The Dimension o f Young Modulus E is shown as;
"\\frac{N}{m^2}=\\frac{kgm}{s^2m^2}=\\frac{kg}{s^2m}"
This means that;
"(\\frac{kg}{s^2m})^\\alpha (\\frac{kg}{m^3})\\beta (m)^\\gamma=\\frac{m}{s}"
Hence; "\\alpha=0.5=-\\beta and \\gamma=0"
Consequently;
"v=\\sqrt\\frac{E}{\\rho}"
i.e "\\rho" is the Density
Answer: "v=\\sqrt\\frac{E}{\\rho}"
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