The angle of \(1'\) (minute of an arc) in radian is nearly equal to:
1. \(2.91 \times 10^{-4} ~\mathrm{rad} \)
2. \(4.85 \times 10^{-4} ~\mathrm{rad} \)
3. \(4.80 \times 10^{-6} ~\mathrm{rad} \)
4. \(1.75 \times 10^{-2} ~\mathrm{rad}\)

The angle of \(1^\circ\) (degree) will be equal to:
(Use \(360^\circ=2\pi\) rad)
1. \(1.034\times10^{-3}\) rad
2. \(1.745\times10^{-2}\) rad
3. \(1.524\times10^{-2}\) rad
4. \(1.745\times10^{3}\) rad

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:
1. \(0.25\) mm
2. \(0.5\) mm
3. \(1.0\) mm
4. \(0.01\) mm

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 \(\mathrm{P}\) MSD in \(\mathrm{N}\) rotations. (\(1\) MSD \(=\) \(\mathrm{1~mm}\).)

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

In an experiment, the height of an object measured by a vernier callipers having least count of \(0.01~\mathrm{cm}\) is found to be \(5.72~\mathrm{cm}\). When no object is there between jaws of this vernier callipers, the reading of the main scale is \(0.1\) cm and the reading of the vernier scale is \(0.3~\mathrm{mm}\). The correct height of the object is:
1. \( 5.72 \mathrm{~cm} \)
2. \( 5.59 \mathrm{~cm} \)
3. \( 5.85 \mathrm{~cm} \)
4. \( 5.69 \mathrm{~cm}\)
 

The main scale of a vernier calliper has \(n\) divisions/cm. \(n\) divisions of the vernier scale coincide with \((n-1)\) divisions of the main scale. The least count of the vernier calliper is:
1. \(\frac{1}{(n+1)(n-1)}\) cm
2. \(\frac{1}{n}\) cm
3. \(\frac{1}{n^{2}}\) cm
4. \(\frac{1}{(n)(n+1)}\) cm