A) Planck's constant
B) gravitational constant
C) dielectric constant
D) None of these
Correct Answer: D
Solution :
Dimensions formula of \[{{(velocity)}^{2}}\div radius\] \[=\frac{{{[{{M}^{0}}L{{T}^{-1}}]}^{2}}}{[{{M}^{0}}L{{T}^{0}}]}=\] accelerationYou need to login to perform this action.
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