JEE PYQ: Units & Measurements - Question ID 00333d0b3ae0 (JEE Main 2020)

ID: 00333d0b3ae0JEE Main 2020Single Correct MCQ
The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale between 3.1 cm and 3.2 cm and 4th VSD coincides with a main scale division. The length of the cylinder is : (VSD is vernier scale division)

Select Option

Step-by-step Explanation

Core Formula & Concept:

In vernier callipers, the least count (LC) is the smallest length that can be measured accurately. It is determined by the difference between one main scale division (MSD) and one vernier scale division (VSD). The key formula is:

Least Count (LC)=Value of 1 MSDValue of 1 VSD\text{Least Count (LC)} = \text{Value of 1 MSD} - \text{Value of 1 VSD}

When the vernier scale is divided into nn divisions that coincide with (n1)(n-1) divisions of the main scale, the least count becomes:

LC=Value of 1 MSDn\text{LC} = \frac{\text{Value of 1 MSD}}{n}

Additionally, the zero error must be accounted for. Zero error occurs when the zero of the vernier scale does not align with the zero of the main scale when the jaws are closed. It can be positive or negative depending on the direction of misalignment.

The actual measurement is calculated as:

Actual Length=Observed Reading±Zero Error\text{Actual Length} = \text{Observed Reading} \pm \text{Zero Error}

Where the sign depends on the nature of the zero error.

--- Step-by-Step Derivation:

Step 1: Determine the Least Count

Given: - Least count of main scale = 1 mm = 0.1 cm - Vernier scale has 10 divisions coinciding with 9 divisions of the main scale

Since 10 VSD = 9 MSD,

1 VSD=910 MSD=910×0.1 cm=0.09 cm1 \text{ VSD} = \frac{9}{10} \text{ MSD} = \frac{9}{10} \times 0.1 \text{ cm} = 0.09 \text{ cm}

Least Count (LC) = 1 MSD – 1 VSD = 0.1 cm – 0.09 cm = 0.01 cm

So, the vernier callipers can measure up to 0.01 cm accuracy.


Step 2: Determine Zero Error

When jaws are touching (zero position): - The 7th division of the vernier scale coincides with a main scale division - Zero of vernier scale lies to the right of the zero of the main scale

This indicates a positive zero error.

Zero error = (Coinciding VSD number) × LC

Zero Error=7×0.01 cm=0.07 cm\text{Zero Error} = 7 \times 0.01 \text{ cm} = 0.07 \text{ cm}

Since the zero of the vernier is to the right of the main scale zero, the instrument reads more than the actual length. Hence, the zero error is positive and must be subtracted from the observed reading.


Step 3: Measure the Length of the Cylinder

When measuring the cylinder: - Zero of vernier scale lies between 3.1 cm and 3.2 cm on the main scale - 4th VSD coincides with a main scale division

Observed Reading = Main Scale Reading + (Coinciding VSD × LC)

Main scale reading just before zero of vernier = 3.1 cm

Observed Reading=3.1 cm+(4×0.01 cm)=3.1 cm+0.04 cm=3.14 cm\text{Observed Reading} = 3.1 \text{ cm} + (4 \times 0.01 \text{ cm}) = 3.1 \text{ cm} + 0.04 \text{ cm} = 3.14 \text{ cm}

Step 4: Apply Zero Error Correction

Actual Length = Observed Reading – Zero Error

Actual Length=3.14 cm0.07 cm=3.07 cm\text{Actual Length} = 3.14 \text{ cm} - 0.07 \text{ cm} = 3.07 \text{ cm}

Conclusion:

The length of the cylinder is 3.07 cm, which corresponds to option C.

--- Common Traps & Exam Tip:

1. Misidentifying Zero Error Sign: Many students confuse whether to add or subtract the zero error. Remember: - If the zero of the vernier is to the right of the main scale zero, the zero error is positive, and must be subtracted from the observed reading. - If to the left, it's negative, and must be added.

2. Incorrect Least Count Calculation: Students often misapply the formula for least count. Always use:

LC=Value of 1 MSDNumber of VSD=1 mm10=0.1 mm=0.01 cm\text{LC} = \frac{\text{Value of 1 MSD}}{\text{Number of VSD}} = \frac{1 \text{ mm}}{10} = 0.1 \text{ mm} = 0.01 \text{ cm}

But ensure the logic matches the physical setup (10 VSD = 9 MSD).

3. Ignoring Units: Mixing mm and cm can lead to decimal errors. Always convert all measurements to the same unit (preferably cm for this question).

Exam Tip: Always sketch the vernier scale alignment for both zero position and measurement. Visualizing the coincidence helps avoid sign errors.

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