JEE PYQ: Units & Measurements - Question ID d1242d4fdc6b (JEE Main 2020)

A screw gauge is a precision instrument used to measure small lengths with high accuracy. It consists of two main scales:

• Main Scale (Linear Scale): This is a fixed scale along the length of the screw gauge, typically marked in millimeters or centimeters.
• Circular Scale (Head Scale): This is a rotating scale with divisions that move past a reference line on the main scale. The number of divisions on the circular scale determines the precision of the measurement.

The key concepts and formulas involved are:

1. Pitch of the Screw Gauge ($p$): The pitch is the distance moved by the screw along the main scale for one complete rotation of the circular scale. Here, $p = 0.1 \text{ cm}$.
2. Least Count ($LC$): The least count is the smallest measurement that can be made with the screw gauge. It is given by: $LC = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{p}{N}$ where $N$ is the number of divisions on the circular scale. Here, $N = 50$.
3. Total Reading: The total thickness ($T$) of the object is calculated as: $T = \text{Main Scale Reading} + (\text{Circular Scale Reading} \times LC)$

The least count determines the precision of the measurement, and the final recorded value must reflect this precision.

Let’s derive the correct recorded thickness step-by-step.

1. Calculate the Least Count ($LC$):
   Given:
   Pitch ($p$) = $0.1 \text{ cm}$
   Number of divisions on circular scale ($N$) = $50$
   Using the formula for least count: $LC = \frac{p}{N} = \frac{0.1 \text{ cm}}{50} = 0.002 \text{ cm}$


2. Interpret the Measurement:
   The question implies that the main scale reading is $2.12 \text{ cm}$ (since all options are around this value), and the circular scale reading is a certain number of divisions. However, the exact circular scale reading is not provided directly. Instead, we infer it from the options.
   
   Let’s assume the circular scale reading is $x$ divisions. Then, the total thickness is: $T = 2.12 \text{ cm} + (x \times 0.002 \text{ cm})$
   The options suggest that $x$ is such that the total thickness is one of the given values. We need to find which option is consistent with the least count of $0.002 \text{ cm}$.


3. Check Consistency of Options:
   The least count is $0.002 \text{ cm}$, so the measurement must be a multiple of $0.002 \text{ cm}$ beyond the main scale reading. Let’s express each option in terms of the main scale reading and the circular scale contribution:
   
   
   • Option A: $2.123 \text{ cm}$
     $2.123 \text{ cm} = 2.12 \text{ cm} + 0.003 \text{ cm}$
     $0.003 \text{ cm}$ is not a multiple of $0.002 \text{ cm}$ (since $0.003 / 0.002 = 1.5$). This is not possible.
   
   
   • Option B: $2.124 \text{ cm}$
     $2.124 \text{ cm} = 2.12 \text{ cm} + 0.004 \text{ cm}$
     $0.004 \text{ cm}$ is a multiple of $0.002 \text{ cm}$ (since $0.004 / 0.002 = 2$). This is possible if the circular scale reading is $2$ divisions.
   
   
   • Option C: $2.125 \text{ cm}$
     $2.125 \text{ cm} = 2.12 \text{ cm} + 0.005 \text{ cm}$
     $0.005 \text{ cm}$ is not a multiple of $0.002 \text{ cm}$ (since $0.005 / 0.002 = 2.5$). This is not possible.
   
   
   • Option D: $2.121 \text{ cm}$
     $2.121 \text{ cm} = 2.12 \text{ cm} + 0.001 \text{ cm}$
     $0.001 \text{ cm}$ is not a multiple of $0.002 \text{ cm}$ (since $0.001 / 0.002 = 0.5$). This is not possible.


4. Conclusion:
   Only Option B ($2.124 \text{ cm}$) is consistent with the least count of $0.002 \text{ cm}$. This corresponds to a circular scale reading of $2$ divisions, giving: $T = 2.12 \text{ cm} + (2 \times 0.002 \text{ cm}) = 2.124 \text{ cm}$

Students often make the following mistakes in such questions:

1. Incorrect Least Count Calculation:
   Some students confuse the pitch with the least count or miscalculate the least count by inverting the formula (e.g., $LC = N / p$ instead of $LC = p / N$). Always remember: $LC = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$


2. Ignoring Precision:
   Students may select an option that does not align with the least count. For example, choosing $2.123 \text{ cm}$ (Option A) because it "looks close" without verifying if it is a multiple of the least count. Always ensure the measurement is consistent with the instrument's precision.


3. Misinterpreting the Main Scale Reading:
   The main scale reading is often assumed to be exact, but students may misread it (e.g., $2.1 \text{ cm}$ instead of $2.12 \text{ cm}$). Pay close attention to the main scale markings.


4. Rounding Errors:
   Some students round off the least count or the final measurement incorrectly. For example, rounding $0.002 \text{ cm}$ to $0.01 \text{ cm}$ would lead to incorrect options. Always use the exact least count value.

Exam Tip:
When solving screw gauge problems, always:

1. Calculate the least count first.
2. Express the total measurement as a sum of the main scale reading and a multiple of the least count.
3. Check which option matches this sum exactly.