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Question: If a ball of steel (density \(\rho = 7.8gc{m^{ - 3}}\)) attains a terminal velocity of \(10cm/s\) wh...

If a ball of steel (density ρ=7.8gcm−3\rho = 7.8gc{m^{ - 3}}) attains a terminal velocity of 10cm/s10cm/s when falling in a tank of water (coefficient of viscosity ηw=8.5×10−4Pa.s{\eta _w} = 8.5 \times {10^{ - 4}}Pa.s) then its terminal velocity in glycerin (σ=1.2gcm−3,ηg=13.2Pa.s\sigma = 1.2gc{m^{ - 3}},{\eta _g} = 13.2Pa.s) would be nearly:
A) 1.6×10−5cms−11.6 \times {10^{ - 5}}cm{s^{ - 1}}
B) 6.25×10−4cms−16.25 \times {10^{ - 4}}cm{s^{ - 1}}
C) 6.45×10−4cms−16.45 \times {10^{ - 4}}cm{s^{ - 1}}
D) 1.5×10−5cms−11.5 \times {10^{ - 5}}cm{s^{ - 1}}

Explanation

Solution

Hint
Terminal velocity is directly proportional to the difference in density of body and fluid and inversely proportional to the coefficient of viscosity. Substitute the data for bodies falling through water and glycerin in two equations. Divide the terminal velocity of the body in glycerin by that in water. Simplify and to get the value of terminal velocity in glycerin.

Complete step-by-step answer

If a spherical ball of radius rr is dropped in a viscous fluid, it is first accelerated and then its acceleration becomes zero and it attains a constant velocity called terminal velocity. It is given by,
v=2r2(ρ−σ)g9ηv = \dfrac{{2{r^2}(\rho - \sigma )g}}{{9\eta }}
v∝(ρ−σ)ηv \propto \dfrac{{(\rho - \sigma )}}{\eta }
Where, ρ\rho is the density of the body, σ\sigma is the density of the fluid and η\eta is the coefficient of viscosity.
Given that,
ρ=7.8gcm−3\rho = 7.8gc{m^{ - 3}}
σw=1gcm−3{\sigma _w} = 1gc{m^{ - 3}}
σg=1.2gcm−3{\sigma _g} = 1.2gc{m^{ - 3}}
ηw=8.5×10−4Pa.s{\eta _w} = 8.5 \times {10^{ - 4}}Pa.s
ηg=13.2Pa.s{\eta _g} = 13.2Pa.s
vw=10cms−1{v_w} = 10cm{s^{ - 1}}
For the ball of steel to fall through water,
vw=7.8−18.5×10−4→(1){v_w} = \dfrac{{7.8 - 1}}{{8.5 \times {{10}^{ - 4}}}} \to (1)
For the ball of steel to fall through glycerin,
vg=7.8−1.213.2→(2){v_g} = \dfrac{{7.8 - 1.2}}{{13.2}} \to (2)
Divide equation (2) by (1).
vgvw=7.8−1.213.2×8.5×10−47.8−1\dfrac{{{v_g}}}{{{v_w}}} = \dfrac{{7.8 - 1.2}}{{13.2}} \times \dfrac{{8.5 \times {{10}^{ - 4}}}}{{7.8 - 1}}
vg=0.625×10−4×10{v_g} = 0.625 \times {10^{ - 4}} \times 10
vg=6.25×10−4{v_g} = 6.25 \times {10^{ - 4}}
Hence the terminal velocity attained by the ball of steel in glycerin is vg=6.25×10−4{v_g} = 6.25 \times {10^{ - 4}} cm/s.
The correct option is B.

Note
In the terminal velocity expression if ρ>σ\rho > \sigma then terminal velocity is positive and body attains constant velocity in downward direction. If ρ<σ\rho < \sigma then terminal velocity is negative and the body attains constant velocity in upward direction. Make sure to compare the two densities else you could end up with undesirable answers (which although would be correct) which won’t be there in the options.