Question

Figure shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction coefficient between the cork and the tube is 0.20. The tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 K. The tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. Let dN denote the magnitude of the normal contact force exerted by a small length dl of the cork along the periphery (see the figure). Assuming that the temperature of the gas is uniform at any instant, calculate $\frac{dN}{dl}$.

Answer 1

$\begin{array}{l}\text{Here,}\\ P_1={10}^5\text{Pa}\\ A=\pi {\left(0.05\right)}^2\\ L=0.2\text{m}\\ V=AL=0.0016{\text{m}}^3\\ T_1=300\text{K}\\ T_2=600\text{K}\\ \mu \text{=0}\text{.20}\\ \text{Applying 5 variable equation of state, we get}\\ \frac{P_{1}V}{T_1}=\frac{P_{2}V}{T_2}\\ \Rightarrow \frac{P_1}{T_1}=\frac{P_2}{T_2}\\ \Rightarrow P_2=\frac{T_2}{T_1}\times P_1=\frac{600}{300}\times {10}^5\\ \Rightarrow P_2=2\times {10}^5\\ \text{Net pressure, P =}{P}_2-P_1\text{=}2\times {10}^5-{10}^5={10}^5\\ \text{Total force acting on the stopper =}PA{\text{=10}}^5\times \pi \times {\left(0.05\right)}^2\\ \text{Applying law of friction, we get}\\ F=\mu N=0.2N\\ \Rightarrow N=\frac{F}{\mu }=\frac{{\text{10}}^5\times \pi \times {\left(0.05\right)}^2}{0.2}\\ \frac{dN}{dl}=\frac{N}{2\pi r}=\frac{{\text{10}}^5\times \pi \times {\left(0.05\right)}^2}{0.2\times 2\pi \times \left(0.05\right)}=0.125\times {10}^5\\ \Rightarrow \frac{dN}{dl}=1.25\times {10}^4\text{ N/m}\end{array}$