The velocity of a small ball of mass $0.3\,g$ and density $8\,g / cc$ when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is $1.3\,g / cc$, then the value of viscous force acting on the ball will be $x \times 10^{-4}\,N$, the value of $x$ is [use $g=10\,m / s ^{2}$ ]
JEE MAIN 2022, Medium
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$F _{ V }+ F _{ B }= mg ( v =$ constant $)$
$F _{ V }= mg - F _{ B }$
$=\rho_{ B } Vg -\rho_{ L } Vg$
$=\left(\rho_{ B }-\rho_{ L }\right) Vg$
$=(8-1.3) \times 10^{+3} \times \frac{0.3 \times 10^{-3}}{8 \times 10^{3}} \times 10$
$=\frac{6.7 \times 0.3}{8} \times 10^{-2} \quad(g=10)$
$=\frac{67 \times 3}{8} \times 10^{-4}=25.125 \times 10^{-4}$
Ans. $25.125$
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