Question

A solid sphere (diameter 6 mm) is rising through oil (mass density $900kg/m^3$, dynamic viscosity $0.7kg/m-s$) at a constant velocity of 1 cm/s. What is the specific weight of the material from which the sphere is made? (Take $g=9.81m/s^2$)

• A. $4.3kN/m^3$
• B. $5.3kN/m^3$
• C. $8.7kN/m^3$
• D. $12.3kN/m^3$

Answer 1

The correct option is B $5.3kN/m^3$

The solid sphere will be acted upon by three forces viz. drag force $\left(F_D\right)$, force of buoyancy $\left(F_B\right)$ and self weight of sphere (W)

$W={\rho }_{s}Vg$
$F_B={\rho }_{0}Vg$
$F_D=C_{D}A\frac{{\rho }_0{v}^2}{2}$
Reynolds Number,
$Re=\frac{{\rho }_{0}vD}{\mu }$
$=\frac{900\times 1\times {10}^{-2}\times 6\times {10}^{-3}}{0.7}$
$=0.07714<0.2\left(OK\right)$
$\therefore$ $C_D=\frac{24}{Re}=\frac{24}{0.07714}=311.12$
Now, from the free body diagram, we have
$W+F_D=F_B$
$\Rightarrow W=F_B-F_D$
$\Rightarrow {\rho }_{s}Vg={\rho }_{0}Vg-C_{D}A\frac{{\rho }_0{v}^2}{2}$
$\Rightarrow w_{s}V={\rho }_{0}Vg-C_D\frac{A{\rho }_0{v}^2}{2}$
$\Rightarrow w_s={\rho }_{0}g-\frac{C_{D}A{\rho }_0{v}^2}{2V}$
$\Rightarrow w_s=900\times 9.81-\frac{[6\times 311.12\times \pi \times (6\times {10}^{-3}{)}^2\times 900\times (1\times {10}^{-2}{)}^2]}{4\times 2\times \pi \times (6\times {10}^{-3}{)}^3}$
$\Rightarrow w_s=5328.9N/m^3=5.3kN/m^3$