Question

A spherical ball of radius \(1 \, \text{mm}\) and density \(10.5 \, \text{g/cc}\) is dropped in glycerine of coefficient of viscosity \(9.8 \, \text{poise}\) and density \(1.5 \, \text{g/cc}\). The viscous force on the ball when it attains constant velocity is \(3696 \times 10^{-x} \, \text{N}\). The value of \(x\) is:
(Given, \(g = 9.8 \, \text{m/s}^2\) and \(\pi = \frac{22}{7}\))

Answer 1

Correct Answer: 7

To find the value of \(x\), we start by calculating the viscous force using Stokes' law: \(F_v = 6\pi \eta r v\), where \(r\) is the radius of the sphere, \(\eta\) is the coefficient of viscosity, and \(v\) is the terminal velocity. At terminal velocity, the forces on the ball are balanced: \(F_g - F_b = F_v\), where \(F_g\) is gravitational force, and \(F_b\) is buoyant force.

First, determine the gravitational force \(F_g\):

\(F_g = \text{Volume} \times \text{Density} \times g = \left(\frac{4}{3}\pi r^3\right)(\rho_{\text{sphere}})(g)\).

For \(r=1\, \text{mm}=0.001 \, \text{m}\), \(\rho_{\text{sphere}}=10.5\, \text{g/cc}=10500\, \text{kg/m}^3\), and \(g=9.8\, \text{m/s}^2\),

\(F_g = \frac{4}{3} \times \frac{22}{7} \times (0.001)^3 \times 10500 \times 9.8\).

Simplifying, \(F_g \approx 4.32 \times 10^{-4}\, \text{N}\).

Next, calculate the buoyant force \(F_b\):

\(F_b = \frac{4}{3} \pi r^3 \rho_{\text{glycerine}} g = \frac{4}{3} \times \frac{22}{7} \times (0.001)^3 \times 1500 \times 9.8\).

Thus, \(F_b \approx 6.168 \times 10^{-5} \, \text{N}\).

At terminal velocity, \(F_g - F_b = F_v \Rightarrow 4.32 \times 10^{-4} - 6.168 \times 10^{-5} \approx 3.7032 \times 10^{-4} \, \text{N}\).

Given that \(F_v = 3696 \times 10^{-x} \,\text{N}\), equate and solve for \(x\):

\(3.7032 \times 10^{-4} = 3696 \times 10^{-x}\)

\(\Rightarrow x = \log_{10}(3696/3.7032 \times 10^{-4})\)

Simplifying, \(x = 7\).

Thus, the value of \(x\) is \(7\), confirming it falls within the expected range [7,7].