# If surface tension(S), Momentent of Inertia (I) and Plank's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be: Option 1) $S^{\frac{1}{2}}I^{\frac{3}{2}}h^{-1}$ Option 2) $S^{\frac{1}{2}}I^{\frac{1}{2}}h^{-1}$ Option 3) $S^{\frac{1}{2}}I^{\frac{1}{2}}h^{0}$ Option 4) $S^{\frac{3}{2}}I^{\frac{1}{2}}h^{-1}$

Surface tension, surface energy -

$\dpi{100} M^{1}L^{0}T^{-2}$

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T =surface tension

$h^{x}T^{y}I^{z}$

$MLT^{-1}=\left ( ML^{2}T^{-1} \right )^{x}\left ( MT^{-2} \right )^{y}\left ( ML^{2} \right )^{z}$

$\left.\begin{matrix} x+y+z=1\\ x+z=\frac{1}{2} \end{matrix}\right\}y=\frac{1}{2}$

$x+2y=1\Rightarrow x+1=1 \Rightarrow x=0 \Rightarrow z=\frac{1}{2}$

$P=h^{0}I^{\frac{1}{2}}s^{\frac{1}{2}}$

Option 1)

$S^{\frac{1}{2}}I^{\frac{3}{2}}h^{-1}$

Option 2)

$S^{\frac{1}{2}}I^{\frac{1}{2}}h^{-1}$

Option 3)

$S^{\frac{1}{2}}I^{\frac{1}{2}}h^{0}$

Option 4)

$S^{\frac{3}{2}}I^{\frac{1}{2}}h^{-1}$

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