Particle in a tube

JEE Advanced 2019 Paper 2, Question 3

A small particle of mass m moving inside a heavy, hollow and straight tube along the tube axis undergoes elastic collision at two ends. The tube has no friction and it is closed at one end by a flat surface while the other end is fitted with a heavy movable flat piston as shown in figure. When the distance of the piston from the closed end is L=L_{0} the particle speed is v=v_{0} . The piston is moved inward at a very low speed V such that V \ll \frac{{\rm d} L}{L} v_{0}, where {\rm d} L is the infinitesimal displacement of the piston. Which of …

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Thermodynamic cycle on a VT diagram

JEE Advanced 2019 Paper 1, Question 9

One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume vs. temperature (VT) diagram. The correct statement(s) is/are:

[R is the gas constant]

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  1. Work done in this thermodynamic cycle (1 \to 2 \to 3 \to 4 \to 1) is |W| = \frac{1}{2} R T_0.
  2. The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
  3. The ratio of heat transfer during processes 1 \to 2 and 2 \to 3 is \left| \frac{Q_{1 \to 2}}{Q_{2 \to 3}} \right| = \frac{5}{3}.
  4. The ratio of heat transfer during processes 1 \to 2 and 3 \to 4 is \left| \frac{Q_{1 \to 2}}{Q_{3 \to 4}} \right| = \frac{1}{2}.

Related article: Thermodynamic processes on an ideal gas

Solution

An ideal gas obeys the relations

(1)   \begin{equation*}   \text{The ideal gas law: } P V = n R T \end{equation*}

(2)   \begin{equation*}   \text{Work done {\it by} the gas: } W = \int_a^b P dV \end{equation*}

(3)   \begin{equation*}   \text{Internal energy of a monatomic gas: } U = \frac{3}{2} n R T \end{equation*}

We …

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Rod heated by wire

JEE Advanced 2019 Paper 1, Question 3

A current carrying wire heats a metal rod. The wire provides a constant power P to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature T in the metal rod changes with time t as

(1)   \begin{equation*}  T(t) = T_0 (1 + \beta t^{\frac{1}{4}}) \end{equation*}

where \beta is a constant with appropriate dimension while T_0 is a constant with dimension of temperature. The heat capacity of the metal is,

  1. \frac{4 P [T(t) - T_0]^3}{\beta^4 T_0^4}
  2. \frac{4 P [T(t) - T_0]^4}{\beta^4 T_0^5}
  3. \frac{4 P [T(t) - T_0]^2}{\beta^4 T_0^3}
  4. \frac{4 P [T(t) - T_0]}{\beta^4 T_0^2}

Solution

The heat capacity of an object is defined by the relation

(2)   \begin{equation*}  \Delta Q = C \Delta T \end{equation*}

where \Delta Q is the heat that the object absorbs and \Delta T is the resulting temperature change of …

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