A student determined Young's modulus of elasticity using the formula Y = (MgL^3)/(4bd^3\delta). The

Q: Why do we add the errors even if there is division in the formula?

In error analysis, we always want to find the maximum possible error. Whether quantities are multiplied or divided, their relative errors always add up to give the maximum relative error of the final result.

Q: Why is the error in 'g' ignored in the calculation?

The problem explicitly states that the value of $g$ is taken to be $9.8 \text{ m/s}^2$ 'without any significant error'. Since it is treated as an exact constant for this experiment, its absolute error $\Delta g$ is zero.

Q: Why do we multiply the error of L and d by 3?

According to the rules of error propagation, if a physical quantity is raised to a power $n$ (like $L^3$ and $d^3$), its relative error is multiplied by $n$. Therefore, the relative error for $L^3$ is $3(\Delta L/L)$.

Q: Can I keep the units in mm and cm instead of converting to meters?

You can, but only if you are extremely careful to ensure that the numerator and denominator of a specific ratio have the exact same unit. For example, for $\Delta b/b$, if you use $0.1 \text{ mm}$ for $\Delta b$, you MUST convert $b$ from $4 \text{ cm}$ to $40 \text{ mm}$ before dividing. Converting everything to standard SI units (meters) is a foolproof way to avoid silly mistakes.

Q: Which quantity requires the most precise measurement in this experiment?

The thickness '$d$'. It has the highest power (3) in the denominator and its measured magnitude is very small. Therefore, even a tiny absolute error in '$d$' gets magnified significantly in the final result.

Question

A student determined Young's modulus of elasticity using the formula $Y = \frac{MgL^3}{4bd^3\delta}$. The value of $g$ is taken to be $9.8 	ext{ m/s}^2$, without any significant error, his observations are as following.

$\begin{array}{|l|l|l|} \hline 	ext{Physical quantity} & 	ext{Least count of the equipment used for measurement} & 	ext{Observed value} \ \hline 	ext{Mass } (M) & 1 	ext{ g} & 2 	ext{ kg} \ 	ext{Length of bar } (L) & 1 	ext{ mm} & 1 	ext{ m} \ 	ext{Breadth of bar } (b) & 0.1 	ext{ mm} & 4 	ext{ cm} \ 	ext{Thickness of bar } (d) & 0.01 	ext{ mm} & 0.4 	ext{ cm} \ 	ext{Depression } (\delta) & 0.01 	ext{ mm} & 5 	ext{ mm} \ \hline \end{array}$

Then, the fractional error in the measurement of $Y$ is

Options:

(A) 0.0083

(B) 0.0155

(C) 0.155

(D) 0.083

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are in a physics lab, conducting an experiment to determine the Young's modulus of a material. You have a rectangular beam supported at its two ends. When you hang a heavy mass $M$ exactly from its center, the beam bends downwards. This downward displacement is called the depression, denoted by $\delta$.

The formula governing this physical phenomenon is given by:

$$Y = \frac{MgL^3}{4bd^3\delta}$$

Here, $L$ is the length of the beam, $b$ is its breadth, and $d$ is its thickness. The acceleration due to gravity, $g$, is a constant and is given as $9.8 	ext{ m/s}^2$ without any significant error. Our goal is to find the maximum possible fractional error in the measurement of $Y$.

### The Master Equation for Error
To find the maximum fractional error in a quantity calculated from a formula involving multiplication and division, we sum up the relative (or fractional) errors of all the individual measured quantities. Crucially, any power to which a quantity is raised becomes a multiplier for its relative error.

Applying this rule to our formula, we get:

$$\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + 3\frac{\Delta L}{L} + \frac{\Delta b}{b} + 3\frac{\Delta d}{d} + \frac{\Delta \delta}{\delta}$$

Notice that the terms for $L$ and $d$ are multiplied by 3 because they appear as $L^3$ and $d^3$ in the original formula. The constant $4$ and the exact value $g$ do not contribute to the error.

### Careful Unit Conversions
Before we plug in the numbers, we must be extremely careful. The least counts (which represent the absolute errors, like $\Delta M$) and the observed values are given in a mix of units: grams, kilograms, millimeters, centimeters, and meters.

**Don't make a silly mistake here!** To calculate a valid ratio, the numerator and denominator must be in the exact same units. The safest approach is to convert everything into standard SI units (meters and kilograms).

Let's break down the substitutions:

**1. Mass ($M$):**
$$\frac{\Delta M}{M} = \frac{1 	ext{ g}}{2 	ext{ kg}} = \frac{10^{-3} 	ext{ kg}}{2 	ext{ kg}} = 0.0005$$

**2. Length ($L$):**
$$3\frac{\Delta L}{L} = 3 	imes \frac{1 	ext{ mm}}{1 	ext{ m}} = 3 	imes \frac{10^{-3} 	ext{ m}}{1 	ext{ m}} = 0.0030$$

**3. Breadth ($b$):**
$$\frac{\Delta b}{b} = \frac{0.1 	ext{ mm}}{4 	ext{ cm}} = \frac{10^{-4} 	ext{ m}}{4 	imes 10^{-2} 	ext{ m}} = 0.0025$$

**4. Thickness ($d$):**
$$3\frac{\Delta d}{d} = 3 	imes \frac{0.01 	ext{ mm}}{0.4 	ext{ cm}} = 3 	imes \frac{10^{-5} 	ext{ m}}{4 	imes 10^{-3} 	ext{ m}} = 0.0075$$

**5. Depression ($\delta$):**
$$\frac{\Delta \delta}{\delta} = \frac{0.01 	ext{ mm}}{5 	ext{ mm}} = \frac{10^{-5} 	ext{ m}}{5 	imes 10^{-3} 	ext{ m}} = 0.0020$$

### Final Calculation
Now, we simply add all these individual fractional errors together to find the total fractional error in Young's modulus:

$$\frac{\Delta Y}{Y} = 0.0005 + 0.0030 + 0.0025 + 0.0075 + 0.0020$$

$$\frac{\Delta Y}{Y} = 0.0155$$

**A quick insight:** If you look closely at the individual terms, the error contributed by the thickness $d$ ($0.0075$) is the largest, making up almost half of the total error! This happens because $d$ is a very small quantity (only $0.4 	ext{ cm}$) and it is raised to the power of 3 in the formula. This teaches us a valuable experimental lesson: quantities that are small and have high powers in the formula must be measured with the utmost precision.

Select Answer:

Visualized Solution

$E x p er im e n tS e t u p$

$F r a c t i o na l E r r or F or m u l a$

$S u b s t i t u t in g M an d L$

$S u b s t i t u t in g b, d, an d δ$

$C a l c u l a t in g I n d i v i d u a l E r r or s$

$T o t a l F r a c t i o na l E r r or$

$E r r or A na l y s i s I n s i g h t$

The Sigma Insight: Errors in Measurement

Analyzing the Setup Imagine you are in a physics lab, conducting an experiment to determine the Young's modulus of a material

The Master Equation for Error To find the maximum fractional error in a quantity calculated from a formula involving multiplication and division, we sum up the relative (or fractional) errors of all the individual measured quantities

Careful Unit Conversions Before we plug in the numbers, we must be extremely careful

Final Calculation

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Select Answer:

Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitutingb,d,andδ

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