A spherical Steel ball of diameter 40 mm and of density 8500kg/m38500kg/m3 is dropped in a large mass of water. The coefficient of drag of the ball in water is given as 0.45. Find the terminal velocity of the ball in water. If the ball is dropped in air, find the increase in terminal velocity of ball. Take density of air = 1.25kg/m3,CD=0.11.25kg/m3,CD=0.1

 A spherical Steel ball of diameter 40 mm and of density 

8500kg/m3$8500kg/m^3$ is dropped in a large mass of water. The coefficient of drag of the ball in water is given as 0.45. Find the terminal velocity of the ball in water. If the ball is dropped in air, find the increase in terminal velocity of ball. Take density of air = 1.25kg/m3,CD=0.1$1.25kg/m^3,C_D=0.1$

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Solution:- 

Given:-

Diameter of steel ball, D=40mm=0.04m

Density of ball, ρS=8500kg/m3=0.45${\rho }_S=8500kg/m^3=0.45$

Let the terminal velocity in water = U1$U_1$

The forces acting on the spherical balls are

(i) Weight, W=ρS×g×[π6×(D3)]$W={\rho }_S\times g\times \left[{\displaystyle \frac{\pi }{6}}\times (D^3)\right]$

W=8500×9.81×[π6×(0.04)3]=2.794N$W=8500\times 9.81\times \left[{\displaystyle \frac{\pi }{6}}\times (0.04{)}^3\right]=2.794N$

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(ii) Buoyant Force

FB=1000×9.81×π6(0.04)3$F_B=1000\times 9.81\times {\displaystyle \frac{\pi }{6}}(0.04{)}^3$

FB=0.3286N$F_B=0.3286N$

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(iii) Drag Force: -

FD=CD×A×ρU22$F_D=C_D\times A\times {\displaystyle \frac{\rho U^2}{2}}$

And ρ=1000$\rho =1000$ for water

FD=CD×π4(D2)×ρU212$F_D=C_D\times {\displaystyle \frac{\pi }{4}}(D^2)\times {\displaystyle \frac{\rho U_1^2}{2}}$

FD=0.45×π4((0.04)2)×1000U212$F_D=0.45\times {\displaystyle \frac{\pi }{4}}((0.04{)}^2)\times {\displaystyle \frac{1000U_1^2}{2}}$

FD=0.2825U21…………..(1)$F_D=0.2825U_1^2\dots \dots \dots \dots ..(1)$

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Using the relation

W=FD+FB$W=F_D+F_B$

2.794=0.2825U21+0.3286$2.794=0.2825U_1^2+0.3286$

U21=2.794−0.32860.2825=8.725$U_1^2={\displaystyle \frac{2.794-0.3286}{0.2825}}=8.725$

∴U1=8.725−−−−√=2.953m/s$\therefore U_1=\sqrt{8.725}=2.953m/s$

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When ball is dropped in air terminal velocity = U2$U_2$

(i) Weight, W=2.794

(ii) Buoyant force

FB=1.25×9.81×π6(0.04)3$F_B=1.25\times 9.81\times {\displaystyle \frac{\pi }{6}}(0.04{)}^3$

FB=0.000411N$F_B=0.000411N$

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(iii) Drag Force: -

FD=CD×A×ρU22$F_D=C_D\times A\times {\displaystyle \frac{\rho U^2}{2}}$

And ρair=1.25${\rho }_{air}=1.25$

FD=CD×π4(D2)×ρU22$F_D=C_D\times {\displaystyle \frac{\pi }{4}}(D^2)\times {\displaystyle \frac{\rho U^2}{2}}$

FD=0.1×π4((0.04)2)×1.25U22$F_D=0.1\times {\displaystyle \frac{\pi }{4}}((0.04{)}^2)\times {\displaystyle \frac{1.25U^2}{2}}$

FD=0.0000785U22…………..(1)$F_D=0.0000785U_2^2\dots \dots \dots \dots ..(1)$

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The buoyant force in Air is (0.00041) while weight of ball is (2.794N). Hence buoyant force is negligible.

Therefore for equilibrium of ball in air,

FD=$F_D=$ weight of the ball

∴0.0000785U22=2.794$\therefore 0.0000785U_2^2=2.794$

U2=2.7940.0000786−−−−−−−−−√$U_2=\sqrt{{\displaystyle \frac{2.794}{0.0000786}}}$

U2=188.67m/s$U_2=188.67m/s$

Therefore Increase in terminal velocity in air

=U2−U1$=U_2-U_1$

=188.67−2.9533$=188.67-2.9533$

=185.717 m/s