IIT-JEE 2007 · Paper 1 · Physics Q18

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PhysicsSingle correct+4 / -1Question 18 of 66

Passage

Paragraph for Question Nos. 17 to 19

A fixed thermally conducting cylinder has a radius $R$ and height $L_0$ . The cylinder is open at its bottom and has a small hole at its top. A piston of mass $M$ is held at a distance $L$ from the top surface, as shown in the figure. The atmospheric pressure is $P_0$ .

(Figure: a vertical cylinder of radius R, closed at the top except for a small hole, open at the bottom; a piston of mass M inside at distance L below the top.)

(The paragraph is not printed in the official archive copy (which reproduces only the questions with their answers); its text is restored from the JEE 2007 question paper.)

While the piston is at a distance $2L$ from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is

\left(\dfrac{2P_0\pi R^2}{\pi R^2P_0+Mg}\right)(2L)

\left(\dfrac{P_0\pi R^2-Mg}{\pi R^2P_0}\right)(2L)

\left(\dfrac{P_0\pi R^2+Mg}{\pi R^2P_0}\right)(2L)

\left(\dfrac{P_0\pi R^2}{\pi R^2P_0-Mg}\right)(2L)

Answer and worked solution

Official answer: D

Sealed gas. Air at $P_0$ occupies length $2L$ . After release the piston settles where the forces balance: atmospheric pressure from below pushes up, the trapped gas pushes down along with the weight:
$PA+Mg=P_0A\implies P=P_0-\frac{Mg}{\pi R^2}.$

Isothermal (conducting cylinder): $P_0(2L)=P\,L'$ :
$L'=\left(\frac{P_0\pi R^2}{P_0\pi R^2-Mg}\right)(2L).$

(D)

Topics:Thermodynamic processes Pressure and Pascal's law

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IIT-JEE 2007 · Paper 1, Physics question 18