Plancks constant has the dimensions as that of

Power

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Energy

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Angular Momentum

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Linear Momentum

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Solution

The correct option is C
Angular Momentum

Explanation for correct option
Step 1: Dimension of Planck constant
Energy can be expressed as, $E = m c^{2} [w h e r e m = m a s s, c = v e l o c i t y o f l i g h t]$
Dimension of mass= $[M^{1} L^{0} T^{0}]$
Dimension of velocity= $[M^{0} L^{1} T^{- 1}]$
Dimension of energy= $[M^{1} L^{0} T^{0}] \times {[M^{0} L^{1} T^{- 1}]}^{2} = [M^{1} L^{0} T^{0}] \times [M^{0} L^{2} T^{- 2}] = [M^{1} L^{2} T^{- 2}]$
Now, frequency is the number of complete wave cycles formed in a unit of time given by, $v = \frac{1}{T} [w h e r e v = f r e q u e n c y o f r a d i a t i o n, T = t i m e p e r i o d]$
Dimension of frequency $v = [M^{0} L^{0} T^{- 1}]$
Planck's constant $h = \frac{E n e r g y}{f r e q u e n c y}$
Dimension of $h = \frac{[M^{1} L^{2} T^{- 2}]}{[M^{0} L^{0} T^{- 1}]} = [M^{1} L^{2} T^{- 1}]$
Step 2: In case of option C
If an object is accelerating around a fixed point, then it also possesses angular momentum.
Angular momentum can be expressed as $L = r p = m v r [w h e r e r = r a d i u s, p = l i n e a r m o m e n t u m]$
Dimension of angular momentum = $[M^{1} L^{2} T^{- 1}]$
So, the dimension of Planck’s constant is the same as the dimension of angular momentum.
The explanation of incorrect options
In case of option A
Power is the rate of doing an activity or work in the minimum possible time. It is the amount of energy transferred or converted per unit time where large power means a large amount of work or energy.
$P = \frac{w o r k}{t i m e}$
$W o r k = F o r c e \times d i s p l a c e m e n t w o r k = (m \times a) \times d i s p l a c e m e n t [A s, F o r c e = m \times a] w o r k = [M^{1} L^{1} T^{- 2}] \times [L]$
Dimension of work = $[M^{1} L^{2} T^{- 2}]$
Dimesion of power = $\frac{[M^{1} L^{2} T^{- 2}]}{[T]} = [M^{1} L^{2} T^{- 3}]$
In case of option B
Energy is equivalent to work done. And as we have discussed above dimension of work = $[M^{1} L^{2} T^{- 2}]$
In case of option D
Linear momentum is defined as the product of the mass (m) of an object and the velocity (v) of the object. This relationship can be described in the form of an equation.
$L i n e a r M o m e n t u m = m v [w h e r e m = m a s s, v = v e l o c t i y]$
Dimension of mass= $[M]$
Dimension of velocity = $\frac{l e n g t h}{t i m e} = [L T^{- 1}]$
Hence, dimension of linear momentum= $[M^{1} L^{1} T^{- 1}]$ .
Thus, option (c) is correct.

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