91Ó°ÊÓ

The given data can be listed below as,

Part (a)

The relation of volume and radius of a sphere is expressed as,

$v=\frac{4}{3}\mathrm{Ï€}{r}^3$

Here v is the volume of sphere, and r is the radius of ball.

Substitute 0.12m for r in the above equation.

$\begin{array}{c}v=\frac{4}{3}\mathrm{Ï€}{\left(0.12\text{″¾}\right)}^3\\ =7.238×{10}^{−3}{\text{″¾}}^{\text{3}}\end{array}$

The percentage of volume submerged is equal to 0.24 time of the volume of sphere. The force required to keep the 76% of ball inside the water will be,

$\begin{array}{l}F=\mathrm{ÒÏ}Vg−\mathrm{ÒÏ}\left(0.24V\right)g\\ F=0.76V\mathrm{ÒÏ}g\\ F=0.76×\frac{4}{3}\mathrm{Ï€}{r}^3×\mathrm{ÒÏ}×g\end{array}$

Here, $\mathrm{ÒÏ}$is the density of water and g is the gravitational acceleration.

Substitute 0.12m for r , 1000kg/m³ for $\mathrm{ÒÏ}$ and 9.81m/s² for g in the above equation.

$\begin{array}{c}F=0.76×\frac{4}{3}\mathrm{Ï€}{\left(0.12\text{″¾}\right)}^3×1000\text{ k²µ}/{\text{m}}^{\text{3}}×9.81\text{″¾}/{\text{s}}^{\text{2}}\\ =53.96\text{ N}\end{array}$

Hence, the required force is 53.96N .

Part (b)

The relation of force in terms of mass and acceleration at the instant is expressed as,

$\begin{array}{c}F=ma\\ a=\frac{F}{m}\end{array}$

Here, m is the mass of ball is equal to the mass of water displaced when ball is floating, a is the acceleration.

The above equation can be written as

$\begin{array}{c}a=\frac{0.76v\mathrm{ÒÏ}g}{0.24v\mathrm{ÒÏ}}\\ a=\frac{0.76g}{0.24}\end{array}$

Substitute 9.81 m/s² for g in the above equation.

$\begin{array}{c}a=\frac{0.76×9.81\text{″¾}/{\text{s}}^{\text{2}}}{0.24}\\ =931.06\text{″¾}/{\text{s}}^{\text{2}}\end{array}$

Hence, the value of acceleration is, 31.06 m/s² .