Question

A uniform plastic sphere of radius r floating almost fully immersed in a fluid of density $\rho$ as shown. The force of the fluid pressure on the lower half of the sphere (excluding the contribution of the atmospheric pressure ) is $\frac{5}{n}\pi r^3\rho g$ Find n.

• A. 2
• B. 3
• C. 4
• D. 7

Answer 1

The correct option is B 3

Solution

REF.Image

Consider a case when only upper

half is inserted in water then,

Case - I

${\overrightarrow{F}}_{net}=$ Buoyant force = $\left(\rho g\right)$ volume

${\overrightarrow{F}}_{net}=\rho g\left(\frac{2}{3}\right)\pi R^3$

Also ${\overrightarrow{F}}_{net}={\overrightarrow{F}}_u-{\overrightarrow{F}}_d$

${\overrightarrow{F}}_u=\left(\rho gR\right)$ (Base Area)

${\overrightarrow{F}}_u=\rho gR\left(\pi R^2\right)$

$=\pi \rho g{R}^3$

So, $\frac{2}{3}\pi gh{R}^3=\pi \rho g{R}^3-{\overrightarrow{F}}_d$

${\overrightarrow{F}}_d=\frac{\pi \rho g{R}^3}{3}$

Case II when whole sphere is

submerged. Then,

${\overrightarrow{F}}_{net}=$ Buogant force = $\left(\rho g\right)$ volume

${\overrightarrow{F}}_{net}=\rho g\left(\frac{4}{3}\right)\pi R^3$

$\Rightarrow {\overrightarrow{F}}_{net}={\overrightarrow{F}}_u^{\prime }-\overrightarrow{F}d$

From case - I, ${\overrightarrow{F}}_d=\frac{\pi }{3}\rho g{R}^3$ so,

$\frac{4}{3}\pi g\rho R^3={\overrightarrow{F}}_u^{\prime }-\frac{\pi }{3}\rho g{R}^3$

$\Rightarrow \frac{5}{3}\pi \rho g{R}^3={\overrightarrow{F}}_u^{\prime }$

Option - B is correct