Question

If surface tension (S), Moment of inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be : -

• A. $S^{\frac{3}{2}} I^{\frac1{2}} h^0$
• B. $S^{\frac1{2}} I^{\frac1{2}} h^0$
• C. $S^{\frac1{2}} I^{\frac1{2}} h^{-1}$
• D. $S^{\frac1{2}} I^{\frac{3}{2}} h^{-1}$

Answer 1

Answer: (B)

$S^{\frac1{2}} I^{\frac1{2}} h^0$

Answer 2

$p =k s^{a}I^{ b}h^{c}$

where $k$ is dimensionless constant
$MLT^{-1} = \left(MT^{-2}\right)^{a}\left(ML^{2}\right)^{b} \left(ML^{2}T^{-1}\right)^{c}$

$a + b + c = 1$

$2 b + 2c = 1$

$-2a - c = -1$

$a = \frac{1}{2} \; \; b = \frac{1}{2} \; \; c = 0$

$\therefore \,S^{\frac1{2}} I^{\frac1{2}} h^0$

Hence, Correct answer is option (B) : $S^{\frac1{2}} I^{\frac1{2}} h^0$.