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

The Correct Option is B

\(p =k s^{a}I^{ b}h^{c}\) 

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) : \(S^{\frac1{2}} I^{\frac1{2}} h^0\).