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← JEE Advanced 2021 · Paper 2
PhysicsNumerical+2Question 7 of 57

JEE Advanced 2021 · Paper 2, Physics question 7

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Question Stem for Question Nos. 7 and 8

A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially the bottle is sealed at atmospheric pressure p0=105p_0 = 10^5 Pa so that the volume of the trapped air is v0=3.3v_0 = 3.3 cc. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure p0+Δpp_0 + \Delta p without changing its orientation. At this pressure, the volume of the trapped air is v0−Δvv_0 - \Delta v.

Let Δv=X\Delta v = X cc and Δp=Y×103\Delta p = Y \times 10^3 Pa.

A sealed soft plastic bottle full of water with a small inverted test tube (open end down, air trapped at its closed top) floating near the top, just below the cap (a Cartesian diver).

The value of XX is ____.

Answer and worked solution

Official answer: 0.29

Volume of glass. 52.5=2\frac{5}{2.5}=2 cc.

When the tube starts to sink. The weight equals the buoyancy on the glass plus the trapped air:
1×(2+v)=5  ⟹  v=3 cc.1\times(2+v)=5\implies v=3\text{ cc}.

Change in air volume. Δv=3.3−3.0=0.3\Delta v=3.3-3.0=0.3 cc, so X=0.3X=0.3.