The dimensional formula of relative density is: 

(1) ML^–3

(2) LT^–1

(3) MLT^–2

(4) Dimensionless

The dimensional formula for young's modulus is 

(1) $M{L}^{-1}{T}^{-2}$

(2) $M^{0}L{T}^{-2}$

(3) MLT^–2

(4) $M{L}^2{T}^{-2}$

Frequency is the function of density $\left(\rho \right)$, length (a) and surface tension (T). Then its value is 

1. k.$\frac{\sqrt{\mathrm{T}}}{{\rho }^{1/2}{\mathrm{a}}^{3/2}}$

2. $k{\rho }^{3/2}{a}^{3/2}/\sqrt{T}$

3. $\frac{{\mathrm{T}}^{3/2}}{{\mathrm{k\rho }}^{1/2}{\mathrm{a}}^{3/2}}$

4. $\frac{T^{3/2}}{{\mathrm{k\rho }}^{1/2}{\mathrm{a}}^{1/2}}$

The dimensions of electric potential are:

(1) $\left[M{L}^2{T}^{-2}{Q}^{-1}\right]$

(2) $\left[ML{T}^{-2}{Q}^{-1}\right]$

(3) $\left[M{L}^2{T}^{-1}Q\right]$

(4) $\left[M{L}^2{T}^{-2}Q\right]$

The dimension of $\frac{R}{L}$ are

(1) T²

(2) T

(3) T^–1

(4) T^–2

The dimensions of shear modulus are

(1) MLT^–1

(2) $M{L}^2{T}^{-2}$

(3) $M{L}^{-1}{T}^{-2}$

(4) $ML{T}^{-2}$

Pressure gradient has the same dimensions as that of:

(1) Velocity gradient

(2) Potential gradient

(3) Energy gradient

(4) None of these

If force (F), length (L) and time (T) are assumed to be fundamental units, then the dimensional formula of the mass will be

(1) $F{L}^{-1}{T}^2$

(2) $F{L}^{-1}{T}^{-2}$

(3) $F{L}^{-1}{T}^{-1}$

(4) $F{L}^2{T}^2$

In the relation, \(y=a \cos (\omega t-k x)\), the dimensional formula for \(k\) will be:
1. \( {\left[M^0 L^{-1} T^{-1}\right]} \)
2. \({\left[M^0 L T^{-1}\right]} \)
3. \( {\left[M^0 L^{-1} T^0\right]} \)
4. \({\left[M^0 L T\right]}\)

"Pascal-Second" has dimension of 

(1) Force

(2) Energy

(3) Pressure

(4) Coefficient of viscosity