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 equilibrium
At terminal speed: viscous force \(=\) weight \(-\) upthrust \(=\frac43\pi r^3(\rho-\sigma)g\).

Step 2: Convert units
\(r=1\) mm \(=10^{-3}\) m. \(\rho-\sigma=10.5-1.5=9\) g/cc \(=9000\) kg/m\(^3\).

Step 3: Calculate
\[ F=\frac43\times\frac{22}{7}\times10^{-9}\times9000\times9.8=3.696\times10^{-4}\text{ N} \]
\(3.696\times10^{-4}=3696\times10^{-7}\), so \(x=7\). Option (C).

Final Answer:
\(x=7\), option (C). \[ \boxed{\text{(C) }7} \]