JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 27)
A Spherical ball of radius 1mm and density 10.5 g/cc is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5 g/cc. Viscous force on the ball when it attains constant velocity is $$3696\times10^{-x}$$ N. The value of $$x$$ is ________.
(Given, g = 9.8 m/s$$^2$$ and $$\pi=\frac{22}{7}$$)
(Given, g = 9.8 m/s$$^2$$ and $$\pi=\frac{22}{7}$$)
Answer
7
Explanation
At state of terminal speed, net force on the ball is zero
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$=\left(\frac{4}{3} \pi R^{3} \rho_{b} g\right)-\left(\frac{4}{3} \pi R^{3} \rho_{l} g\right)$
$=\frac{4}{3} \pi R^{3}\left(\rho_{b}-\rho_{l}\right) g$
$=\frac{4}{3} \times \frac{22}{7} \times\left(10^{-3}\right)^{3}\left[9 \times 10^{3}\right] \times 9.8$
$=3696 \times 10^{-7}$
$\therefore x=7$
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