Figure 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 pf the gas is uniform at any instant, calculate(dN)/(dl).

Explanation

Frictional force = ` mu N`
` Let the cork move to a distance = dl`
` :. Workdone by frictional force = muN dl `
Before that the work will not start, that
means volume remain constant.
` rArr ((P_1)/(T_1))= ((P_2)/(T_2))`
` rArr (1/300) = ( P_2 / 600)`
` rArr (P_2) = 2 atm `
` :. Extra pressure = 2 atm - 1 atm `
` =1 atm `
` Workdone by 1 atm (Adl)`
` mu N dl = [1 atm][Adl]`
` N= (1 xx (10^5) xx pi ((5 xx (10^-2))^2)/0.2)`
` = (1 xx (10^5) xx pi xx 25 xx (10^-4)/0.2)`
Total circumference of work
` = 2 pi r (dN/dl)= (N/2 pi r)`
` = (1 xx (10^5) xx pi xx 25 xx (10^-4)/0.2 xx 2 pi r )`
` (1 xx (10^5) xx 25 xx (10^-4)/ 0.2 xx 2 xx 5 xx (10^-2))`
` = 1.25 xx (10^4) N/m`.