Paragraph: A small spherical monoatomic ideal gas bubble (γ=(5)/(3)) is trapped inside a liquid of density _1 (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles

Correct answer: B. T_0 ((p_0+ l g(H-y))/(p_0+ l g H))^ (2)/(5) As there is no exchange of heat. Therefore, process is adiabatic. Applying, array rlrl & T p^ (1-γ)/(γ) = constant & T_2 p_2^ (1-γ)/(γ) =T_1 p_1^ (1-γ)/(γ) or & T_2 & =T_1((p_1)/(p_2))^ (1-γ)/(γ) =T_1((p_2)/(p_1))^ (γ-1)/(γ) array Substituting the values we have, aligned T_2 & =T_0[(p_0+ (H-y)^T)/(p_0+ H)]^ (5 / 3-1)/(5 / 3) & =T_0[(p_0+ (H-y))/(p_0+ H)]^ 2 / 5 aligned correct option is (b).

JEE Advanced 2008Paper 1

Exam: JEE Advanced
Year: 2008
Paper: Paper 1
Subject: Physics
Chapter: Mechanical Properties of Fluids

JEE Advanced 2008 · Paper 1 · Physics · Mechanical Properties of Fluids

Paragraph: A small spherical monoatomic ideal gas bubble...

Paragraph: A small spherical monoatomic ideal gas bubble

\left(\gamma=\frac{5}{3}\right)

is trapped inside a liquid of density

\rho_1

(see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains

n

moles of gas. The temperature of the gas when the bubble is at the bottom is

T_0

, the height of the liquid is

H

and the atmospheric pressure is

p_0

(Neglect surface tension). Paragraph: A small spherical monoatomic ideal gas bubble (γ=(5)/(3)) is trapped inside a liquid of density _1 (see

Question: When the gas bubble is at height

y

from the bottom, its temperature is

Options

AT_0 $\left(\frac{p_0+\rho \lg H}{p_0+\rho \lg y}\right)^{\frac{2}{5}}$
BT_0 $\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{2}{5}}$
CT_0 $\left(\frac{p_0+\rho l g H}{p_0+\rho \lg y}\right)^{\frac{3}{5}}$
DT_0 $\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{3}{5}}$

Correct answer

B. T_0 $\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{2}{5}}$

Step-by-step solution

As there is no exchange of heat. Therefore, process is adiabatic. Applying,

\begin{array}{rlrl} & T p^{\frac{1-\gamma}{\gamma}}=\text{ constant } \\ \therefore \quad & T_2 p_2^{\frac{1-\gamma}{\gamma}}=T_1 p_1^{\frac{1-\gamma}{\gamma}} \\ \text{ or } & T_2 & =T_1\left(\frac{p_1}{p_2}\right)^{\frac{1-\gamma}{\gamma}}=T_1\left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}} \end{array}

Substituting the values we have,

\begin{aligned} T_2 & =T_0\left[\frac{p_0+\rho \lg (H-y)^T}{p_0+\rho \lg H}\right]^{\frac{5 / 3-1}{5 / 3}} \\ & =T_0\left[\frac{p_0+\rho \lg (H-y)}{p_0+\rho \lg H}\right]^{2 / 5} \end{aligned}

\therefore

correct option is (b).

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