Question 40 of 192

The velocity of a small ball of mass $0.3g$ and density $8g{cc}^{-1}$ when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is $1.3g{cc}^{-1}$, then the value of viscous force acting on the ball will be $x\times {10}^{-4}N$, the value of $x$ is _____ .
[use $g=10m{s}^{-2}$]

Your answerCheck answer

Show answer and solution▾

Answer: 25

Let ${F}_{V}$ be the viscous force and ${F}_{B}$ be the buoyant force acting on the ball.
​​​​​​​
Then, when body moves with constant velocity,
${F}_{V}+{F}_{B}=Mg$
Or, ${F}_{V}=Mg-{F}_{B}$
${F}_{V}={\rho }_{B}Vg-{\rho }_{L}Vg$ $=\left({\rho }_{B}-\right)\left({\rho }_{L}\right)Vg$, where, $V$ is the volume of small ball.
$=\left(8-1.3\right)\times {10}^{3}\times \dfrac{0.3\times {10}^{-3}}{8\times {10}^{3}}\times 10$
$=\dfrac{6.7\times 0.3}{8}\times {10}^{-2}$
$=\dfrac{67\times 3}{8}\times {10}^{-4}=25.125\times {10}^{-4}=25\times {10}^{-4}N$
Thus, the value of $x=25$.