Plancks' constant has the dimensions of

Options

A. linear momentum
B. angular momentum
C. energy
D. power

Answer

B

Explanation

`E = hv`
`rArr h =` Planck's constant `= ("Energy"(E))/("frequency"(v))`
`:. [h] = (E)/(v) = ([ML^(2)T^(-2)])/([T^(-1)]) = [ML^(2)T^(-1)]`
(a) Linear momentum = Mass `xx` velocity or `p = m xx v = [M] [LT^(-1)] = [MLT^(-1)]`
(b) Angular momentum = Moment of inertia `xx` angular velocity `[ :' I = mr^(2)]`
`:. [L] = [M] [L^(2)] [T^(-1)] = [ML^(2)T^(-1)]`
(c) Energy `[E] = [ML^(2)T^(-2)]`
(d) Power = Force `xx` velocity
or `P = F xx v`
`:. [P] = [MLT^(-2)] [LT^(-1)] = [ML^(2)T^(-3)]`
Hence, option (b) is correct.
NOTE
According to homogeneity of dimensions, the dimensions of all the terms in a physical expression should be same .e.g. in the physical expression `s = ut + 1//2 at^(2)`, the dimensions of s,ut and `1//2 at^(2)` all are same.