Question

A pop-gun consists of a cylindrical barrel in cross-section \[3\,{\text{c}}{{\text{m}}^2}\] closed at one end by a cork and has well fit piston at the other. If the pistons is pushed slowly in, the cork is finally ejected, giving a pop, the frequency of which is found to be \[512\,{\text{Hz}}\]. Assuming that the initial distance between the cork and the piston was \[25\,{\text{cm}}\] and there is no leakage of air, calculate the force required to eject the cork (in \[{\text{kg}} \cdot {\text{wt}}\])[Atmospheric pressure=\[{10^5}\,{\text{kg/m}} \cdot {{\text{s}}^2}\], \[v = 240\,{\text{m/s}}\]]
A. $1.5$
B. $3$
C. $6$
D. $8$

Answer 1

Hint: Use the formula for fundamental frequency in an organ pipe. Using this formula calculate the decreased distance between the piston and cork when the piston is pushed. Use the ideal gas equation to calculate the increased pressure inside the barrel when the piston is pushed. Also use the formula for pressure in terms of force and calculate the force required to eject the cork.

Formulae used:
The fundamental frequency \[n\] in an organ pipe is
\[n = \dfrac{v}{{4l}}\] …… (1)
Here, \[v\] is the velocity of the wave and \[l\] is length of the organ pipe.
The ideal gas equation is
\[PV = nRT\] …… (2)
Here, \[P\] is pressure of gas, \[V\] is volume of gas, \[n\] is number of moles of gas, \[R\] is gas constant and \[T\] is temperature of the gas.
The pressure \[P\] acting on an object is
\[P = \dfrac{F}{A}\] …… (3)
Here, \[F\] is a force acting on the object and \[A\] is the area on which the force is applied.

Complete step by step solution:
We have given that the cross sectional area of the cylindrical barrel is \[3\,{\text{c}}{{\text{m}}^2}\].
\[A = 3\,{\text{c}}{{\text{m}}^2}\]
The fundamental frequency of the sound of the cork after ejecting from the barrel is \[512\,{\text{Hz}}\].
\[n = 512\,{\text{Hz}}\]
The initial distance between the cork and the piston which is the length of the cylindrical barrel for the production of the sound is \[25\,{\text{cm}}\].
\[{l_1} = 25\,{\text{cm}}\]
The atmospheric pressure is \[{10^5}\,{\text{kg/m}} \cdot {{\text{s}}^2}\].
\[{P_0} = {10^5}\,{\text{kg/m}} \cdot {{\text{s}}^2}\]
The piston in the cylindrical barrel pushes slowly which increases the pressure inside of the barrel and the cork ejects out of the barrel. Hence, the distance between the cork and the barrel decreases.

Let us first calculate the changed distance \[{l_2}\] between the cork and piston.Rewrite equation (1) for the fundamental frequency of the sound of the cork for the changed distance.
\[n = \dfrac{v}{{4{l_2}}}\]
\[ \Rightarrow {l_2} = \dfrac{v}{{4n}}\]
Substitute \[340\,{\text{m/s}}\] for \[v\] and \[512\,{\text{Hz}}\] for \[n\] in the above equation.
\[ \Rightarrow {l_2} = \dfrac{{340\,{\text{m/s}}}}{{4\left( {512\,{\text{Hz}}} \right)}}\]
\[ \Rightarrow {l_2} = \dfrac{{85}}{{512}}\,{\text{m}}\]
This is the new distance between the cork and piston in the barrel.

Initially, the pressure inside and out the barrel was the equal to atmospheric pressure. But when the piston is pushed, the pressure inside the barrel increases.Since the number of moles, gas constant and temperature of the gas in the barrel is the product of pressure and volume is constant for the gas in the barrel.
\[PV = {\text{constant}}\]
Rewrite the above relation for the initial and final condition.
\[{P_1}{V_1} = {P_2}{V_2}\]
Substitute \[A{l_1}\] for \[{V_1}\] and \[A{l_2}\] for \[{V_2}\] in the above equation.
\[{P_1}A{l_1} = {P_2}A{l_2}\]
\[ \Rightarrow {P_1}{l_1} = {P_2}{l_2}\]
\[ \Rightarrow {P_2} = \dfrac{{{P_0}{l_1}}}{{{l_2}}}\]

Substitute \[25\,{\text{cm}}\] for \[{l_1}\] and \[\dfrac{{85}}{{512}}\,{\text{m}}\] for \[{l_2}\] in the above equation.
\[ \Rightarrow {P_2} = \dfrac{{{P_0}\left( {25\,{\text{cm}}} \right)}}{{\dfrac{{85}}{{512}}\,{\text{m}}}}\]
\[ \Rightarrow {P_2} = \dfrac{{{P_0}\left( {25 \times {{10}^{ - 2}}\,{\text{m}}} \right)}}{{\dfrac{{85}}{{512}}\,{\text{m}}}}\]
\[ \Rightarrow {P_2} = \dfrac{{128{P_0}}}{{\,85}}\]
This is the increased pressure inside the barrel. But the pressure outside the barrel is equal to atmospheric pressure.Hence, there is a pressure difference between the pressure inside the barrel and outside the pressure which is given by
\[{P_2} - {P_0} = \dfrac{F}{A}\]
\[ \Rightarrow F = \left( {{P_2} - {P_0}} \right)A\]

Substitute \[\dfrac{{128{P_0}}}{{\,85}}\] for \[{P_2}\] and \[3\,{\text{c}}{{\text{m}}^2}\] for \[A\] in the above equation.
\[ \Rightarrow F = \left( {\dfrac{{128{P_0}}}{{\,85}} - {P_0}} \right)\left( {3\,{\text{c}}{{\text{m}}^2}} \right)\]
\[ \Rightarrow F = \left( {\dfrac{{128}}{{\,85}} - 1} \right){P_0}\left( {3\,{\text{c}}{{\text{m}}^2}} \right)\]
Substitute \[{10^5}\,{\text{kg/m}} \cdot {{\text{s}}^2}\] for \[{P_0}\] in the above equation.
\[ \Rightarrow F = \left( {\dfrac{{128}}{{\,85}} - 1} \right)\left( {{{10}^5}\,{\text{kg/m}} \cdot {{\text{s}}^2}} \right)\left( {3 \times {{10}^{ - 4}}\,{{\text{m}}^2}} \right)\]
\[ \Rightarrow F = 15\,{\text{N}}\]
\[ \therefore F = 1.5\,{\text{kg}} \cdot {\text{wt}}\]
Therefore, the force required to eject the cork is \[1.5\,{\text{kg}} \cdot {\text{wt}}\].

Hence, the correct option is A.

Note: The students should not forget to convert the units of initial distance between the piston and cork and the cross sectional area of the cork in the SI system of the units as all the physical quantities used are in the SI system of units. The students should also not forget we have asked to calculate the force required to eject the cork in kilogram weight and not in newton.