Explanation
According to the principle of dimensional homogenity
Dimension of v = dimension of at = dimension of $$\displaystyle \frac {b} {t +c} $$
Dimension of a = $$\displaystyle \frac {[v]} {[t]} = \frac {LT^{-2}} {T} = [LT^{-2}] $$
Dimension of b = $$\displaystyle [v] [t] = [LT^{-1}T] = [L] $$
Dimension of c = $$\displaystyle [t] = [T] $$
$$\textbf{Explanation:}$$
$$\bullet$$The quantity that has same unit will have same dimension representation in terms of MLTAK where M is mass, L is length, T is time , A is ampere and K is for kelvin (temperature).
$$\bullet$$Gravitational force and coulomb force are types of force which can be measured in Newton. Both have same unit. So, there dimensions will be same as well.
$$\bullet$$Similarly, coefficient of friction and number of molecules in a container both are unitless and kinetic energy of freely falling body and potential energy of compressed spring both are measure in joule. So, they have same dimensions.
$$\bullet$$But latent heat is energy required to change the state of a unit mass while specific heat is energy required to change to change unit mass by 1°. So, unit of latent heat is joule/kg while unit of specific heat is joule/°c. As unit of both quantity are different, their dimensions will be different.
$$Answer:$$
Hence, option A is the correct answer.
$$F = \dfrac {G.M.M}{L^{2}} = MLT^{-2}$$
$$G = \dfrac {MLT^{-2} L^{2}}{M^{2}} = M^{-1} L^{3} T^{-2}$$
$$F = Bqv \rightarrow F = \dfrac {\phi}{A} qv$$
$$\phi = \dfrac {F . A}{qv} = \dfrac {MLT^{-2} L^{2}}{ATLT^{-1}} = ML^{+2} A^{-1} T^{-2}$$
$$B = \mu ni; F = Bqv$$
$$F = \mu ni qv \Rightarrow \mu = \dfrac {F}{niqv}$$
$$\mu = \dfrac {MLT^{-2}}{\dfrac {1}{L} A.AT.LT^{-1}}$$ [n is no. of turns per unit length]
$$= MLA^{-2} T^{-2}$$
$$F = \dfrac {1}{4\pi \epsilon_{0}} \cdot \dfrac {q_{1} q_{2}}{r^{2}} \Rightarrow \epsilon_{0} = \dfrac {q_{1}q_{2}}{4\pi Fr^{2}}$$
$$\dfrac {AT \cdot AT}{MLT^{-2} L^{2}} = \dfrac {A^{2}T^{2}}{ML^{3} T^{-2}} = M^{-1} L^{-3} A^{2} T^{4}$$
$$\left [C\right ] = \left [\dfrac {Q}{V}\right ] = \left [\dfrac {Q^{2}}{W}\right ] = \left [M^{-1} L^{-2} T^{2} Q^{2}\right ]$$
Impedance is same as resistance but in ac circuit
$$\therefore$$ Dimension of impedance $$= \dfrac {\text {dimension of voltage}}{\text {dimension of current}}$$ $$= \dfrac {\left [V\right ]}{\left [I\right ]} = \dfrac {\left [ML^{2} T^{-3} I^{-1}\right ]}{I} = \left [ML^{2} T^{-3} I^{-2}\right ]$$
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