9.8 = 1 kg mass
98 N = ?
\(\text{Mass of the object }=\frac{98\times1}{9.8}=10\ kg\)

Pitch scale reading is = 1 mm
Head scale coincidence = 68
Thickness of 5 rupee coin = PSR + (HSC x L.C)
= 1 +(68 x 0.01)
= 1 + 0.68
= 1.68 mm
Thickness of five rupee coin is = 1.68 mm

Main scale reading = 7 x 10 = 70 mm.(∴ 1 cm = 10 mm)
Vernier coincidence = 6
Least count = 0.1 mm.
Zero error = Nil
The diameter of the ball = MSR + (V.C x L.C) ± ZE
= 70 mm + (6 x 0.1)
= 70 mm + 0.6 = 70.6 mm
radius of the ball = 70.6/2 (∴ r = d/2)
35.3 mm

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.

The thickness of coin (one rupee) is finding with the help of screw gauge:
1. Determine the pitch, least count and the zero error of the screw gauge.
2. Place the coin between two studs.
3. Rotate the head until the coin is held firmly but no tightly with the help of the ratchat.
4. Note the reading of a pitch scale crossed by the head scale and the head scale (PSR) division that coincides with the pitch scale axis (HSC).
5. The thickness of the coin is given by PSR + (HSR + ZC) x LC. Repeat the experiment for the different position of the coin.
6. Tabulate the reading.
7. The average of the reading given the thickness of the coin.

The thickness of one rupee coin is = 1.33 mm.