IIT-JEE 2007 Linked Comprehension type Questions (MCQ) on Properties of Matter

Among the IIT-JEE 2007 physics questions there were 4 Linked Comprehension Type questions, each paper carrying two of them. Each Linked Comprehension Type question contained a paragraph based on which three multiple choice questions were asked. Here is the question on properties of matter:

Paragraph for the multiple choice questions:

A fixed thermally conducting cylinder has a radius R and height L₀. The cylinder is open at the 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 P₀.

Questions:

(1) The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be

(A) P₀ (B) P₀/2 (C) P₀/2 + Mg/πR² (D) P₀/2 –Mg/πR²

The correct option is (A) since the air will enter the cylinder through the hole at the top and will keep the pressure at P_०

(2) 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

(A)[2P₀πR²/(πR²P₀ +Mg)](2L)

(B) [(P₀πR²–Mg)/πR²P₀](2L)

(C)[(P₀πR² +Mg)/πR²P₀](2L)

(D) [P₀πR²/( πR²P₀ –Mg)](2L)

If P is the pressure inside (the region between the piston and the top of the cylinder), the weight of the piston is balanced by the thrust (P₀ – P)πR² arising due to the pressure difference between the top and the bottom of the cylinder. Therefore,

(P₀ –P)πR² = Mg from which P = P₀ –Mg/πR² = (πR²P₀ –Mg)/πR²

Now, substitute this value of P in the relation, P₀×2L = P×L’, where L’ is the final distance of the piston from the top. [Note that this equation follows from Boyle’s law].

Thus, P₀×2L = [(πR²P₀ –Mg)/ πR²] × L' from which

L' = [P₀πR²/( πR²P₀ –Mg)](2L).

(3) 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 water is ρ. In equilibrium, the height H of the water column in the cylinder satisfies

(A) ρg(L₀–H)² +P₀(L₀–H)+ L₀P₀ = 0

(B) ρg(L₀–H)² –P₀(L₀–H)– L₀P₀ = 0

(C) ρg(L₀–H)² +P₀(L₀–H)– L₀P₀ = 0

(D) ρg(L₀–H)² –P₀(L₀–H)+ L₀P₀ = 0

The correct option is obtained by applying Boyle’s law. The initial pressure of air column of length L₀ is the atmospheric pressure P₀. The final pressure of the air column of length (L₀–H) is [(P₀+(L₀–H)ρg] since the pressure of air inside the cylinder exceeds the atmospheric pressure by the hydrostatic pressure exerted by water column of height (L₀–H).

Applying Boyle's law, P₀L₀ = [(P₀ +(L₀–H)ρg](L₀–H).

Rearranging this you will get the equation in option (c).