Given that K = energy, V = velocity, T = time. If they are chosen as fundamental units, then dimensi

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1.Units, Dimensions and Measurement

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Given that $K =$ energy, $V =$ velocity, $T =$ time. If they are chosen as the fundamental units, then what is dimensional formula for surface tension?

A$[K\,{V^{ - 2}}{T^{ - 2}}]$

B$[K^2\,{V^2}{T^{ - 2}}]$

C$[K^2\,{V^{ - 2}}{T^{ - 2}}]$

D$[K\,{V^2}{T^2}]$

Solution

$\begin{array}{l}
Surface\,tension\, = \frac{F}{\ell } = \frac{F}{\ell } \cdot \frac{\ell }{\ell } \cdot \frac{{{T^2}}}{{{T^2}}}\\
\left( {As,F.\ell = K(energy);\frac{{{T^2}}}{{{\ell ^2}}} = {V^{ – 2}}} \right)\\
Therefore,\,surface\,tension = \left[ {K{V^{ – 2}}{T^{ – 2}}} \right]
\end{array}$

Standard 11

Physics

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