A light-year is a unit of: 

A screw gauge has the least count of \(0.01~\text{mm}\) and there are \(50\) divisions in its circular scale. The pitch of the screw gauge is:

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:

We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71 s, and 2.80 s. The average absolute error and percentage error, respectively, are:

1.  0.22 s and 4%

2.  0.11 s and 4%

3.  4 s and 0.11%

4.  5 s and 0.22%

In a vernier calliper, \(N\) divisions of vernier scale coincide with (\(N\text-1\)) divisions of the main scale (in which the length of one division is \(1\) mm). The least count of the instrument should be:
1. \(N~\text{mm}\)
2. \((N-1)~\text{mm}\)
3. \(\frac{1}{10N}~\text{cm}\)
4. \(\frac{1}{(N-1)}~\text{mm}\)

The main scale reading is \(-1\) mm when there is no object between the jaws. In the vernier calipers, \(9\) main scale division matches with \(10\) vernier scale divisions. Assume the edge of the Vernier scale as the '0' of the vernier. The thickness of the object using the defected vernier calipers will be:

         
1. \(12.2~\text{mm}\)
2. \(1.22~\text{mm}\)
3. \(12.3~\text{mm}\)
4. \(12.4~\text{mm}\)

A screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading: \(0\) mm
Circular scale reading: \(52\) divisions 
Given that \(1\) mm on the main scale corresponds to \(100\) divisions on the circular scale, the diameter of the wire that can be inferred from the given data is:

A student measured the diameter of a small steel ball using a screw gauge of least count \(0.001\) cm. The main scale reading is \(5\) mm and zero of circular scale division coincides with \(25\) divisions above the reference level. If the screw gauge has a zero error of \(-0.004\) cm, the correct diameter of the ball is:

The pitch of a screw gauge is \(1~\)mm and there are \(100\) divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads \(1\) mm and \(47\)^th division on the circular scale coincides with the reference line. The length of the wire is \(5.6\) cm. Curved surface area (in cm²) of the wire in appropriate number of significant figures will be:

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