In a screw gauge, $5$ complete rotations of the screw cause it to move a linear distance of $0.25\, cm$. There are $100$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $4$ main scale divisions and $30$ circular scale divisions. Assuming negligible zero error, the thickness of the wire is :

Question

The area enclosed between the parabola $ y^2 = 4x $ and the line $ y = 2x - 4 $ is:

Answer 1

To solve the problem, we should first understand the components of a screw gauge and how measurements are calculated:

Pitch Calculation: The pitch of the screw gauge is defined as the distance moved by the spindle per complete rotation. Given that 5 complete rotations of the screw move it a linear distance of 0.25\, \text{cm}, we can calculate the pitch as follows: \text{Pitch} = \frac{0.25\, \text{cm}}{5} = 0.05\, \text{cm}
Least Count of Screw Gauge: The least count of the screw gauge is calculated by dividing the pitch by the number of divisions on the circular scale. Here, it is given that there are 100 circular scale divisions: \text{Least Count} = \frac{\text{Pitch}}{\text{Number of Circular Scale Divisions}} = \frac{0.05\, \text{cm}}{100} = 0.0005\, \text{cm}
Measurement Reading: The formula for calculating the total measurement using a screw gauge is: \text{Total Reading} = (\text{Main Scale Reading}) + (\text{Circular Scale Reading} \times \text{Least Count}) Given: Main Scale Reading = 4 main scale divisions and Circular Scale Reading = 30 circular scale divisions. Therefore: \text{Total Measurement} = 4\, \text{cm} + (30 \times 0.0005\, \text{cm}) = 4\, \text{cm} + 0.015\, \text{cm} = 4.015\, \text{cm}
Convert Measurement: Now, subtract the main scale divisions from the total, as per the reading formula: \text{Thickness of Wire} = (4 \times \text{Pitch}) + (30 \times \text{Least Count}) = (4 \times 0.05\, \text{cm}) + (30 \times 0.0005\, \text{cm}) = 0.20\, \text{cm} + 0.015\, \text{cm} = 0.2150\, \text{cm}

The correct thickness of the wire, accounting for main and circular scale readings, is 0.2150\, \text{cm}. Thus, the answer is 0.2150\, \text{cm}.

Answer 2

To solve the problem, we should first understand the components of a screw gauge and how measurements are calculated:

Pitch Calculation: The pitch of the screw gauge is defined as the distance moved by the spindle per complete rotation. Given that 5 complete rotations of the screw move it a linear distance of 0.25\, \text{cm}, we can calculate the pitch as follows: \text{Pitch} = \frac{0.25\, \text{cm}}{5} = 0.05\, \text{cm}
Least Count of Screw Gauge: The least count of the screw gauge is calculated by dividing the pitch by the number of divisions on the circular scale. Here, it is given that there are 100 circular scale divisions: \text{Least Count} = \frac{\text{Pitch}}{\text{Number of Circular Scale Divisions}} = \frac{0.05\, \text{cm}}{100} = 0.0005\, \text{cm}
Measurement Reading: The formula for calculating the total measurement using a screw gauge is: \text{Total Reading} = (\text{Main Scale Reading}) + (\text{Circular Scale Reading} \times \text{Least Count}) Given: Main Scale Reading = 4 main scale divisions and Circular Scale Reading = 30 circular scale divisions. Therefore: \text{Total Measurement} = 4\, \text{cm} + (30 \times 0.0005\, \text{cm}) = 4\, \text{cm} + 0.015\, \text{cm} = 4.015\, \text{cm}
Convert Measurement: Now, subtract the main scale divisions from the total, as per the reading formula: \text{Thickness of Wire} = (4 \times \text{Pitch}) + (30 \times \text{Least Count}) = (4 \times 0.05\, \text{cm}) + (30 \times 0.0005\, \text{cm}) = 0.20\, \text{cm} + 0.015\, \text{cm} = 0.2150\, \text{cm}

The correct thickness of the wire, accounting for main and circular scale readings, is 0.2150\, \text{cm}. Thus, the answer is 0.2150\, \text{cm}.

The Correct Option is B