Question

If time (t), velocity (v), and angular momentum (l) are taken as the fundamental units. Then the dimension of mass (m) in terms of t, v and l is :

• A. \([t^{-1}v^{1}l^{-2}]\)
• B. \([t^{1}v^{2}l^{-1}]\)
• C. \([t^{-2}v^{-1}l^{1}]\)
• D. \([t^{-1}v^{-2}l^{1}]\)

Hint:

Write \(m = t^a v^b l^c\) and equate the powers of M, L and T on both sides.

Answer 1

The Correct Option is D

Let \([m] = [t]^a[v]^b[l]^c\), where \([t] = T\), \([v] = LT^{-1}\), \([l] = ML^2T^{-1}\) (angular momentum).
\(M^1L^0T^0 = T^a (LT^{-1})^b (ML^2T^{-1})^c = M^{c}L^{b+2c}T^{a-b-c}\).
M: \(c = 1\). L: \(b + 2c = 0\Rightarrow b = -2\). T: \(a - b - c = 0\Rightarrow a = b + c = -1\).
So \([m] = [t^{-1}v^{-2}l^{1}]\).
Answer: (4) \([t^{-1}v^{-2}l^{1}]\)