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. \(7.65\)

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. \(12.4~\text{mm}\)

The main scale of a vernier callipers reads 10 mm in 10 divisions. 10 divisions of Vernier scale coincide with 9 divisions of the main scale. When the two jaws of the callipers touch each other, the fifth division of the vernier coincides with 9 main scale divisions and the zero of the vernier is to the right of zero of main scale. When a cylinder is tightly placed between the two jaws, the zero of vernier scale lies slightly behind 3.2 cm and the fourth vernier division coincides with a main scale division. The diameter of the cylinder is.

(1)  3.10 cm

(2)  3.8 cm

(3)  3.09 cm

(4)  -3.09 cm

The dimensions of ${\left({\mathrm{\mu }}_0{\mathrm{\epsilon }}_0\right)}^{-1/2}$ are:

1.  $\left[{\mathrm{L}}^{-1}\right]$ $\mathrm{T}$

2.  $\left[{\mathrm{LT}}^{-1}\right]$

3.  $\left[{\mathrm{L}}^{-1/2}\right]$ ${\mathrm{T}}^{1/2}$

4.  $\left[{\mathrm{L}}^{-1/2}\right]$ ${\mathrm{T}}^{-1/2}$

Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, length L, time T, and current I, would be:

1. $\left[{\mathrm{ML}}^2{\mathrm{T}}^{-3}{\mathrm{I}}^{-1}\right]$

2. $\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]$

3. $\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}{\mathrm{I}}^{-1}\right]$

4. $\left[{\mathrm{ML}}^2{\mathrm{T}}^{-3}{\mathrm{I}}^{-2}\right]$