = LMass of cylinder = MCross Sectional area of cylinder = AStep 2: Formula Used:$k{x}_{o}+{F}_{B}=Mg$Here, $k{x}_{o}$ is stress on spring, ${F}_{B}$ is buoyant force, $M$ is mass, and $g$ is acceleration due to gravity.Step 3: Calculating extension xoThe figure shows the various forces acting on the cylinder.$Mg$ is the weight of the cylinder, ${F}_{B}$ is buoyant force, and $k{x}_{o}$ is stress on spring, ${x}_{o}$ is the extension in the string, and $k$ is spring constant.At equilibrium, Upward forces = Downward forces$k{x}_{o}+{F}_{B}=Mg$ ------------- (i)${F}_{B}$ is the force due to buoyancy which is equal to the weight of the liquid displaced.${F}_{B}=\sigma ·\frac{L}{2}·A·g$Putting the value of ${F}_{B}$ in equation (i)$k{x}_{o}+\sigma ·\frac{L}{2}·A·g=Mg\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{x}_{o}=\frac{Mg-\sigma ·\frac{L}{2}·A·g}{k}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{x}_{o}=\frac{Mg}{k}\left(1-\frac{\sigma LA}{2M}\right)$

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