[1990]
(a) $[MLT^{-2}]$
(b) $[ML^{-1} T^{2}]$
(c) $[ML^{-1} T^{-2}]$
(d) $[MLT^{2}]$
Correct Answer: 28.(c)
Solution:
So dimensional formula
$ =\frac{[MLT^{-2}]}{[L^{2}]}=[ML^{-1} T^{-2}] $
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