A body floats with frac{1}{n} of its volume o | vedclass

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(D) Let $V$ be the total volume of the body, $d$ be its density, and $\sigma$ be the density of water.For a floating body, the weight equals the buoya

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$t \propto \sqrt{n-1}$

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(D) Let $V$ be the total volume of the body, $d$ be its density, and $\sigma$ be the density of water.For a floating body, the weight equals the buoyant force: $Vdg = V(1 - \frac{1}{n})\sigma g$.Thus, $d = (\frac{n-1}{n})\sigma$.When the body is submerged, the net upward force $F_{net}$ is $F_B - mg = V\sigma g - Vdg = V\sigma g - V(\frac{n-1}{n})\sigma g = V\sigma g (1 - \frac{n-1}{n}) = V\sigma g (\frac{1}{n})$.The acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{V\sigma g / n}{Vd} = \frac{\sigma g / n}{(\frac{n-1}{n})\sigma} = \frac{g}{n-1}$.Using the kinematic equation $h = \frac{1}{2}at^2$, we have $h = \frac{1}{2} (\frac{g}{n-1}) t^2$.Solving for $t$, we get $t = \sqrt{\frac{2h(n-1)}{g}}$.Therefore, $t \propto \sqrt{n-1}$.

$A$ body floats with $\frac{1}{n}$ of its volume outside of water. If the body is pushed to a depth $h$ inside the water and released, it will come to the surface after time $t$. Then:

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