Explanation
Given,
$$\theta =0.03316\,\,rad$$
Also $$b = \mathrm { AB } =$$ diameter of earth $$= 1.276 \times 10 ^ { 7 } \mathrm { m }$$
Now d $$=\dfrac{b}{\theta }=\dfrac{1.276\times {{10}^{7}}}{0.03316\,}=3.84\times {{10}^{8}}\text{m}$$
Hence, distance between earth and moon is $$3.84\times {{10}^{8}}\text{m}$$
Let $$X$$ be the quantity represented.
$$X = \dfrac 12 CV^2$$
$$X = \dfrac 12 \dfrac QV \times V^2$$
$$X = \dfrac12 Q.V$$
$$[X] = [Q][V]=[IT][ML^2T^{-3}I^{-1}]=[ML^2T^{-2}]$$
$$Latent\, heat (L)=\dfrac{Heat}{Mass}$$ . . . . . . (1)
The dimensional formula of mass $$=[M^1L^0T^0]$$. . . . (2)
Also, the dimensions of heat $$=$$ dimensions of energy $$=$$ dimensions of work
Since, $$work = force\times displacement$$ . . . . (3)
And, the dimensional formula of,
Displacement $$= [M^0L^1T^0] $$. . . (4)
Force $$= m \times a = [M^1L^1T^{-2}]$$ . . . (5)
On substituting equation (4) and (5) in equation (3) we get,
Work $$=[M^1L^1T^{-2}]\times [L^{1}]$$
Therefore, the dimensions of work or heat $$=[M^1L^2T^{-2}]$$ . . . . (6)
On substituting equation (2) and (6) in equation (1) we get,
Or, $$L = \dfrac{[M^1L^2T^{-2}] }{ [M^1L^0T^0]}=[M^0L^2T^{-2}]$$
Therefore, latent heat is dimensionally represented as $$[M^0L^2T^{-2}]$$
Please disable the adBlock and continue. Thank you.