Length is defined as the distance between two points. The SI unit of length is metre.

Upper jaws.

Correct reading = 3.55 cms.

Celsius = (F – 32) × \(\frac{5}{9}\)= (104 - 32) ×\(\frac{5}{9}\) = 72 × \(\frac{5}{9}\) = 40°C.

No. I cannot find the diameter of a thin wire by using the ruler.

M.S. reading = 3.8 cm
V.S.R = 5^th division
least count = 0.01 cm
Length = P.S.R + (V.S.C x L.C)
= 3.8 + (5 x 0.01) = 3.8 + 0.05 = 3.85
Length of the iron cylinder = 3.85 cm

Given: P.S.R = 1.5 mm, H.S.R = 20 ; L.C = 0.01 mm
Z.E = + 0.05 mm ; Zero correction = - 0.05 mm
Diameter of iron nail = P.S.R + (H.S.R + L.C) + Z.C
= 1.5 + (20 x 0.01) - 0.05
= 1.5 + 0.20 - 0.05 = 1.65 mm
Diameter of iron nail = 1.65 mm

Mass m = 5 kg
Acceleration due to gravity g = 9.8 ms^-1
Weight W = m x g = 5 x 9.8 = 49 N
Therefore the weight of an object is 49 N

The value is one smallest main scale division is divided by total number of vernier scale division is called the least count of vernier caliper.
\( \text { Least count } =\frac{\text { The value of one smallest main scale division }}{\text { Totalnumber of vernier scale division }}\)
\( \\ \text { L.C } =\frac{1 \mathrm{~mm}}{10}=0.1 \mathrm{~mm}=0.01 \mathrm{~cm}\)

The measuring unit of the thickness of a plastic carry bag is micron.

(i) Determine the pitch, the least count and the zero error of the screw gauge.
(ii) Place the wire between the two slides.
(iii) Rotate the head until the wire is held firmly but not tightly with the help of ratchet.
(iv) Note the P.S.R and H.S.R
(v) The diameter of the wire is given by
     P.S.R + (H.S.C x L.C) ± Z.C,
(vi) Repeat the experiment for different portions of the wire.
(vii) Tabulate the readings.
(viii) The average of the last column reading gives the diameter of the wire.

(ix) The average gives the correct reading.

Case (a): Zero of the vernier scale and zero of the main scale are coinciding with each other. So there is no zero error.
Case (b): The zero of vernier scale is shifted to the right from the zero of the main scale. It is positive error. The 8th division of vernier scale coincides with one of the main scale divisions. So the positive error = (8×0.01cm) = 0.08 cm.
Case (c): The zero of vernier scale is shifted to the left from the zero of main scale. It is negative error. The 4th division of vernier scale (6th from backward) coincides with one of the main scale divisions. So the negative zero error = ‒(6×0.01cm)= ‒0.06 cm.