The velocity of a small ball of mass $0.3 \mathrm{~g}$ and density $8 \mathrm{~g} / \mathrm{cc}$ when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is $1.3 \mathrm{~g} / \mathrm{cc}$, then the value of viscous force acting on the ball will be $x \times 10^{-4} \mathrm{~N}$, The value of $x$ is _________. [use $\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]$
Answer (integer)
25
Solution
$F_{\mathrm{V}}+F_B=m g(v=$ constant $)$
<br/><br/>$F_V=m g-F_B$
<br/><br/>$=\rho_{\mathrm{B}} \mathrm{Vg}-\rho_{\mathrm{L}} \mathrm{Vg}$
<br/><br/>$=\left(\rho_{\mathrm{B}}-\rho_{\mathrm{L}}\right) \mathrm{Vg}$
<br/><br/>$=(8-1.3) \times 10^{+3} \times \frac{0.3 \times 10^{-3}}{8 \times 10^3} \times 10$
<br/><br/>$=\frac{6.7 \times 0.3}{8} \times 10^{-2} \quad(\mathrm{~g}=10)$
<br/><br/>$=\frac{67 \times 3}{8} \times 10^{-4}=25.125 \times 10^{-4}$
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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