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(D) The $SI$ unit of length is $1 \ m$.Given that the new unit of length is $x \ m$.Therefore,$1 \ m = \frac{1}{x}$ new units.We need to express the a

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$\frac{1}{x^2}$

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(D) The $SI$ unit of length is $1 \ m$.Given that the new unit of length is $x \ m$.Therefore,$1 \ m = \frac{1}{x}$ new units.We need to express the area of $1 \ m^2$ in terms of the new unit.Area $= 1 \ m^2 = 1 \ m 	imes 1 \ m$.Substituting the value of $1 \ m$ in terms of the new unit:Area $= (\frac{1}{x} 	ext{ new units}) 	imes (\frac{1}{x} 	ext{ new units}) = \frac{1}{x^2} 	ext{ (new units)}^2$.Thus,the magnitude is $\frac{1}{x^2}$.

The $SI$ unit of length is 'metre'. Suppose we adopt a new unit of length which equals $x$ metre. Then,the area of $1 \ m^2$ expressed in terms of the new unit has a magnitude of:

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