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QJEEM23Phf6ce: JEE Main 2023 Physics - Class 11 JEEM23Phf6ce Solution: A Spherical ball of radius 1mm and density 10.5 g/cc is dropped …

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 3696×10−x3696\times10^{-x} N. The value of xx is ________.

(Given, g = 9.8 m/s 2^2 and π=227\pi=\frac{22}{7} )

Step-by-Step Verified Solution

At state of terminal speed, net force on the ball is zero

JEE Main 2023 (Online) 24th January Evening Shift Physics - Properties of Matter Question 84 English Explanation

∴Fv=w−FB\therefore \quad F_{v}=w-F_{B}

=(43πR3ρbg)−(43πR3ρlg)=\left(\frac{4}{3} \pi R^{3} \rho_{b} g\right)-\left(\frac{4}{3} \pi R^{3} \rho_{l} g\right)

=43πR3(ρb−ρl)g=\frac{4}{3} \pi R^{3}\left(\rho_{b}-\rho_{l}\right) g

=43×227×(10−3)3[9×103]×9.8=\frac{4}{3} \times \frac{22}{7} \times\left(10^{-3}\right)^{3}\left[9 \times 10^{3}\right] \times 9.8

=3696×10−7=3696 \times 10^{-7}

∴x=7\therefore x=7

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