Question

A spherical ball of radius \(1\) mm 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 \(3696\times 10^{-x}\) N. The value of \(x\) is
(Given, \(g = 9.8\text{ m}/\text{s}^2\) and \(π = \frac{22}{7}\))

• A. \(5\)
• B. \(6\)
• C. \(7\)
• D. \(8\)

Hint:

At constant velocity the viscous force equals the weight minus the buoyant force.

Answer 1

The Correct Option is C

Step 1: Use Stokes law
Terminal speed $v=\frac{2r^2(\rho-\sigma)g}{9\eta}$ with $\eta=9.8$ poise $=0.98$ Pa s: $v=\frac{2\times10^{-6}\times9000\times9.8}{9\times0.98}=0.02$ m/s.

Step 2: Force
$F=6\pi\eta rv=6\times\frac{22}{7}\times0.98\times10^{-3}\times0.02=3.696\times10^{-4}$ N $=3696\times10^{-7}$ N. So $x=7$.

Final Answer:
Option (C). \[ \boxed{\text{(C)}} \]