Physics - Unit & Dimension Question with Solution | TestHub

We should check it dimensionally

Let$\mu =K{\text{F}}^{\text{a}}{\text{M}}^{\text{b}}{\text{L}}^{\text{c}}$

$\left[\mu \right]=\left[{\text{F}}^{\text{a}}{\text{M}}^{\text{b}}{\text{L}}^{\text{c}}\right]$

$=\left[{\left({\text{MLT}}^2\right)}^{\text{a}}{\text{M}}^{\text{b}}{\text{L}}^{\text{c}}\right]$

$\left[{\text{M}}^0{\text{L}}^0{\text{T}}^{-1}\right]=\left[{\text{M}}^{\text{a}+\text{b}}{\text{L}}^{\text{a}+\text{c}}{\text{T}}^{-2\text{a}}\right]$

$\text{a}+\text{b}=0 \Rightarrow \text{b}=-\text{a}$

$\text{a}+\text{c}=0 \Rightarrow \text{c}=-\text{a}$

$-2\text{a}=-1$

$\Rightarrow \text{a}=\frac{1}{2}$

$\text{b}=-\frac{1}{2}$

$\text{c}=-\frac{1}{2}$

$\Rightarrow \mu =K{\text{F}}^{\frac{1}{2}}{\text{M}}^{-\frac{1}{2}}{\text{L}}^{-\frac{1}{2}}$

$\mu =\text{K}\sqrt{\frac{\text{F}}{\text{ML}}}$

So only option (iv) satisfies it dimensionally.