In certain vernier callipers, \(25\) divisions on the vernier scale have the same length as \(24\) divisions on the main scale. One division on the main scale is \(1\) mm long. The least count of the instrument is:

1.	\(0.04\) mm	2.	\(0.01\) mm
3.	\(0.02\) mm	4.	\(0.08\) mm

One centimeter on the main scale of Vernier calliper is divided into ten equal parts. If 10 divisions of Vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is:

(1)  0.01 cm

(2)  0.02 cm

(3)  0.05 cm

(4)  0.07 cm

A Screw Guage gives the following readings when used to measure the diameter of a wire.
Main scale reading = 0.0 mm
Circular scale reading = 52 divisions
Given that: 1 mm on the main scale corresponds to 100 divisions of the circular scale.
The diameter of the wire from the above data is:

1.  0.026 cm

2.  0.005 cm

3.  0.52 cm

4.  0.052 cm

Two full turns of the circular scale of gauge cover a distance of \(1\) mm on scale. The total number of divisions on the circular scale is \(50.\) Further, it is found that screw gauge has a zero error of \(-0.03\) mm. While measuring the diameter of a thin wire a student notes the main scale reading of \(3\) mm and the number of circular scale division in line, with the main scale as \(35.\) The diameter of the wire is:
1. \(3.32\) mm
2. \(3.73\) mm
3. \(3.67\) mm
4. \(3.38\) mm

The circular division of shown screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball is 

(1)  2.25 mm

(2)  2.20 mm

(3)  1.20 mm

(4)  1.25 mm

The pitch of a screw gauge is 1.0 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and the 47^th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the wire's curved surface area (in cm²) in an appropriate number of significant figures.

1.  2.4 cm²

2.  2.56 cm²

3.  2.6 cm²

4.  2.8 cm²

Consider a screw gauge without any zero error. What will be the final reading corresponding to the final state as shown?
It is given that the circular head translates \(P\) MSD in \({N}\) rotations. (\(1\) MSD \(=\) \(1~\text{mm}\).)

              
1. \( \left(\frac{{P}}{{N}}\right)\left(2+\frac{45}{100}\right) \text{mm} \)
2. \( \left(\frac{{N}}{{P}}\right)\left(2+\frac{45}{{N}}\right) \text{mm} \)
3. \(P\left(\frac{2}{{N}}+\frac{45}{100}\right) \text{mm} \)
4. \( \left(2+\frac{45}{100} \times \frac{{P}}{{N}}\right) \text{mm}\)

A screw gauge has some zero error but its value is unknown. We have two identical rods. When the first rod is inserted in the screw, the state of the instrument is shown by diagram (I). When both the rods are inserted together in series then the state is shown by the diagram (II). What is the zero error of the instrument? \(1~\text{msd}= 100~\text{csd}=1~\text{mm}\)

Find the zero correction in the given figure. 

1. \(0.4\) mm

2. \(0.5\) mm

3. \(-0.5\) mm

4. \(-0.4\) mm

Find the thickness of the wire. The least count is \(0.01~\text{mm}\). The main scale reads (in mm):

                       
1. \(7.62\)
2. \(7.63\)
3. \(7.64\)
4. \(7.65\)