Question

7.
If force (F). acceleration (a) and time (1) are
considered to be fundamental quantities, then the
dimensional formula of surface tension is
(1) [Faf
(2) (Fa-1-1]
(3) (Fa-4-2] (4) [Fa-142]​

Answer 1

$\displaystyle\large\boxed {\sf {[\sigma]=\left[F\ a^{-1}\ t^{-2}\right]}}$

We know that,

$\displaystyle\longrightarrow\sf{Surface\ Tension=\dfrac {Force}{Length}}$

$\displaystyle\longrightarrow\sf{\sigma=\dfrac {F}{L}\quad\quad\dots (1)}$

But length is equivalent to acceleration multiplied by the second power of time, since acceleration is the second derivative of displacement (length dimensionally) with respect to time, i.e.,

$\displaystyle\longrightarrow\sf{L\equiv at^2}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf{\sigma\equiv\dfrac {F}{at^2}}$

$\displaystyle\longrightarrow\sf{\sigma\equiv Fa^{-1}t^{-2}}$

Therefore the dimension of surface tension will be,

$\displaystyle\longrightarrow\sf {\underline {\underline {[\sigma]=\left[F\ a^{-1}\ t^{-2}\right]}}}$