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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 $\mathrm{x \times 10^{-4} \mathrm{~N}}$ , The value of $\mathrm{x}$ is______________. [use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ]
Option: 1
25

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{m=0.3g=3\times 10^{-4}kg}$

$\mathrm{d=8\frac{g}{cc}=8000\frac{kg}{m^{3}}}$

$\mathrm{Volume \: of\: ball=\frac{m}{d}=\frac{3\times 10^{-4}}{8\times 10^{3}}=\frac{3}{8}\times 10^{-7}}$

$\mathrm{V=\frac{4}{3}\pi r^{3}=\frac{3}{8}\times 10^{-7}}$

At terminal velocity,

$\mathrm{ B+F_v =m g }$

$\mathrm{F_v =m g-\rho v g }$

$\mathrm{ =3 \times 10^{-3}-1300 \times \frac{3}{8} \times 10^{-7} \times 10 }$

$\mathrm{ =3 \times 10^{-3}-\frac{39}{8} \times 10^{-4} }$

$\mathrm{ =(3-0.4875) \times 10^{-3} }$

$\mathrm{ =2.5125 \times 10^{-3} }$

$\mathrm{ F_v =25.125 \times 10^{-4} }$

$\mathrm{x \simeq 25}$

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Ritika Harsh

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