Question

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 $25 \,gm / cc$ Initially the bottle is sealed at atmospheric pressure $p _{0}=10^{5} \,Pa$ so that the volume of the trapped air is $v _{0}=33\, cc$ 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 $p _{0}+\Delta p$ without changing its orientation At this pressure, the volume of the trapped air is $v _{0}-\Delta v$ Let $\Delta v = X \,\,cc$ and $\Delta p = Y \times 10^{3} Pa$ The value of $X$ is _______

Answer 1

Correct Answer: 0.3

Given:
- Density of water = 1 gm/cc
- Mass of test-tube = 5 gm
- Density of glass = 25 gm/cc
- Initial volume of trapped air, v₀ = 33 cc
- Initial pressure, p₀ = 10⁵ Pa
- Final pressure, p = p₀ + Δp = (10⁵ + Y × 10³) Pa
- Final air volume = v = v₀ − X cc
- At this point, the test-tube just begins to sink

Step 1: Volume of the glass
Volume of glass = mass / density = 5 gm / 25 gm/cc = 0.2 cc

Step 2: Sinking condition
The test-tube starts to sink when its total weight equals the buoyant force (i.e., when total volume submerged = 5 cc).
This total volume submerged = volume of glass + volume of trapped air = 0.2 + (33 − X)
Set this equal to 5 cc:
0.2 + (33 − X) = 5 ⇒ 33 − X = 4.8 ⇒ X = 33 − 4.8 = 28.2 cc

Step 3: Use Boyle’s Law
Since temperature is constant: p × v = constant ⇒ p₀ × v₀ = (p₀ + Δp) × (v₀ − X)
Now plug in values:
10⁵ × 33 = (10⁵ + Y × 10³) × (33 − X)
We are given: Y = 10 ⇒ Δp = 10 × 10³ = 10⁴ Pa
So:
(10⁵ + 10⁴) = 1.1 × 10⁵ Pa
Equation becomes:
1.1 × 10⁵ × (33 − X) = 10⁵ × 33 ⇒ (33 − X) = (10⁵ × 33) / (1.1 × 10⁵) = 33 / 1.1 = 30
Therefore:
X = 33 − 30 = 0.30 cc