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

The determination of the value of acceleration due to gravity (g) by simple pendulum method employs the formula,
  \(g=4\pi^2\frac{L}{T^2}\)
The expression for the relative error in the value of 'g' is:
1.  \(\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\Big(\frac{\Delta T}{T}\Big)\)
2.  \(\frac{\Delta g}{g}=4\pi^2\Big[\frac{\Delta L}{L}-2\frac{\Delta T}{T}\Big]\)
3.  \(\frac{\Delta g}{g}=4\pi^2\Big[\frac{\Delta L}{L}+2\frac{\Delta T}{T}\Big]\)
4.  \(\frac{\Delta g}{g}=\frac{\Delta L}{L}-2\Big(\frac{\Delta T}{T}\Big)\)

A Screw Guage gives the following readings when used to measure the diameter of a wire.
Main scale reading = 0.0 mm
Circular scale reading = 52 divisions
Given that: 1 mm on the main scale corresponds to 100 divisions of the circular scale.
The diameter of the wire from the above data is:

1.  0.026 cm

2.  0.005 cm

3.  0.52 cm

4.  0.052 cm

If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities, then find the dimensions of energy.

1. [F][A][T^-1]

2. [F][A^-1][T]

3. [F][A][T]

4. [F][A][T²]

If E and G, respectively, denote energy and gravitational constant, then $\frac{\mathrm{E}}{\mathrm{G}}$ has the dimensions of:

1. $\left[\mathrm{M}\right]\left[{\mathrm{L}}^0\right]\left[{\mathrm{T}}^0\right]$

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

3. $\left[{\mathrm{M}}^2\right]\left[{\mathrm{L}}^{-1}\right]\left[{\mathrm{T}}^0\right]$

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

A screw gauge has least count of 0.01 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