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Figure shows a cylindrical tube of radius 5cm and length 20cm. It is closed by 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 temperature 300 K. The tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. Let $d N$ denote the magnitude of normal contact force exerted by small length $d l$ of the cork along the periphery. Assuming that the temperature of gas is uniform at any instant, the value of $\frac{d N}{d t} = 25 α \times 10^{β} .$ Find the value of $α + β .$

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Detailed Solution

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Volume of the cylinder remains constant $\frac{P_{0}}{T_{0}} = \frac{P}{2 T_{0}}$
$\Rightarrow P = 2 P_{0}$
Now, $P A = P_{0} A + f$
$\Rightarrow (P - P_{0}) A = f \Rightarrow (2 P_{0} - P_{0}) A = \int μ \frac{d N}{d l} = \frac{d N}{d l} \times μ \times 2 π r \Rightarrow \frac{d N}{d l} = \frac{P_{0} \times π r^{2}}{μ \times 2 π \times r} = \frac{P_{0} r}{2 μ} = \frac{10^{5} \times 5 \times 10^{- 2}}{2 \times 0.2} = 125 \times 10^{2} N / m$

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Figure shows a cylindrical tube of radius 5cm and length 20cm. It is closed by 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 temperature 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 normal contact force exerted by small length dl of the cork along the periphery. Assuming that the temperature of gas is uniform at any instant, the value of dNdt=25α×10β. Find the value of α+β.