A small wooden ball of density rho is immerse | vedclass

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(B) The net upward force acting on the body is $F = \sigma V g - \rho V g$.Using Newton's second law,$ma = F$,where $m = \rho V$:$\rho V a = V g(\sigm

Vedclass · Accepted Answer

$\left( \frac{\sigma}{\rho} - 1 \right) h$

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(B) The net upward force acting on the body is $F = \sigma V g - \rho V g$.Using Newton's second law,$ma = F$,where $m = \rho V$:$\rho V a = V g(\sigma - \rho)$$a = g \left( \frac{\sigma}{\rho} - 1 \right)$.The velocity $v$ of the ball when it reaches the surface after traveling distance $h$ is given by $v^2 = 2ah$:$v^2 = 2g \left( \frac{\sigma}{\rho} - 1 \right) h$.Once the ball leaves the water,it moves under gravity. Let $H$ be the height it reaches above the surface. By conservation of energy,the kinetic energy at the surface equals the potential energy at height $H$:$\frac{1}{2} m v^2 = m g H$$H = \frac{v^2}{2g} = \frac{2g \left( \frac{\sigma}{\rho} - 1 \right) h}{2g}$$H = \left( \frac{\sigma}{\rho} - 1 \right) h$.

$A$ small wooden ball of density $\rho$ is immersed in water of density $\sigma$ to a depth $h$ and then released. The height $H$ above the surface of water up to which the ball will jump out of water is

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