JEE Advance - Physics (2021 - Paper 2 Online - No. 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 atmosphere pressure p0 = 105 Pa so that the volume of the trapped air is v0 = 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 + $$\Delta$$p without changing its orientation. At this pressure, the volume of the trapped air is v0 $$-$$ $$\Delta$$v. Let $$\Delta$$v = X cc and $$\Delta$$p = Y $$\times$$ 103 Pa.
The value of Y is _______________.
The value of Y is _______________.
Answer
10
Explanation
When the tube just sinks/floats, then average density = density of water
$${{Mass} \over {total\,volume}}$$ = 1 gm/cc
$$ \Rightarrow {{5gm} \over {total\,volume}}$$ = 1 gm/cc
$$\Rightarrow$$ Total volume = 5 cc
$$\Rightarrow$$ Volume of tube + final volume of air in the tube = 5 cc
$$ \Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$$
$$\Rightarrow$$ Vf = 5 $$-$$ 2 = 3 cc
$$\Rightarrow$$ $$\Delta$$V = 0.3 cc
For isothermal process,
$${p_i}{V_i} = {p_f}{V_f} \Rightarrow {p_f} = {10^5} \times {{3.3} \over 3}$$
pf = 1.1 $$\times$$ 105
pf $$-$$ pi = 1.1 $$\times$$ 105 $$-$$ 105
= 0.1 $$\times$$ 105 = 10 $$\times$$ 103 Pa $$\Rightarrow$$ Y = 10
$${{Mass} \over {total\,volume}}$$ = 1 gm/cc
$$ \Rightarrow {{5gm} \over {total\,volume}}$$ = 1 gm/cc
$$\Rightarrow$$ Total volume = 5 cc
$$\Rightarrow$$ Volume of tube + final volume of air in the tube = 5 cc
$$ \Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$$
$$\Rightarrow$$ Vf = 5 $$-$$ 2 = 3 cc
$$\Rightarrow$$ $$\Delta$$V = 0.3 cc
For isothermal process,
$${p_i}{V_i} = {p_f}{V_f} \Rightarrow {p_f} = {10^5} \times {{3.3} \over 3}$$
pf = 1.1 $$\times$$ 105
pf $$-$$ pi = 1.1 $$\times$$ 105 $$-$$ 105
= 0.1 $$\times$$ 105 = 10 $$\times$$ 103 Pa $$\Rightarrow$$ Y = 10
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