Correct option is (B)

To determine the dimensions of $\frac{b^2}{a}$, let's start by identifying the dimensions of each term in the equation $F=a{x}^2+b{t}^{\frac{1}{2}}$, where $F$ represents force, $x$ represents distance, and $t$ represents time.

The dimension of force ($F$) is given by [MLT^-2], where $M$ is mass, $L$ is length, and $T$ is time.

The term $a{x}^2$ has the same dimension as force, so:

$[a]=\frac{[F]}{[x{]}^2}=\frac{ML{T}^{-2}}{L^2}=M{L}^{-1}{T}^{-2}$

The term $b{t}^{\frac{1}{2}}$ also has the same dimension as force, which gives:

$[b]=\frac{[F]}{[t{]}^{\frac{1}{2}}}=\frac{ML{T}^{-2}}{T^{\frac{1}{2}}}=ML{T}^{-\frac{5}{2}}$

Now, to find $\frac{b^2}{a}$, we substitute the dimensions of $b$ and $a$:

$\left[\frac{b^2}{a}\right]=\frac{\left(M^2{L}^2{T}^{-5}\right)}{\left(M{L}^{-1}{T}^{-2}\right)}=[M^1{L}^3{T}^{-3}]$

Therefore, the dimensions of $\frac{b^2}{a}$ are $\left[{\mathrm{ML}}^3{\mathrm{T}}^{-3}\right]$, which corresponds to mass times length cubed per time cubed.

F

Correct option is (D)

$\begin{array}{rl}& F=\eta A\frac{dv}{dy}\\ & \left[ML{T}^{-2}\right]=\eta \left[L^2\right]\left[T^{-1}\right]\\ & \eta =\left[M{L}^{-1}{T}^{-1}\right]\\ & S.T=\frac{F}{\ell }=\frac{\left[ML{T}^{-2}\right]}{[L]}=\left[M{L}^0{T}^{-2}\right]\\ & L=mvr=\left[M{L}^2{T}^{-1}\right]\\ & K.E=\frac{1}{2}I{\omega }^2=\left[M{L}^2{T}^{-2}\right]\end{array}$

Correct option is (A)

$\begin{array}{rl}& [\mathrm{h}]={\mathrm{ML}}^2{\mathrm{T}}^{-1}\\ & [\mathrm{L}]={\mathrm{ML}}^2{\mathrm{T}}^{-1}\\ & [\mathrm{P}]={\mathrm{MLT}}^{-1}\\ & [\tau ]={\mathrm{ML}}^2{\mathrm{T}}^{-2}\end{array}$

(Here $\mathrm{h}$ is Planck's constant, $\mathrm{L}$ is angular momentum, $\mathrm{P}$ is linear momentum and $\tau$ is moment of force)