Question

shows a cylindrical tube of radius $5cm$ and length $20cm$. 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 $1atm$ and a temperature of $300K$. The tube is slowly heated and it is found that the cork pops out when the temperature reaches $600K$. 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 $dN/dl$

Answer 1

Frictional force $=\mu N$

Let the cork moves to a distance $=dl$

$\therefore$ Work done by frictional force $=\mu Ndl$

Before that, the work will not start that means volume remains constant

So, using the boyle's law:

$\Rightarrow \frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow \frac{1}{300}=\frac{P_2}{600}\Rightarrow P_2=2atm$

$\therefore$ Extra Pressure $=2atm-1atm=1atm$

Work done by atmospheric pressure$\mu Ndl=\left[1atm\right]\left[Adl\right]$

$N=\frac{1\times {10}^5\times (5\times {10}^{-2}{)}^2}{2}$

$=\frac{1\times {10}^5\times \pi \times 25\times {10}^{-5}}{2}$

Total circumference of work

$=2\pi r\frac{dN}{dl}=\frac{N}{2\pi r}$

$=\frac{1\times {10}^5\times \pi \times 25\times {10}^{-5}}{0.2\times 2\pi r}$

$=\frac{1\times {10}^5\times 25\times {10}^{-5}}{0.2\times 2\times 5\times {10}^5}=1.25\times {10}^{4}N/M$