Units, Dimensional Analysis & Significant Figures - College Physics

1. A physics equation contains the term $\frac{k x^2}{m}$ where $k$ is a spring constant, $x$ is displacement, and $m$ is mass. What are the dimensions of this expression?
   Answer: $[L^2T^{-2}]$ representing specific energy (energy per unit mass) When analyzing dimensional expressions in physics, you need to determine the fundamental dimensions by examining each component systematically. Start by identifying the dimensions of each variable in the expression $\frac{kx^2}{m}$. The spring constant $k$ has dimensions of force per displacement, which gives us $[MLT^{-2}]/[L] = [MT^{-2}]$. The displacement $x$ has dimensions $[L]$, so $x^2$ has dimensions $[L^2]$. Mass $m$ has dimensions $[M]$. Combining these: $\frac{kx^2}{m} = \frac{[MT^{-2}][L^2]}{[M]} = [L^2T^{-2}]$ This dimensional combination $[L^2T^{-2}]$ represents energy per unit mass, also called specific energy. You can verify this by noting that energy has dimensions $[ML^2T^{-2}]$, so energy divided by mass gives $[L^2T^{-2}]$. Looking at the wrong answers: Choice B ($[MLT^{-2}]$) represents force, but our expression doesn't have the mass dimension in the numerator needed for force. Choice C ($[ML^2T^{-2}]$) represents total energy, but we're missing the mass dimension since we divided by $m$. Choice D ($[MT^{-2}]$) isn't a standard physical quantity and doesn't match our calculated dimensions. Study tip: When checking dimensions, always work systematically through each variable, and remember that dividing by mass often converts extensive quantities (like total energy) into intensive quantities (like specific energy or energy per unit mass).
2. A calculated result gives $0.02496$. If this must be rounded to 2 significant figures and expressed in scientific notation, which is correct?
   Answer: $2.5 \times 10^{-2}$ using standard rounding rules for the third digit When working with significant figures and scientific notation, you need to apply proper rounding rules while maintaining the correct format. Scientific notation expresses numbers as a coefficient between 1 and 10 multiplied by a power of 10. Starting with $0.02496$, you first convert to scientific notation: $2.496 \times 10^{-2}$. To round to 2 significant figures, you look at the third digit (9) to determine how to round the second digit. Since 9 ≥ 5, you round the 4 up to 5, giving you $2.5 \times 10^{-2}$. This makes choice A correct. Choice B ($2.4 \times 10^{-2}$) represents truncation rather than proper rounding. Truncation simply cuts off digits without considering whether to round up or down, which violates standard rounding conventions used in scientific work. Choice C ($0.025$) technically has the right numerical value rounded to 2 significant figures, but it's not in scientific notation as the question requires. Scientific notation specifically demands the format $a \times 10^n$ where $1 ≤ a < 10$. Choice D ($25 \times 10^{-3}$) is mathematically equivalent to the correct answer but violates scientific notation format because the coefficient (25) is not between 1 and 10. Remember: proper scientific notation requires both correct rounding to the specified significant figures AND the standard format with a single non-zero digit before the decimal point. Don't let equivalent mathematical expressions fool you if they don't follow the required format.
3. A student measures three quantities: $A = 12.0$ (exact), $B = 3.42 \pm 0.02$, and $C = 0.156 \pm 0.003$. What is the result of $A \times B \times C$ expressed with appropriate significant figures?
   Answer: $6.40$ because B has 3 significant figures and limits the calculation When working with measurements that have uncertainties, you need to apply significant figure rules to determine the precision of your final answer. The key principle is that your result can't be more precise than your least precise measurement. Let's calculate $A \times B \times C = 12.0 \times 3.42 \times 0.156$. Since A is exact, it doesn't limit precision. For B = 3.42 ± 0.02, you have 3 significant figures. For C = 0.156 ± 0.003, you also have 3 significant figures (the leading zeros don't count as significant). The calculation gives: $12.0 \times 3.42 \times 0.156 = 6.39984$ Since both measured quantities (B and C) have 3 significant figures, your answer should have 3 significant figures: 6.40. Now let's examine why the other choices are wrong. Choice A incorrectly states that C has only 2 significant figures - but 0.156 actually has 3 significant figures since leading zeros aren't significant. Choice C suggests keeping an extra digit from intermediate steps, but significant figure rules apply to the final answer based on the precision of your measurements, not computational steps. Choice D keeps too many digits, ignoring the fundamental principle that your answer can't be more precise than your input data. Study tip: When counting significant figures, remember that leading zeros (like in 0.156) are never significant - they're just placeholders. Focus on the uncertainty values (±) to confirm how many digits are actually meaningful in each measurement.
4. A rectangular sheet of metal has measured dimensions $25.4 \pm 0.1$ cm by $15.2 \pm 0.1$ cm. What is the area of the sheet expressed with the correct number of significant figures?
   Answer: $386 \text{ cm}^2$ with uncertainty of approximately $\pm 5 \text{ cm}^2$ When dealing with measurements that have uncertainties, you need to apply error propagation rules to determine both the final result and how many significant figures are meaningful. First, calculate the area: $25.4 \times 15.2 = 386.08 \text{ cm}^2$. Now for the uncertainty: when multiplying quantities with uncertainties, you add the relative uncertainties. The relative uncertainty for length is $\frac{0.1}{25.4} = 0.0039$ and for width is $\frac{0.1}{15.2} = 0.0066$. The total relative uncertainty is $0.0039 + 0.0066 = 0.0105$, giving an absolute uncertainty of $386.08 \times 0.0105 \approx 4 \text{ cm}^2$. Since your uncertainty is about $\pm 4-5 \text{ cm}^2$, it affects the ones place, making any decimal places meaningless. Therefore, the area should be reported as $386 \text{ cm}^2$ with uncertainty $\pm 5 \text{ cm}^2$, making choice A correct. Choice B incorrectly keeps a decimal place that's smaller than the uncertainty range. Choice C uses scientific notation correctly but isn't necessary here since we can clearly show the appropriate precision with $386$. Choice D falls into the common trap of reporting all calculator digits—just because your calculator shows more digits doesn't mean they're significant given your measurement uncertainty. Study tip: When measurements have uncertainties, always calculate the propagated uncertainty first, then round your final answer so that uncertain digits aren't reported as if they were reliable. The uncertainty determines your significant figures, not the original measurements alone.
5. A measurement is recorded as $0.00450$ kg. If this measurement is converted to grams and expressed in scientific notation, which form correctly maintains the same number of significant figures?
   Answer: $4.50 \times 10^0$ g to maintain three significant figures from the original When working with significant figures and unit conversions, you need to preserve the precision indicated by the original measurement throughout your calculation. The measurement $0.00450$ kg contains three significant figures. The digits 4, 5, and 0 are all significant because the trailing zero after the decimal point indicates measured precision to the ten-thousandths place. Converting to grams: $0.00450 \text{ kg} \times 1000 \text{ g/kg} = 4.50 \text{ g}$. In scientific notation, this becomes $4.50 \times 10^0$ g, maintaining all three significant figures from the original measurement. Option A is incorrect because $4.50 \times 10^0$ already shows three significant figures clearly - there's no need for additional explanation about "showing" them. Option B demonstrates a fundamental misunderstanding of significant figures. The trailing zero in $0.00450$ kg is significant because it appears after the decimal point and after other non-zero digits, indicating the measurement's precision. Removing it in $4.5 \times 10^0$ reduces the precision to only two significant figures. Option C incorrectly adds an extra zero that wasn't in the original measurement, creating $4.500 \times 10^0$ with four significant figures instead of three, which overstates the precision. Option D correctly maintains the three significant figures (4, 5, and 0) from the original measurement in the proper scientific notation format. Remember: trailing zeros after a decimal point are always significant, and unit conversions should never change the number of significant figures in your measurement.
6. A physics student reports a measurement as $6.050 \times 10^{-3}$ kg. How many significant figures does this measurement contain?
   Answer: Four significant figures, because trailing zeros after the decimal point are significant When you encounter scientific notation with trailing zeros, you need to understand the rules for significant figures to determine what information the measurement actually conveys. In the measurement $6.050 \times 10^{-3}$ kg, all four digits (6, 0, 5, and 0) are significant. The key principle here is that trailing zeros after a decimal point are always significant because they indicate the precision of the measurement. The scientist who recorded this value chose to include that final zero, meaning they measured to the thousandths place and found the value to be exactly 6.050, not 6.05. Looking at the incorrect options: Choice A incorrectly states that trailing zeros after a decimal point aren't significant - this is false and contradicts a fundamental rule of significant figures. Choice B suggests only three significant figures, which would ignore the final zero, but this zero is meaningful since it appears after the decimal point. Choice C gives the right number (four) but for the wrong reason - not all digits in scientific notation are automatically significant; the rules still apply to the coefficient. Choice D correctly identifies four significant figures and provides the accurate reasoning: trailing zeros after the decimal point are indeed significant because they indicate measurement precision. Remember this pattern: zeros after the decimal point and after non-zero digits are always significant. They tell you how precisely the measurement was made. When you see $6.050 \times 10^{-3}$, think of it as 0.006050 - that trailing zero matters and counts as a significant figure.
7. The universal gravitational constant $G$ has units of $\text{N}\cdot\text{m}^2/\text{kg}^2$. Which expression correctly shows these units in terms of fundamental SI base units?
   Answer: $\text{m}^3\cdot\text{kg}^{-1}\cdot\text{s}^{-2}$ When working with dimensional analysis problems, you need to break down compound units into their fundamental SI base units: meters (m), kilograms (kg), and seconds (s). Start with the given units for G: $\text{N}\cdot\text{m}^2/\text{kg}^2$. The key is recognizing that a Newton is not a fundamental unit—it's derived. From Newton's second law, $F = ma$, so $1 \text{ N} = 1 \text{ kg}\cdot\text{m}\cdot\text{s}^{-2}$. Substitute this into the expression for G: $\frac{\text{N}\cdot\text{m}^2}{\text{kg}^2} = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}^2}{\text{kg}^2} = \frac{\text{kg}\cdot\text{m}^3\cdot\text{s}^{-2}}{\text{kg}^2}$ Simplifying: $\text{kg}^{1-2}\cdot\text{m}^3\cdot\text{s}^{-2} = \text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2} = \text{m}^3\cdot\text{kg}^{-1}\cdot\text{s}^{-2}$ This matches option B. Option A incorrectly keeps both kg terms without properly canceling, showing $\text{kg}^1$ and $\text{kg}^{-2}$ separately instead of combining them. Option C has the wrong time dimension ($\text{s}^{-1}$ instead of $\text{s}^{-2}$), missing the fact that force involves acceleration, not velocity. Option D has the wrong spatial dimension ($\text{m}^2$ instead of $\text{m}^3$), forgetting that the original expression already contained $\text{m}^2$ before adding the meter from the Newton. Study tip: Always convert derived units (like N, J, W) to fundamental SI units first, then carefully track exponents when multiplying and dividing. Practice recognizing that $\text{N} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}$—this conversion appears frequently in physics problems.
8. Students measure the period of a pendulum five times and obtain: $2.14$ s, $2.16$ s, $2.13$ s, $2.15$ s, $2.17$ s. How should they report the average period?
   Answer: $2.15$ s because all measurements have the same number of decimal places When reporting experimental measurements, you must follow significant figures rules to avoid overstating your precision. The key principle is that your final result cannot be more precise than your least precise measurement. Looking at the data: 2.14 s, 2.16 s, 2.13 s, 2.15 s, 2.17 s, all measurements are reported to the hundredths place (two decimal places). The average calculation gives 10.75 ÷ 5 = 2.15 s. Since all your original measurements have the same precision (hundredths), your average should be reported with the same precision: 2.15 s. Option A is correct because it maintains the appropriate number of decimal places consistent with the original measurements. Option B (2.150 s) is wrong because adding an extra decimal place implies greater precision than your instruments actually provided - you cannot create precision that wasn't in your original data. Option C (2.1 s) is incorrect because it unnecessarily reduces precision; while repeated measurements don't improve instrument precision, they do help reduce random error, and you should report to the same precision as your measurements. Option D (2.15 ± 0.02 s) includes uncertainty analysis, which is scientifically valuable but wasn't requested in the question about how to report "the average period." Study tip: In significant figures problems, your answer's precision should match your least precise measurement. You cannot gain decimal places through calculations - mathematics doesn't create physical precision that your instruments lack.
9. The kinematic equation $x = x_0 + v_0 t + \frac{1}{2}at^2$ is dimensionally correct. If position $x$ has units of meters and time $t$ has units of seconds, what are the correct units for the acceleration $a$?
   Answer: $\text{m/s}^2$ because each term must have the same units as position When working with kinematic equations, dimensional analysis is your key tool for checking correctness and understanding physical relationships. The principle is that every term in an equation must have the same dimensions for the equation to be physically meaningful. Let's examine the kinematic equation $x = x_0 + v_0 t + \frac{1}{2}at^2$. Since this equals position $x$, every term on the right side must have units of meters. The first term $x_0$ is already in meters. For the second term $v_0 t$, if velocity has units m/s and time has units s, then $v_0 t$ gives us (m/s)(s) = m, which is correct. Now for the acceleration term $\frac{1}{2}at^2$. This must also equal meters. We have $t^2$ with units of $s^2$, so for the entire term to have units of meters, acceleration $a$ must have units that, when multiplied by $s^2$, give meters. Therefore: $a \times s^2 = m$, which means $a = m/s^2$. This confirms answer B is correct. Looking at the wrong answers: A suggests m/s, but while this correctly identifies acceleration as rate of change of velocity, it ignores the dimensional requirement of this specific equation. C proposes $m^2/s$, which would give $m^2 s$ when multiplied by $t^2$ - completely wrong dimensions. D suggests $s/m^2$, which would yield $s^3/m^2$ when combined with $t^2$ - again, wrong dimensions. Remember: in dimensional analysis problems, work backwards from what the equation must equal, then determine what units each variable needs to make the math work.
10. A student measures the diameter of a sphere as $12.4 \pm 0.2$ cm. Using the formula $V = \frac{4}{3}\pi r^3$, what is the volume of the sphere expressed with the correct number of significant figures?
    Answer: $998 \text{ cm}^3$ When dealing with measurements and calculated quantities in physics, significant figures reflect the precision of your data and must be carefully tracked through calculations. First, let's calculate the volume. The diameter is $12.4 \pm 0.2$ cm, so the radius is $r = 6.2$ cm. Using $V = \frac{4}{3}\pi r^3$: $V = \frac{4}{3}\pi (6.2)^3 = \frac{4}{3}\pi (238.328) = 997.8 \text{ cm}^3$ Now for significant figures: your measurement $12.4 \pm 0.2$ cm has 3 significant figures. When you cube the radius in the volume formula, uncertainties get magnified significantly. The uncertainty analysis shows that your final answer should have only 3 significant figures, and given the size of the uncertainty, the last digit is not reliable. Answer A ($998 \text{ cm}^3$) correctly rounds to 3 significant figures and appropriately reflects the precision limitation of the original measurement. Answer B ($997.8 \text{ cm}^3$) shows 4 significant figures, which implies false precision given your measurement uncertainty. Answer C ($998.0 \text{ cm}^3$) also shows 4 significant figures by explicitly including the decimal and trailing zero. Answer D ($9.98 \times 10^2 \text{ cm}^3$) maintains 3 significant figures but doesn't round properly - it should be $9.98 \times 10^2 = 998$, making this equivalent to answer A but in an unnecessarily complex form. Remember: when measurements have uncertainties, your final calculated result can't be more precise than your least precise input measurement allows.
11. A student calculates $\frac{(4.2 \times 10^3)(1.8 \times 10^{-2})}{(3.0 \times 10^1)}$ and wants to express the result with proper significant figures. What is the correct answer?
    Answer: $2.5$ because the calculation is limited by the two-significant-figure quantities When you encounter scientific calculations involving multiple measurements, the key principle is that your final answer can only be as precise as your least precise measurement. This is fundamental to significant figures in physics. Let's work through this calculation step by step. First, compute the numerical result: $\frac{(4.2 \times 10^3)(1.8 \times 10^{-2})}{(3.0 \times 10^1)} = \frac{4.2 \times 1.8 \times 10^1}{3.0 \times 10^1} = \frac{7.56}{3.0} = 2.52$ Now for the crucial step: determining significant figures. Looking at your given values, you have 4.2 (2 sig figs), 1.8 (2 sig figs), and 3.0 (2 sig figs). Since all measurements have exactly two significant figures, your final answer must also have two significant figures, giving you 2.5. Choice A is correct because the rule for multiplication and division is that your result should have the same number of significant figures as the measurement with the fewest significant figures. Choice B incorrectly assumes you should keep extra precision from intermediate calculations, but significant figure rules apply to the final answer based on input precision. Choice C makes the common mistake of thinking computational steps require extra digits in the final result—they don't. Choice D inappropriately rounds to just one significant figure, which loses precision unnecessarily since all your measurements support two significant figures. Remember: in multiplication and division, count significant figures in each measurement and match the smallest count in your final answer. This prevents false precision in scientific work.
12. The speed of light $c$ has units of m/s. Planck's constant $h$ has units of J·s. What are the dimensions of the quantity $\frac{hc}{\lambda}$ where $\lambda$ is wavelength?
    Answer: $[ML^2T^{-2}]$ representing energy of a photon When you encounter dimensional analysis problems in physics, especially involving fundamental constants like Planck's constant and the speed of light, you're likely dealing with quantum mechanics relationships. The key is to systematically work through the units. Let's analyze $\frac{hc}{\lambda}$ step by step. Planck's constant $h$ has units of J·s, which in base dimensions is $[ML^2T^{-2}][T] = [ML^2T^{-1}]$. The speed of light $c$ has units $[LT^{-1}]$. Wavelength $\lambda$ has units $[L]$. Therefore: $\frac{hc}{\lambda} = \frac{[ML^2T^{-1}][LT^{-1}]}{[L]} = \frac{[ML^3T^{-2}]}{[L]} = [ML^2T^{-2}]$ This is indeed the dimension of energy, and specifically represents the energy of a photon with wavelength $\lambda$, from Einstein's relation $E = hf = \frac{hc}{\lambda}$. Answer A is correct. Answer B gives momentum dimensions $[MLT^{-1}]$, but photon momentum is $\frac{h}{\lambda}$, not $\frac{hc}{\lambda}$. Answer C represents power $[ML^2T^{-3}]$, which would require an additional factor of time⁻¹. Answer D gives force dimensions $[MT^{-2}]$, which is missing a length dimension and doesn't match our calculation. Remember: when working with Planck's constant, you're usually dealing with quantum energy relationships. Practice converting Joules to base dimensions $[ML^2T^{-2}]$ to make these problems more straightforward.
13. A student performs calculations and obtains the result $2.847 + 15.2 - 0.0036 = ?$ What is the correct answer when expressed with appropriate significant figures?
    Answer: $18.0$ because the least precise measurement determines significant figures When working with significant figures in addition and subtraction, you need to focus on decimal places rather than the total number of significant figures. The rule is straightforward: your final answer should have the same number of decimal places as the measurement with the fewest decimal places. Let's examine each number in the calculation $2.847 + 15.2 - 0.0036$: - $2.847$ has 3 decimal places - $15.2$ has 1 decimal place - $0.0036$ has 4 decimal places Since $15.2$ has the fewest decimal places (1), your final answer must be rounded to 1 decimal place. The raw calculation gives $18.0436$, which rounds to $18.0$. Option A correctly identifies that $18.0$ is the answer because the least precise measurement ($15.2$ with 1 decimal place) determines the precision of the result. Option B ($18.04$) incorrectly keeps 2 decimal places, going beyond what the data supports. Option C ($18$) makes the common error of thinking $15.2$ has no decimal places—it actually has one decimal place shown. The zero after the decimal point in $18.0$ is significant and must be included. Option D ($18.0436$) ignores significant figure rules entirely by keeping all calculated digits. Remember this key distinction: for addition and subtraction, count decimal places and match the fewest; for multiplication and division, count total significant figures and match the fewest. Many students confuse these rules, so practice identifying which operation you're dealing with first.
14. An equation for terminal velocity is given as $v_t = \sqrt{\frac{2mg}{\rho A C_D}}$, where $m$ is mass, $g$ is gravitational acceleration, $\rho$ is fluid density, $A$ is cross-sectional area, and $C_D$ is a dimensionless drag coefficient. What must be the dimensions of $\rho$?
    Answer: $[ML^{-3}]$ to make the equation dimensionally consistent Dimensional analysis is a powerful tool in physics that ensures equations make physical sense. When you encounter a formula with multiple variables, you can check its validity by verifying that both sides have the same dimensions. To find the dimensions of ρ, start with what you know. Terminal velocity $v_t$ has dimensions $[LT^{-1}]$. The left side of the equation therefore has dimensions $[LT^{-1}]$. For dimensional consistency, the expression under the square root must have dimensions $[L^2T^{-2}]$. Let's analyze the numerator and denominator separately. The numerator $2mg$ has dimensions $[M][LT^{-2}] = [MLT^{-2}]$. The denominator contains $\rho AC_D$. Since $C_D$ is dimensionless and area $A$ has dimensions $[L^2]$, the denominator has dimensions $[\rho][L^2]$. For the fraction to yield $[L^2T^{-2}]$, we need: $\frac{[MLT^{-2}]}{[\rho][L^2]} = [L^2T^{-2}]$ Solving for ρ: $[\rho] = \frac{[MLT^{-2}]}{[L^2][L^2T^{-2}]} = \frac{[MLT^{-2}]}{[L^4T^{-2}]} = [ML^{-3}]$ Choice A is correct: $[ML^{-3}]$ represents mass per unit volume, which is indeed the physical meaning of density. Choice B gives mass per area, not volume. Choice C incorrectly includes time dependence, while choice D represents volume per mass (density's inverse). Remember: dimensional analysis not only checks equations but often reveals the physical meaning of quantities. Density should always be mass per unit volume: $[ML^{-3}]$.
15. Which of the following physical quantities has the same dimensions as $\frac{\text{force} \times \text{time}}{\text{mass}}$?
    Answer: Velocity, representing the momentum change per unit mass When you encounter dimensional analysis problems, your goal is to break down each physical quantity into its fundamental dimensions and see what matches. Let's analyze the given expression $\frac{\text{force} \times \text{time}}{\text{mass}}$. Force has dimensions of $[MLT^{-2}]$ (mass × acceleration), time is $[T]$, and mass is $[M]$. So our expression becomes: $\frac{[MLT^{-2}] \times [T]}{[M]} = \frac{[MLT^{-1}]}{[M]} = [LT^{-1}]$ This gives us dimensions of length per time, which is exactly the dimension of velocity. You can also think about this physically: $\frac{\text{force} \times \text{time}}{\text{mass}} = \frac{\text{impulse}}{\text{mass}}$. Since impulse equals change in momentum ($\Delta p = m\Delta v$), dividing by mass gives you the change in velocity. Looking at the wrong answers: Choice A (acceleration) has dimensions $[LT^{-2}]$, not $[LT^{-1}]$. Choice C (power per unit mass) would have dimensions $[L^2T^{-3}]$, since power is energy per time. Choice D (energy per unit mass per unit time) also gives $[L^2T^{-3}]$, the same as choice C. The correct answer is B - velocity represents momentum change per unit mass, which matches our dimensional analysis. Study tip: In dimensional analysis problems, always convert to fundamental dimensions $[M]$, $[L]$, and $[T]$ first. Don't get distracted by the physical descriptions in the answer choices until you've done the math.
16. The equation $v = \sqrt{\frac{2gh}{m}}$ is proposed for the velocity of an object. Which statement about this equation is correct?
    Answer: The equation is dimensionally incorrect because mass appears in the denominator When you encounter a proposed physics equation, dimensional analysis is your first line of defense to check if it makes physical sense. This means verifying that both sides of the equation have the same units. Let's analyze the dimensions of $v = \sqrt{\frac{2gh}{m}}$. Velocity $v$ has dimensions of $[LT^{-1}]$. On the right side, $g$ has dimensions $[LT^{-2}]$, height $h$ has dimensions $[L]$, and mass $m$ has dimensions $[M]$. So the expression under the square root becomes: $\frac{[LT^{-2}][L]}{[M]} = \frac{[L^2T^{-2}]}{[M]} = [ML^{-1}T^{-2}]$ Taking the square root gives $[M^{1/2}L^{-1/2}T^{-1}]$, which doesn't match velocity's dimensions of $[LT^{-1}]$. The presence of mass in the denominator creates this dimensional mismatch. Answer B correctly identifies this flaw. The equation is dimensionally incorrect specifically because mass appears in the denominator, making the units incompatible with velocity. Answer A is wrong because we've shown the equation is dimensionally incorrect, ruling out any physical interpretation. Answer C misses the real issue—multiplying $g$ and $h$ actually works fine dimensionally ($[LT^{-2}] \times [L] = [L^2T^{-2}]$), giving energy per unit mass. Answer D is incorrect because the equation isn't dimensionally correct in the first place. Study tip: Always check dimensions first when evaluating physics equations. If the dimensions don't match, the equation is automatically wrong regardless of how reasonable it might look conceptually.

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