During Searle's experiment, zero of the vernier scale lies between 3.20 x 10^-2 m and 3.25 x 10^-2 m

Q: Why is the error in extension $\Delta l$ taken as LC and not 2 LC?

Strictly speaking, if two independent measurements are subtracted, the absolute errors add up ($\Delta l = 2\text{LC}$). However, in many standard JEE problems, the error in the measured extension is conventionally taken as the least count of the instrument ($\Delta l = \text{LC}$) unless specified otherwise. Always check the options or follow this convention if 2 LC isn't applicable.

Q: Why did we ignore the errors in length $L$, area $A$, and force $F$?

The problem provides these values as exact numbers without any given uncertainties. In error analysis problems, if an uncertainty is not provided for a quantity, we assume it is perfectly measured and contributes zero error to the final result.

Q: What happens to the main scale reading (MSR) when calculating the extension?

Since the zero of the vernier scale lies between the same two main scale marks ($3.20 \times 10^{-2}$ m and $3.25 \times 10^{-2}$ m) for both the initial and final readings, the MSR is identical for both. When we subtract the initial reading from the final reading to find the extension, the MSR completely cancels out.

Q: Do we need to use the actual value of the least count ($1.0 \times 10^{-5}$ m) in the calculation?

No, it's not strictly necessary! As you saw in the final step, the extension $l$ is expressed as $25 \times \text{LC}$ and the error $\Delta l$ is $\text{LC}$. When we take the ratio $\frac{\Delta l}{l}$, the LC term cancels out entirely, leaving just $\frac{1}{25}$.

Q: What is Searle's experiment used for?

Searle's experiment is a classic laboratory method used to determine the Young's modulus of the material of a wire. It involves hanging weights to stretch the wire and precisely measuring that tiny extension using a micrometer or vernier scale.

Question

During Searle's experiment, zero of the vernier scale lies between $3.20 	imes 10^{-2}	ext{ m}$ and $3.25 	imes 10^{-2}	ext{ m}$ of the main scale. The $20^{	ext{th}}$ division of the vernier scale exactly coincides with one of the main scale divisions. When an additional load of $2	ext{ kg}$ is applied to the wire, the zero of the vernier scale still lies between $3.20 	imes 10^{-2}	ext{ m}$ and $3.25 	imes 10^{-2}	ext{ m}$ of the main scale but now the $45^{	ext{th}}$ division of vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $2	ext{ m}$ and its cross-sectional area is $8 	imes 10^{-7}	ext{ m}^2$. The least count of the vernier scale is $1.0 	imes 10^{-5}	ext{ m}$. The maximum percentage error in the Young's modulus of the wire is

Sigma Percentile · Accepted Answer

### The Setup: Searle's Experiment

Imagine you are in a physics lab performing Searle's experiment. You have a long, thin metallic wire suspended from a rigid support. When you hang a weight from it, the wire stretches. This tiny stretch, or extension, is what we need to measure to calculate the Young's modulus of the material.

The formula for Young's modulus is given by:
$$Y = \frac{FL}{Al}$$
where $F$ is the applied force, $L$ is the original length, $A$ is the cross-sectional area, and $l$ is the extension.

### Pinpointing the Culprit: Where is the Error?

In any experiment, measurements are never perfect. However, in this specific problem, the values for the load ($2	ext{ kg}$), the original length ($2	ext{ m}$), and the cross-sectional area ($8 	imes 10^{-7}	ext{ m}^2$) are given as exact constants. There are no uncertainties provided for them.

This means we can treat them as perfectly accurate. The only quantity that introduces an error into our calculation of $Y$ is the extension $l$, which we are measuring using a vernier scale.

Therefore, the maximum percentage error in Young's modulus is directly equal to the percentage error in the extension:
$$\frac{\Delta Y}{Y} 	imes 100 = \frac{\Delta l}{l} 	imes 100$$

### Decoding the Vernier Scale

To find the extension $l$, we need to take two readings: an initial reading before the extra load is added, and a final reading after.

The problem states that for both readings, the zero of the vernier scale lies between $3.20 	imes 10^{-2}	ext{ m}$ and $3.25 	imes 10^{-2}	ext{ m}$ on the main scale. This is a crucial piece of information! It tells us that the Main Scale Reading (MSR) is exactly the same for both measurements.

Let's write out the readings:
$$	ext{Initial Reading} = 	ext{MSR} + 20 	imes 	ext{LC}$$
$$	ext{Final Reading} = 	ext{MSR} + 45 	imes 	ext{LC}$$

The extension $l$ is simply the difference between these two readings. When we subtract them, the MSR beautifully cancels out:
$$l = (45 - 20) 	imes 	ext{LC} = 25 	imes 	ext{LC}$$

### The Final Calculation

Now, what about the error in this extension, $\Delta l$? In standard JEE conventions for this experiment, the maximum error in the measured extension is taken to be the least count of the instrument itself. So, $\Delta l = 	ext{LC}$.

Let's plug these into our percentage error formula:
$$\frac{\Delta Y}{Y} 	imes 100 = \frac{	ext{LC}}{25 	imes 	ext{LC}} 	imes 100$$

Notice how the actual numerical value of the least count ($1.0 	imes 10^{-5}	ext{ m}$) isn't even needed! The $	ext{LC}$ terms cancel out perfectly:
$$\frac{\Delta Y}{Y} 	imes 100 = \frac{1}{25} 	imes 100 = 4\%$$

And there we have it. The maximum percentage error in the Young's modulus is exactly **4%**.

Enter Numerical Value:

Visualized Solution

$Y = \frac{F L}{A l}$

$\frac{Δ Y}{Y} \times 100 = \frac{Δ l}{l} \times 100$

$l = Final - Initial$

$l = 25 \times LC$

$\frac{Δ Y}{Y} \times 100 = 4%$

The Sigma Insight: Young's Modulus

The Setup

Pinpointing the Culprit

Decoding the Vernier Scale

The Final Calculation