PHYSICS CH 1 Measurement — Science STD 9 — Question

(i)Determine the pitch, of the least count and zero error of the screw gauge.

• Pitch of the screw =  Distance moved by the pitch  No. of rotations by Head scale 
• Least count (LC) = 0.01 mm
• Zero error:
  Positive zero error (ZE) = + (n × LC)mm = + (n × 0.01) mm
  ∴ Zero correction (ZC) = – (n × 0.01) mm
  Negative zero error (ZE) = – (100 – n) × LC mm
  ∴ Zero correction (ZC) = (100 – n) × LC mm

(ii) Place the teacup between the two studs.

(iii) Rotate the head until the tea cup is held firmly but not tightly, with the help of ratchat.

(iv) Note the reading of the pitch scale crossed by the head scale (PSR) and the head scale
the division that coincides with the pitch scale axis (HSC).

(v) The thickness of the teacup is given by PSR + CHSR (Corrected HSR). Repeat the experiment for different positions of the teacup.

(vi) Tabulate the readings.

(vii) The average of the last column reading gives the thickness of the tea cup.

Thickness of the teacup = ……….. mm