Screw Gauge

- Screw gauge
- Wire
- A sheet of paper
- An irregular lamina
- A centimetre graph paper
- A pointed pencil

- Determine the pitch and least count of the screw gauge using the equations (1) and (2) respectively..
- Bring the anvil and screw in contact with each other and find the zero error. Do it three times and record them. If there is no zero error, then record ‘zero error nil’.
- Move the screw away from the anvil and place the lead shot and move the screw towards the anvil using the ratchet head. Stop when the ratchet slips without moving the screw.
- Note the number of divisions on the pitch scale that is visible and uncovered by the edge of the cap. The reading N is called the pitch scale reading(PSR)
- Note the number (n) of the division of the circular scale lying over the reference line.
- Repeat steps 4 and 5 after rotating the lead shot by 900 for measuring the diameter in a perpendicular direction. Record the observations in the tabular column.
- Find total reading using the equation 3 and apply zero correction in each case.
- Take the mean of different values.

Note: Place the other objects like, wire, glass plate etc between the screw and the anvil and follow the above procedure to find the measurement.

- We can select the least count of the screw gauge from the 'Select screw gauge' drop down list.
- The object of our choice can be chosen from the 'Select object' drop down list.
- Click on the object to place it between the screw and the anvil to find its measurement. To remove the object, click the object on the left menu once more.
- Click on the arrows seen on the screw head to tighten it till the screw touches the object.
- Note the PSR and HSR values.
- Find the total reading and enter the value in the text box provided.
- Click on the check button to verify the answer.
- The 'Reset' button can be used to redo the experiment.

1 Linear Scale Division, LSD = 1 mm

Number of full rotations given to screw =4

Distance moved by the screw = 4mm

Hence , pitch p= = 1mm

Number of divisions on circular scale=100

Hence, least count, L.C == 0.01 mm= 0.001 cm

(i) zero error = --------------mm

(ii) zero error = ---------------mm

(iii) zero error = ----------------mm

Mean zero error, e= ------------mm

Mean zero correction , c= -e = -------mm

Object Placed | Pitch Scale Reading (N) mm | HeadScale Reading | Total Reading | ||

No of circular divisions on reference line(n) | Value [n x L.C]mm | Observed D_{0}=N+n mm |
Corrected D=D_{0} + c mm |
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L ead shot |
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Wire |
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Glass Plate | |||||

Irregular Lamina | |||||

Mean Diameter of the lead shot=----------cm

Mean Diameter of the wire=---------cm

Mean length of the wire=----------cm

Volume of the wire, =------------cm^{3}

Thickness of the glass plate=--------cm

Thickness of irregular lamina=--------cm

Area, A= -----------------------cm^{2}

Volume of irregular lamina, V= A x t =------------cm^{3}

Diameter of the lead shot=----------cm

The volume of the given wire is ---- cm^{3}

The thickness of given sheet is ------- ---cm

The volume of given lamina is* = ....... *cm^{3}

**Cite this Simulator:**