Question

1 If force (F), velocity (V) and acceleration (A) are taken as the fundamental units instead of mass, length and time, then which of the following gives the
dimensions of Planck's constant in terms of F. V and A?
A. FV2A-1
B. F1V4 Aº
CEVA2
D. FVA​

Answer 1

The dimensions of Planck's constant are [FV³A⁻²]

Explanation:

We know that Planck's constant

$h=\frac{E}{\nu}$

Where E is energy and $\nu$ is frequency

We have to write the above in terms of Force F, velocity V and acceleration V

We know that unit of energy is same as work and work is nothing but displacement multiplied by force

Thus,

$\frac{\text{Unit of Energy}}{\text{Unit of frequency}}=\frac{\text{Unit of force}\times\text{Unit of displacement}}{\text{1/Unit of time}}$

$\implies \frac{\text{Unit of Energy}}{\text{Unit of frequency}}=F\times \text{Unit of displacement}\times\text{Unit of Time}$

$\implies \frac{\text{Unit of Energy}}{\text{Unit of frequency}}=F\times \frac{\text{Unit of velocity}\times\text{Unit of velocity}}{\text{Unit of acceleration}}\times\frac{\text{Unit of velocity}}{\text{Unit of acceleration}}$

$\implies \text{Dimensions of Planck's constant}=F\times\frac{V^2}{A}\times\frac{V}{A}$

$\implies \text{Dimensions of Planck's constant}=[FV^3A^{-2}]$

Hope this answer is helpful.

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