Question

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 $\frac{dN}{dl} = 25\alpha \times 10^{\beta}$. Find the value of $\alpha + \beta$.

• A. 4.6
• B. 4.16
• C. 3.6
• D. 1.6

Answer 1

4.6

Answer 2

The final pressure $P_2$ when the temperature reaches $T_2 = 600$ K can be found using Gay-Lussac's Law, as the volume is constant: $P_2 = P_1 \frac{T_2}{T_1}$ Given $P_1 = 1 \text{ atm}$ and $T_1 = 300 \text{ K}$, $T_2 = 600 \text{ K}$: $P_2 = 1 \text{ atm} \times \frac{600 \text{ K}}{300 \text{ K}} = 2 \text{ atm}$ The normal contact force $dN$ exerted by a small length $dl$ of the cork along the periphery is due to the internal pressure $P_2$ acting on the area element $dA$. This area element can be approximated as $dA = L \times dl$, where $L$ is the length of the cork and $dl$ is the arc length along the periphery. Thus, the force $dN$ is given by: $dN = P_2 \times dA = P_2 \times L \times dl$ The quantity $\frac{dN}{dl}$ is therefore: $\frac{dN}{dl} = P_2 \times L$ We are given the radius $r = 5 \text{ cm}$ and length $L = 20 \text{ cm} = 0.20 \text{ m}$. We need to convert the pressure to Pascals: $1 \text{ atm} \approx 1.013 \times 10^5 \text{ Pa}$. So, $P_2 = 2 \text{ atm} \approx 2 \times 1.013 \times 10^5 \text{ Pa} = 2.026 \times 10^5 \text{ Pa}$. Now, calculate $\frac{dN}{dl}$: $\frac{dN}{dl} = (2.026 \times 10^5 \text{ Pa}) \times (0.20 \text{ m}) = 0.4052 \times 10^5 \text{ N/m} = 4.052 \times 10^4 \text{ N/m}$ We are given that $\frac{dN}{dl} = 25\alpha \times 10^{\beta}$. So, $4.052 \times 10^4 = 25\alpha \times 10^{\beta}$. To find integer values for $\alpha$ and $\beta$ that fit the format, we can approximate $P_2$ to $2 \times 10^5 \text{ Pa}$. Then, $\frac{dN}{dl} = (2 \times 10^5 \text{ Pa}) \times (0.20 \text{ m}) = 0.4 \times 10^5 \text{ N/m} = 4 \times 10^4 \text{ N/m}$. Now, $4 \times 10^4 = 25\alpha \times 10^{\beta}$. Let's choose $\beta = 3$. $4 \times 10^4 = 25\alpha \times 10^3$ $40 = 25\alpha$ $\alpha = \frac{40}{25} = \frac{8}{5} = 1.6$. Therefore, $\alpha = 1.6$ and $\beta = 3$. The value of $\alpha + \beta = 1.6 + 3 = 4.6$.