Nursing Elites

1. During an isothermal (constant temperature) expansion, what is the work done by the gas on the surroundings?

• A. Positive and equal to the change in internal energy.
• B. Zero.
• C. Negative and equal to the change in internal energy.
• D. Positive and greater than the change in internal energy.

Correct answer: D

Rationale: In an isothermal expansion, the temperature remains constant, meaning there is no change in internal energy. However, the gas still does work on the surroundings as it expands, and this work is positive. Since internal energy does not change, the correct answer is D, 'Positive and greater than the change in internal energy.' Choice A is incorrect because the work done is not equal to the change in internal energy. Choice B is incorrect as work is done during the expansion. Choice C is incorrect since the work done is not negative during an isothermal expansion.

2. Certain non-Newtonian fluids exhibit shear thickening behavior. In this case, the fluid's viscosity:

• A. Remains constant with increasing shear rate
• B. Decreases with increasing shear rate (shear thinning)
• C. Increases with increasing shear rate
• D. Depends solely on the applied pressure

Correct answer: C

Rationale: When a non-Newtonian fluid exhibits shear thickening behavior, its viscosity increases with increasing shear rate. This means that as more force is applied to the fluid, its resistance to flow also increases, resulting in a higher viscosity. This phenomenon is opposite to shear thinning, where viscosity decreases with increasing shear rate. Therefore, in the case of shear thickening behavior, the correct answer is that the fluid's viscosity increases with increasing shear rate. Choices A, B, and D are incorrect because shear thickening behavior specifically involves an increase in viscosity with increasing shear rate, not remaining constant, decreasing, or depending on applied pressure.

3. A 60-watt lightbulb is powered by a 110-volt power source. What is the current being drawn?

• A. 0.55 amperes
• B. 1.83 amperes
• C. 50 amperes
• D. 6,600 amperes

Correct answer: A

Rationale: To calculate the current being drawn, use the formula I = P / V, where I is the current, P is the power in watts, and V is the voltage. Substituting the given values, I = 60 / 110 ≈ 0.55 amperes. Therefore, the current being drawn by the 60-watt lightbulb is approximately 0.55 amperes. Choice B, 1.83 amperes, is incorrect as it does not match the calculated value. Choices C and D, 50 amperes and 6,600 amperes, are significantly higher values and do not align with the expected current draw of a 60-watt lightbulb powered by a 110-volt source.

4. A common example of a shear-thinning (non-Newtonian) fluid is:

• A. Water
• B. Ketchup
• C. Air
• D. Alcohol

Correct answer: B

Rationale: The correct answer is B: Ketchup. Shear-thinning fluids become less viscous under stress. Ketchup is an example of a shear-thinning fluid because its viscosity decreases when it is shaken or squeezed, allowing it to flow more easily. Choice A, Water, is a Newtonian fluid with a constant viscosity regardless of stress. Choice C, Air, is also a Newtonian fluid. Choice D, Alcohol, does not exhibit shear-thinning behavior; it typically has a constant viscosity as well.

5. What is the diameter of a loop if its radius is 6 meters?

• A. 6 m
• B. 12 m
• C. 18 m
• D. 36 m

Correct answer: B

Rationale: The diameter of a loop is calculated by multiplying the radius by 2. Since the radius is 6 meters, the diameter is 6 × 2 = 12 meters. Therefore, the correct answer is 12 meters. Choice A (6 m) is the radius, not the diameter. Choices C (18 m) and D (36 m) are incorrect as they do not reflect the correct calculation for determining the diameter of a loop.

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