Question

The distance moved by the screw of a screw gauge is \(2 \mathrm{~mm}\) in four rotations and there are 50 divisions on its cap. When nothing is put between its jaws, \(30^{\text {th }}\) division of circular scale coincides with reference line, with zero of circular scale lying above the reference line. When a plate is placed between the jaws, main scale reads 2 division and circular scale reads 20 division. Thickness of plate is : (1) \(1.5 \mathrm{~mm}\) (2) \(1.2 \mathrm{~mm}\) (3) \(1.4 \mathrm{~mm}\) (4) \(1.6 \mathrm{~mm}\)

Short answer

Calculated thickness seems to be 1.90 mm, which does not match the given options.

Step-by-step solution

Determine the least count

The least count of a screw gauge is calculated using the formula: \[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} \] where Pitch is the distance moved by the screw in one rotation. Here, Pitch = \( \frac{2 \text{ mm}}{4 \text{ rotations}} = 0.5 \text{ mm} \). Hence, \[ \text{Least Count} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm} \].

Calculate the zero error

The zero error can be calculated by noting the reading when nothing is placed between the jaws. Here, the 30th division of the circular scale coincides with the reference line. Therefore, the zero error is: \[ \text{Zero Error} = +30 \times \text{Least Count} \]. With the Least Count as \( 0.01 \text{ mm} \), \[ \text{Zero Error} = +30 \times 0.01 \text{ mm} = +0.30 \text{ mm} \].

Read the main and circular scales with the plate

When the plate is placed between the jaws, the main scale reads 2 divisions and the circular scale reads 20 divisions. The value from the circular scale is then: \[ \text{Circular Scale Reading} = 20 \times \text{Least Count} = 20 \times 0.01 \text{ mm} = 0.20 \text{ mm} \]

Calculate the thickness of the plate

The total reading is the sum of the main scale reading, the circular scale reading, and corrected by the zero error. Hence, \[ \text{Total Thickness} = \text{Main Scale Reading} + \text{Circular Scale Reading} - \text{Zero Error} \]\[ = 2 \text{ mm} + 0.20 \text{ mm} - 0.30 \text{ mm} = 1.90 \text{ mm} \].

Identify the correct option

Comparing the calculated thickness (\(1.90 \text{ mm}\)) with the given options, none match directly. Therefore, review each step carefully to ensure accurate understanding of any possible errors or misinterpretations.

Key concepts

least count

The least count of a screw gauge is one of its most important factors. It tells us the smallest measurement that the screw gauge can reliably read. To find the least count, use this simple formula: \[ \text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} \] Here, the 'pitch' represents how far the screw moves per rotation. Let's take an example: if the screw moves 2 mm in four rotations, then the pitch is: \[ \text{Pitch} = \frac{2 \text{ mm}}{4 \text{ rotations}} = 0.5 \text{ mm} \] Now, let's consider the circular scale has 50 divisions. Plugging these values in, we get: \[ \text{Least Count} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm} \] This calculation tells us that the smallest measurable unit using this screw gauge is 0.01 mm, which can help in determining very fine measurements accurately.

zero error

Zero error occurs when the screw gauge does not read zero when nothing is between its jaws. This can lead to inaccurate measurements if not accounted for. To calculate the zero error, follow these steps: If you see the 30th division aligning with the reference line when the jaws are empty, then you have a zero error. Using our example, the zero error is: \[ \text{Zero Error} = +30 \times \text{Least Count} \] Inserting our previously calculated least count of 0.01 mm: \[ \text{Zero Error} = +30 \times 0.01 \text{ mm} = +0.30 \text{ mm} \] This positive zero error must be subtracted from your future measurements to get accurate readings. So, always remember to check for zero error before making any measurements and adjust your final reading by this amount.

main scale reading

The main scale reading is straightforward. It shows how many full units the object spans on the main scale of the screw gauge. For instance, if you place a plate and notice that the main scale reads 2 divisions, it means the width of the plate covers 2 mm. This reading forms the base for any further detailed calculation. Never overlook the main scale reading, as this forms the primary value to which you will add the minor, more precise calculations from the circular scale reading.

circular scale reading

While the main scale reading gives a coarse measurement, the circular scale reading provides fine tuning. To find this value, multiply the division on the circular scale by the least count. For example, if the 20th division is aligning on the circular scale when the plate is placed: \[ \text{Circual Scale Reading} = 20 \times 0.01 \text{ mm} = 0.20 \text{ mm} \] This small adjustment ensures you capture fine details. Adding this to the main scale reading gives a more accurate measurement. Just don't forget to account for any zero error in the final calculation!