$(a)$. A cube of side $5\ cm$ is immersed in water and then in saturated salt solution. In which case will it experience a greater buoyant force. If each side of the cube is reduced to $4\ cm$ and then immersed in water, what will be the effect on the buoyant force experienced by the cube as compared to the first case for water. Give reason for each case.
$(b)$. A ball weighing $4\ kg$ of density $4000\ kg-m^{-3}$ is completely immersed in water of density $10^3\ kg/m^3$. Find the force of buoyancy on it. [Given $g=10\ ms^{-2}$]

$(a)$. We know that salt solution has a greater density than that water. So, the cube will experience a greater buoyant force in a saturated salt solution.

We know that, buoyant force$=V\rho g$

Here, $V\rightarrow$volume

$\rho\rightarrow$ density

$g\rightarrow$gravitational acceleration

If each side of the cube is reduced to $4\ cm$ from $5\ cm$ and then immersed in water, then the volume of the cube will also decrease. So, buoyant force will also decrease.

$(b)$. As given, mass of the ball $m=4\ kg$

Density of the ball $\rho=4000\ kg-m^{-3}$

Density of water $\sigma=10^3\ kg-m^{-3}$

Volume of the ball $V=\frac{mass\ of\ the\ ball}{density\ of\ the\ ball}$

$=\frac{4}{4000\ kg-m^{-3}}$

$=10^{-3}\ m^3$

Therefore, bouyant force $F_b=V\sigma g$

$=10^{-3}\ m^3\times 10^3\ kg-m^{-3}\times g=10\ ms^{-2}$

$=10\ N$

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Updated on: 10-Oct-2022

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