A) \[\frac{c}{g}\]
B) \[\frac{c}{p}\]
C) \[pcg\]
D) \[\frac{{{c}^{2}}}{g}\]
Correct Answer: D
Solution :
[d] \[\frac{{{c}^{2}}}{g}=\frac{{{L}^{2}}{{T}^{-2}}}{L{{T}^{-2}}}=[L]\]
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