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JEE Advanced 2007Paper 1
Exam
JEE Advanced
Year
2007
Paper
Paper 1
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2007 · Paper 1 · Physics · Thermodynamics

Paragraph: P_ 14-16 : Paragraph for Questions Nos. 14 to 16 A fixed...

Paragraph: P14−16P_{14-16} : Paragraph for Questions Nos. 14 to 16 A fixed thermally conducting cylinder has a radius RR and height L0L_0. The cylinder is open at its bottom and has a small hole at its top. A piston of mass MM is held at a distance LL from the top surface as shown in the figure. The atmospheric pressure is p0p_0.Paragraph: P_ 14-16 : Paragraph for Questions Nos. 14 to 16 A fixed thermally conducting cylinder has a radius R and
Question: The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is ρ\rho. In equilibrium, the height HH of the water column in the cylinder satisfiesParagraph: P_ 14-16 : Paragraph for Questions Nos. 14 to 16 A fixed thermally conducting cylinder has a radius R and

Options

  1. Aρg(L0−H)2+p0(L0−H)+L0p0=0\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)+L_0 p_0=0
  2. Bρg(L0−H)2−p0(L0−H)−L0p0=0\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)-L_0 p_0=0
  3. Cρg(L0−H)2+p0(L0−H)−L0p0=0\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0
  4. Dρg(L0−H)2−p0(L0−H)+L0p0=0\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)+L_0 p_0=0

Correct answer

C. ρg(L0−H)2+p0(L0−H)−L0p0=0\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0

Step-by-step solution

p1=p2p0+ρg(L0−H)=pNow,applying\begin{gathered} p_1=p_2 \\ p_0+\rho g \left( L_0-H\right)=p \end{gathered} Now, applying p_1 V_1=p_2 V_2 for theairinsidethecylinder,wehavep0(L0)=p(L0−H)<imgsrc="/api/proxy−image?url=httpsthe air inside the cylinder, we have p_0 \left(L_0\right)=p \left(L_0-H\right)<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_advanced%252F7oQA8KNhtTUSHFHvt9EQSIUJcVfMbFX9a2U9386X-5c.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2F7oQA8KNhtTUSHFHvt9EQSIUJcVfMbFX9a2U9386X-5c.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2F7oQA8KNhtTUSHFHvt9EQSIUJcVfMbFX9a2U9386X-5c.original.fullsize.png?alt=media"> \\ ∴p=p0L0L0−HSu\therefore \quad p=\frac{p_0 L_0}{L_0-H}Substituting in Eq.(i),wehavep0+ρg(L0−H )=p0L0L0−HEq. (i), we have p_0+\rho g\left(L_0-H\ \right)=\frac{p_0 L_0}{L_0-H}orρg(L0−H)2+p0(L0−H)−p0L0=0∴\rho g\left(L_0-H\right)^2+p_0 \left( L_0-H\right)-p_0 L_0=0\therefore Option (c) is correct.
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