A) \[q>p\]
B) \[q<p\]
C) \[pq=1\]
D) \[{{x}^{2}}=4ay\]
Correct Answer: C
Solution :
\[\pi \] But at equilibrium \[{{h}^{2}}\] \[h\] \[{{h}^{\frac{3}{2}}}\] \[{{h}^{\frac{1}{2}}}\] \[\text{5 }\times \text{1}{{0}^{\text{8}}}\]
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