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  • Solve! 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:

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}

Answers (1)

best_answer

 

Surface tension, surface energy -

M^{1}L^{0}T^{-2}

-

 

 

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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