Question

if force F and acceleration a and time t are considered to be fundamental quantities then the dimension formula surface tension​

Answer 1

[σ]=[F a

−1

t

−2

]

We know that,

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

Length

Force

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

L

F

…(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}⟶L≡at

2

Then (1) becomes,

\displaystyle\longrightarrow\sf{\sigma\equiv\dfrac {F}{at^2}}⟶σ≡

at

2

F

\displaystyle\longrightarrow\sf{\sigma\equiv Fa^{-1}t^{-2}}⟶σ≡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]}}}⟶

[σ]=[F a

−1

t

−2

]

Answer 2

Answer:

the energy is $FAT^{2}$

Explanation:

$Energy =F^{\alpha} A^{\beta} T^{y} \\M^{1} L^{2} T^{-2} =(MLT^{-2})^{\alpha } (LT^{-2} )^{\beta } (T)^{y} \\M^{1} L^{2} T^{-2} =M^{a} L^{a} +^{\beta } T^{-2\alpha } ^{-2 \beta } +y\\\alpha =1\\\alpha +\beta =2\implies\beta =1\\-2\alpha -2\beta +y=-2\\\implies y=2\\\implies Energy =F^{1} A^{1} T^{2}$

HOPE IT HELPS