JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 22)
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
25
Explanation
$F_{\mathrm{V}}+F_B=m g(v=$ constant $)$
$F_V=m g-F_B$
$=\rho_{\mathrm{B}} \mathrm{Vg}-\rho_{\mathrm{L}} \mathrm{Vg}$
$=\left(\rho_{\mathrm{B}}-\rho_{\mathrm{L}}\right) \mathrm{Vg}$
$=(8-1.3) \times 10^{+3} \times \frac{0.3 \times 10^{-3}}{8 \times 10^3} \times 10$
$=\frac{6.7 \times 0.3}{8} \times 10^{-2} \quad(\mathrm{~g}=10)$
$=\frac{67 \times 3}{8} \times 10^{-4}=25.125 \times 10^{-4}$
$F_V=m g-F_B$
$=\rho_{\mathrm{B}} \mathrm{Vg}-\rho_{\mathrm{L}} \mathrm{Vg}$
$=\left(\rho_{\mathrm{B}}-\rho_{\mathrm{L}}\right) \mathrm{Vg}$
$=(8-1.3) \times 10^{+3} \times \frac{0.3 \times 10^{-3}}{8 \times 10^3} \times 10$
$=\frac{6.7 \times 0.3}{8} \times 10^{-2} \quad(\mathrm{~g}=10)$
$=\frac{67 \times 3}{8} \times 10^{-4}=25.125 \times 10^{-4}$
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