Amount of solar energy received on the earth's surface per unit area per unit time is defined a solar constant. Dimension of solar constant is:

We begin with the verbal definition: the solar constant is “the amount of solar energy received on the earth’s surface per unit area per unit time.” In symbols, if we denote energy by $$E$$, area by $$A$$ and time by $$t$$, then the solar constant $$S$$ can be written as

$$S=\dfrac{E}{A \, t}.$$

Now we translate each physical quantity into its fundamental dimensions using the M L T system.

• Energy has the dimensional formula of work. By definition of work, $$ \text{Work} = \text{Force}\times\text{Displacement}. $$ First, force is mass times acceleration, and acceleration is length per time squared, so $$ \text{Force} \;:\; M\,L\,T^{-2}. $$ Multiplying by displacement (length $$L$$) gives $$ \text{Energy (or Work)} \;:\; M\,L^{2}\,T^{-2}. $$

• Area is length squared, so $$ A \;:\; L^{2}. $$

• Time is simply $$ t \;:\; T. $$

Substituting these dimensional forms into the expression for $$S$$, we obtain

$$ \begin{aligned} [S] &= \dfrac{[E]}{[A]\,[t]} \\[4pt] &= \dfrac{M\,L^{2}\,T^{-2}}{L^{2}\,T}. \end{aligned} $$

Now we cancel common powers of length and add the time exponents:

• The $$L^{2}$$ in the numerator and the $$L^{2}$$ in the denominator cancel completely, leaving $$L^{0}$$.

• For time, we have $$T^{-2}$$ in the numerator and $$T^{+1}$$ in the denominator. Subtracting the exponents,

$$T^{-2-1}=T^{-3}.$$

Hence the reduced dimensional formula becomes

$$[S] = M\,L^{0}\,T^{-3}.$$

This matches Option B in the given list.

Hence, the correct answer is Option B.