The dimensional formula for young's modulus is 

1. $M{L}^{-1}{T}^{-2}$

2. $M^{0}L{T}^{-2}$

3. MLT^–2

4. $M{L}^2{T}^{-2}$

The dimensions of shear modulus are

1. MLT^–1

2. $M{L}^2{T}^{-2}$

3. $M{L}^{-1}{T}^{-2}$

4. $ML{T}^{-2}$

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. \({\left[M^0 L T\right]}\)

The position of a body with acceleration \(a\) is given by \(x= Ka^{m}t^{n}\) (assume \(t\) to be time). The values of \(m\) and \(n\) will be:
1. \(m=1,~n=1\)
2. \(m=1,~n=2\)
3. \(m=2,~n=1\)
4. \(m=2,~n=2\)

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. $F{L}^2{T}^2$

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. $\left[M{L}^2{T}^{-2}Q\right]$

"Pascal-Second" has dimension of 

1. Force

2. Energy

3. Pressure

4. Coefficient of viscosity

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

1. T²

2. T

3. T^–1

4. T^–2

Pressure gradient has the same dimensions as that of:

1. Velocity gradient

2. Potential gradient

3. Energy gradient

4. None of these