Nursing Elites

1. A system undergoes an isobaric process (constant pressure). In this process, the work done (W) by the system is:

• A. Zero, if the volume change (ΔV) is zero.
• B. Positive and equal to the pressure multiplied by the volume change (W = PΔV).
• C. Negative and equal to the pressure multiplied by the volume change.
• D. Independent of the pressure or volume change.

Correct answer: B

Rationale: In an isobaric process (constant pressure), the work done is given by the formula W = PΔV, where P is the pressure and ΔV is the change in volume. If the volume does not change, the work done is zero, not negative. Choice A is incorrect as it states the work done is zero when the volume change is zero, which is the correct condition for zero work. Choice C is incorrect as it incorrectly suggests that the work done is negative in an isobaric process. Choice D is incorrect as the work done in an isobaric process is indeed dependent on the volume change and pressure.

2. According to the Clausius inequality, for a cyclic process involving heat transfer between a system and its surroundings at a single constant temperature (T), the following inequality must hold true:

• A. There is no relationship between heat transfer and temperature in a cyclic process.
• B. ∫ dQ/T ≥ 0
• C. ∫ Q/T = constant
• D. ∫ dQ/T ≤ 0

Correct answer: D

Rationale: The Clausius inequality states that for a cyclic process involving heat transfer at a single constant temperature, the integral of heat transfer divided by temperature (∫ dQ/T) must be less than or equal to zero. This inequality reflects the irreversibility of natural processes. Choice A is incorrect as there is a direct relationship between heat transfer and temperature in the Clausius inequality. Choice B is incorrect because the integral of dQ/T must be less than or equal to zero, not greater than or equal to zero. Choice C is incorrect because the integral of Q/T is not a constant in a cyclic process involving heat transfer at a single constant temperature.

3. A plucked guitar string makes 80 vibrations in one second. What is the period?

• A. 0.0125 s
• B. 0.025 s
• C. 0.125 s
• D. 0.25 s

Correct answer: B

Rationale: The period is the time taken for one complete vibration of the guitar string. To find the period, you need to take the reciprocal of the frequency. Since the string makes 80 vibrations in one second, the period is 1/80 = 0.0125 seconds (or 0.025 s). Choice A is incorrect because it is the reciprocal of 80. Choice C is incorrect as it is 10 times the reciprocal of 80. Choice D is incorrect as it is 100 times the reciprocal of 80.

4. A constant force is exerted on a stationary object. In this scenario, work is:

• A. Performed
• B. Not performed
• C. Partially performed
• D. Inconclusive without further information

Correct answer: B

Rationale: Work is only done when a force causes displacement. Since the object is stationary, no displacement occurs, and therefore, no work is performed. Choice A is incorrect because work requires both force and displacement. Choice C is incorrect as there is no partial work - work is either done or not done. Choice D is incorrect as the scenario provided is clear - the object is stationary, so no work is being performed.

5. A 10-kg object moving at 5 m/s has an impulse acted on it causing the velocity to change to 15 m/s. What was the impulse that was applied to the object?

• A. 10 kg⋅m/s
• B. 15 kg⋅m/s
• C. 20 kg⋅m/s
• D. 100 kg⋅m/s

Correct answer: D

Rationale: Impulse is the change in momentum of an object. The initial momentum is calculated as 10 kg × 5 m/s = 50 kg⋅m/s, and the final momentum is 10 kg × 15 m/s = 150 kg⋅m/s. The change in momentum (impulse) is 150 kg⋅m/s - 50 kg⋅m/s = 100 kg⋅m/s. Therefore, the impulse applied to the object is 100 kg⋅m/s. Choices A, B, and C are incorrect because they do not reflect the correct calculation of the impulse based on the change in momentum of the object.

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