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 dNdl.
Figure

Answer 1

The value of $\frac{dN}{dl}$  is $1.25 \times 10^{4} \frac{N}{m}$

Explanation:

Given Data

A cylindrical Tube with

Radius = 5 cm

Length (L)= 20 cm = 0.2 m

Friction coefficient $\mu$ = 0.20

Pressure = 1 atm

Temperature $T_1$= 300 K

Temperature $T_2$ = 600 K

A = π (0.05)²

Volume = Area × Length

Volume (V) = 0.0016 m³

To Calculate - dNdl

When we apply five variable gas equations, we get

$\begin{aligned}&\frac{P_{1} V}{T_{1}}=\frac{P_{2} V}{T_{2}}\\&\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\end{aligned}$

$\begin{aligned}&P_{2}=\frac{T_{2}}{T_{1}} \times P_{1}\\&=\frac{600}{300} \times 10^{5}\end{aligned}$

$\mathrm{P}_{2}=2 \times 10^{5}$

$\text { Total pressure } P=P_{2}-P_{1}=2 \times 10^{5}-10^{5}=10^{5}$

$\text {The sum force that works on the stopper }=\mathrm{PA}=10^{5} \times \pi \times(0.05)^{2}$

Applying friction law, we get

$\mathrm{F}=\mu \mathrm{N}=0.2 \mathrm{N}$

$N=\frac{F}{\mu}$

$N=\frac{10^{5} \times \pi \times(0.05)^{2}}{0.2}$

$\frac{\mathrm{d} \mathrm{N}}{\mathrm{dl}}=\frac{\mathrm{N}}{2 \pi \mathrm{r}}=\frac{10^{5} \mathrm{x} \pi \times(0.05)^{2}}{0.2 \times 2 \pi \times(0.05)}=0.125 \times 10^{5}$

$\frac{d N}{d l}=1.25 \times 10^{4} \frac{N}{m}$

Therefore the value of $\frac{dN}{dl}$ is $1.25 \times 10^{4} \frac{N}{m}$ where at any instant the temperature of the gas is uniform.

Answer 2

Answer:

1.25 .......

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