The position of a particle at time \(t\) is given by the relation \({x}({t})=\left(\frac{{v}_0}{\alpha}\right)\left(1-{e}^{-\alpha {t}}\right)\), where \(v_0\)${\mathrm{}}_{}$ is a constant and \(\alpha >0\). The dimensions of \(v_0\) and \(\alpha\) are respectively:
1. \(\left[M^0L^{1}T^{-1}\right]~\text{and}~\left[T^{-1}\right]\)
2. \(\left[M^0L^{1}T^{0}\right]~\text{and}~\left[T^{-1}\right]\)
3. \(\left[M^0L^{1}T^{-1}\right]~\text{and}~\left[LT^{-1}\right]\)
4. \(\left[M^0L^{1}T^{-1}\right]~\text{and}~\left[T\right]\)

If force (\(F\)), velocity (\(\mathrm{v}\)), and time (\(T\)) are taken as fundamental units, the dimensions of mass will be:

If dimensions of critical velocity \({v_c}\) of a liquid flowing through a tube are expressed as \(\eta^{x}\rho^yr^{z}\), where \(\eta, \rho~\text{and}~r\) are the coefficient of viscosity of the liquid, the density of the liquid, and the radius of the tube respectively, then the values of \({x},\) \({y},\) and \({z},\) respectively, will be:

Planck's constant (\(h\)), speed of light in the vacuum (\(c\)), and Newton's gravitational constant (\(G\)) are the three fundamental constants. Which of the following combinations of these has the dimension of length?

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. \(12.4~\text{mm}\)

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 certain vernier callipers, \(25\) divisions on the vernier scale have the same length as \(24\) divisions on the main scale. One division on the main scale is \(1\) mm long. The least count of the instrument is:

In a vernier calliper, \(N\) divisions of vernier scale coincide with (\(N\text-1\)) divisions of the main scale (in which the length of one division is \(1\) mm). The least count of the instrument should be:
1. \(N~\text{mm}\)
2. \((N-1)~\text{mm}\)
3. \(\frac{1}{10N}~\text{cm}\)
4. \(\frac{1}{(N-1)}~\text{mm}\)

For the expression, \(10^{(at+3)}\), the dimensions of \(a\) will be:
1. \(\left[M^0L^0T^{0}\right]\)
2. \(\left[M^0L^0T^{1}\right]\)
3. \(\left[M^0L^0T^{-1}\right]\)
4. None of these