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Step-by-Step Solution
Step 1: Identify the Forces on the Ball
When the ball is dropped into glycerine and acquires a constant velocity, the net force on it must be zero. This means:
F_{\mathrm{V}} + F_{B} = mg
where
F_{\mathrm{V}} is the viscous force,
F_{B} is the buoyant force,
m is the mass of the ball, and
g is the acceleration due to gravity.
Step 2: Express the Viscous Force in Terms of Known Quantities
Rearranging gives:
F_{\mathrm{V}} = mg - F_{B} .
The buoyant force F_{B} equals the weight of the displaced fluid:
F_{B} = \rho_{L} \, V \, g,
where \rho_{L} is the density of the liquid (glycerine) and V is the volume of the ball. Thus:
F_{\mathrm{V}} = mg - \rho_{L}\,V\,g.
But m = \rho_{B}\,V , where \rho_{B} is the density of the ball. Hence:
F_{\mathrm{V}} = \rho_{B}\,V\,g - \rho_{L}\,V\,g = (\rho_{B} - \rho_{L}) \, V \, g.
Step 3: Calculate the Volume of the Ball
The volume of the ball V can be found from
m = \rho_{B} \, V , so:
V = \frac{m}{\rho_{B}}.
Given:
m = 0.3 \,\mathrm{g} = 0.3\times10^{-3}\,\mathrm{kg}, \quad \rho_{B} = 8\,\mathrm{g/cc} = 8\times10^{3}\,\mathrm{kg/m^3}.
Therefore:
V = \frac{0.3 \times 10^{-3}}{8 \times 10^{3}} \,\mathrm{m^3}.
Step 4: Plug in the Values and Compute
Now substitute \rho_{B} = 8\times10^{3}\,\mathrm{kg/m^3} ,
\rho_{L} = 1.3\times10^{3}\,\mathrm{kg/m^3} ,
V = \dfrac{0.3 \times 10^{-3}}{8 \times 10^{3}} \,\mathrm{m^3} , and g = 10\,\mathrm{m/s^2} into
F_{\mathrm{V}} = (\rho_{B} - \rho_{L}) V g :
F_{\mathrm{V}} = \bigl(8 - 1.3\bigr)\times 10^{3} \times \frac{0.3 \times 10^{-3}}{8 \times 10^{3}} \times 10.
Simplifying step by step:
\rho_{B} - \rho_{L} = 6.7 \times 10^{3}\,\mathrm{kg/m^3}
V = \frac{0.3 \times 10^{-3}}{8 \times 10^{3}}\,\mathrm{m^3}
Multiply by g = 10\,\mathrm{m/s^2}
Performing the arithmetic carefully gives:
F_{\mathrm{V}} = \frac{6.7 \times 0.3}{8} \times 10^{-2}
= \frac{67 \times 3}{8} \times 10^{-4}
= 25.125 \times 10^{-4}\,\mathrm{N}.
Step 5: Interpret the Final Result
We see that the viscous force can be written as 25.125 \times 10^{-4}\,\mathrm{N} . If we denote this as x \times 10^{-4}\,\mathrm{N} , then x = 25.125 .
