Question

if force F speed V and mass M are taken as fundamental quantities then dimensional formula of length is? ​

Answer 1

Explanation:

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Answer 2

Given:

Force (F) , speed (V) and mass (M) are taken as fundamental quantities .

To find:

Dimensional formula of length (L)

Concept:

In case of a Dimensional analysis of this question , we have to represent the given fundamental quantities in such a manner that the net dimension becomes equal to that of length.

Calculation:

Let length (L) be represented in the following way :

$\therefore \: \: L \propto \: {F}^{x} \times {V}^{y} \times {M}^{z}$

Now , representing each of the fundamental quantities in form of mass, length and time :

$= > \: \: L \propto \: { \{ML {T}^{ - 2} \} }^{x} \times { \{L {T}^{ - 1} \} }^{y} \times {M}^{z}$

$= > L \propto \: {M}^{(x + z)} \times {L}^{(x + y)} \times {T}^{( - 2x - y)}$

Now, comparing both sides of the equation , we get the following :

$1) \: x + y = 1 \\ 2) \: x + z = 0 \\ 3) \: - 2x - y = 0$

After solving the 3 equations , we have:

$1) \: x = - 1 \\ 2) \: y = 2 \\ 3) \: z = 1$

Now putting the values of x , y , z :

$\therefore \: \: L \propto \: {F}^{( - 1)} \times {V}^{2} \times {M}^{1}$

$= > \: \: L \propto \: \dfrac{ {V}^{2} \times {M}^{1} }{F}$

Considering a dimensionless constant k ;

$= > \: \: L = k \bigg \{ \: \dfrac{ {V}^{2} \times {M}^{1} }{F} \bigg \}$

So the final answer is :

$\boxed{ \red{ \huge{ \bold{ L = k \bigg \{ \: \dfrac{ {V}^{2} \times {M}^{1} }{F} \bigg \}}}}}$