When the circular scale of a screw gauge completes \(2\) rotations, it covers \(1\) mm over the pitch scale. The total number of circular scale divisions is \(50\). The least count of the screw gauge in metres is:
1. \(10^{-4}\)
2. \(10^{-5}\)
3. \(10^{-2}\)
4. \(10^{-3}\)

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:
1. \(0.26\) cm 
2. \(0.052\) cm 
3. \(0.52\) cm 
4. \(0.026\) cm 

A screw gauge has the least count of \(0.01~\mathrm{mm}\) 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

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 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

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:
1. \(0.521\) cm
2. \(0.525\) cm
3. \(0.053\) cm
4. \(0.529\) cm