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Heat and Thermodynamics question

2009 · Shift 2 · Q48
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Heat and Thermodynamics question

2009 · Shift 2 · Q48

JEE AdvancedPhysicsHeat and ThermodynamicsMCQ+3 / −1

Column II gives certain systems undergoing a process. Column I suggests changes in some of the parameters related to the system. Match the statements in Column I to the appropriate process(es) from Column II:

Column I Column II
(A) The energy of the system is increased. (P) System : A capacitor, initially uncharged.
Process : It is connected to a battery.
(B) Mechanical energy is provided to the system, which is converted into energy of random motion of its parts. (Q) System : A gas in an adiabatic container filled with an adiabatic piston.
Process : The gas is compressed by pushing the piston.
(C) Internal energy of the system is converted into its mechanical energy. (R) System : A gas in a rigid container.
Process : The gas gets cooled due to colder atmosphere surrounding it.
(D) Mass of the system is decreased. (S) System : A heavy nucleus, initially at rest.
Process : The nucleus fissions into two fragments of nearly equal masses and some neutrons are emitted.
(T) System : A resistive wire loop.
Process : The loop is placed in a time varying magnetic field perpendicular to its plane.

  1. A
    (A) →\to→(P), (S), (T); (B) →\to→(Q); (C) →\to→(S); (D) →\to→(S), (R)
  2. B
    (A) →\to→(P), (Q), (T); (B) →\to→(Q); (C) →\to→(S); (D) →\to→(S)
  3. C
    (A) →\to→(P), (Q), (T); (B) →\to→(T); (C) →\to→(S); (D) →\to→(S)
  4. D
    (A) →\to→(S), (Q), (T); (B) →\to→(Q); (C) →\to→(S); (D) →\to→(S)
View written solutionFree

Correct answer: B

This is a matching question where we need to associate processes from Column II with the physical changes described in Column I.

Step-by-step analysis of each process in Column II:

  1. Process (P): A capacitor, initially uncharged, is connected to a battery.

    • When an uncharged capacitor is connected to a battery, the battery does work to move charges onto the capacitor plates. This process stores electrical potential energy in the capacitor.
    • The energy stored in the capacitor increases from zero to U=12CV2U = \frac{1}{2}CV^2U=21​CV2, where C is the capacitance and V is the voltage of the battery.
    • Thus, the energy of the system (the capacitor) is increased.
    • This matches statement (A).
  2. Process (Q): A gas in an adiabatic container is compressed by pushing the piston.

    • The container is adiabatic, which means there is no heat exchange with the surroundings (Q=0Q = 0Q=0).
    • The gas is compressed by a piston, which means external mechanical work is done on the gas (Won>0W_{\text{on}} > 0Won​>0).
    • According to the First Law of Thermodynamics, the change in internal energy is ΔU=Q+Won\Delta U = Q + W_{\text{on}}ΔU=Q+Won​.
    • Since Q=0Q=0Q=0, we have ΔU=Won\Delta U = W_{\text{on}}ΔU=Won​. Because work is done on the gas, ΔU>0\Delta U > 0ΔU>0. The internal energy of the system increases. The internal energy of a gas is the kinetic energy of its randomly moving molecules.
    • This matches statement (A) (The energy of the system is increased).
    • It also matches statement (B) (Mechanical energy is provided to the system, which is converted into energy of random motion of its parts).
  3. Process (R): A gas in a rigid container gets cooled.

    • The container is rigid, so its volume is constant (isochoric process). This means no work is done by or on the gas (W=∫PdV=0W = \int P dV = 0W=∫PdV=0).
    • The gas is cooled, which means heat flows out of the system (Q<0Q < 0Q<0).
    • From the First Law of Thermodynamics, ΔU=Q−W\Delta U = Q - WΔU=Q−W. With W=0W=0W=0, we get ΔU=Q<0\Delta U = Q < 0ΔU=Q<0. The internal energy of the system decreases.
    • The mass of the gas remains constant (ignoring negligible relativistic effects).
    • This process does not match any of the statements in Column I.
  4. Process (S): A heavy nucleus at rest fissions into two fragments and some neutrons.

    • In nuclear fission, the total mass of the products (fragments and neutrons) is less than the mass of the original heavy nucleus. This difference in mass is called the mass defect.
    • Therefore, the mass of the system is decreased. This matches statement (D).
    • According to Einstein's mass-energy equivalence (E=mc2E=mc^2E=mc2), the lost mass is converted into a large amount of energy. This energy is released primarily as the kinetic energy of the fission fragments and neutrons.
    • The rest mass energy of the nucleus is a form of internal energy. This internal energy is converted into the kinetic energy of the fragments, which is a form of mechanical energy.
    • This matches statement (C) (Internal energy of the system is converted into its mechanical energy).
  5. Process (T): A resistive wire loop is placed in a time-varying magnetic field.

    • According to Faraday's Law of Induction, a time-varying magnetic flux through the loop induces an electromotive force (EMF).
    • This EMF drives an electric current in the resistive wire.
    • The flow of current through the resistor causes Joule heating (P=I2RP = I^2RP=I2R), dissipating energy as heat.
    • This heat increases the temperature and thus the internal energy (thermal energy) of the wire loop.
    • The energy is supplied by the external time-varying magnetic field.
    • Therefore, the energy of the system (the wire loop) is increased. This matches statement (A).

Summary of Matches:

  • (A) The energy of the system is increased. -> Matches with (P), (Q), and (T).
  • (B) Mechanical energy is provided to the system, which is converted into energy of random motion of its parts. -> Matches with (Q).
  • (C) Internal energy of the system is converted into its mechanical energy. -> Matches with (S).
  • (D) Mass of the system is decreased. -> Matches with (S).

Matching with Options:

Let's check the given options based on our derived matches:

  • A: (A) →\to→(P), (S), (T); (B) →\to→(Q); (C) →\to→(S); (D) →\to→(S), (R) - Incorrect because (A) should include (Q) and (D) should not include (R).
  • B: (A) →\to→(P), (Q), (T); (B) →\to→(Q); (C) →\to→(S); (D) →\to→(S) - This perfectly matches our analysis.
  • C: (A) →\to→(P), (Q), (T); (B) →\to→(T); (C) →\to→(S); (D) →\to→(S) - Incorrect because (B) matches (Q), not (T).
  • D: (A) →\to→(S), (Q), (T); (B) →\to→(Q); (C) →\to→(S); (D) →\to→(S) - Incorrect because (A) should include (P).

The correct option is B.

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