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

In an experiment, the height of an object measured by a vernier callipers having least count of \(0.01~\text{cm}\) is found to be \(5.72~\text{cm}\). 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 \(0.3~\text{mm}\). The correct height of the object is:
1. \( 5.72 ~\text{cm} \)
2. \( 5.59~\text{cm} \)
3. \( 5.85~\text{cm} \)
4. \( 5.69~\text{cm} \)

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