Question

If force (F), momentum (P) and mass (M) are fundamental quantities in system of units, then dimensional formula for length (L) and time (T) will be​

Answer 1

Answer:

Hey here is ur answer ;-

LαF^aM^bT^c

.......

.......

α = Alpha

Hope this will help you....

Have a Nice Day

Answer 2

Given:

Force (F), momentum (P) and mass (M) are fundamental quantities in system of units.

To find:

Dimensional formula for length (L) and time (T) will be ?

Calculation:

Let length be expression as follows :

$\rm \therefore \: L \propto \: {F}^{x} \: {P}^{y} \: {M}^{z}$

$\rm \implies \: L \propto \: { \bigg(ML{T}^{ - 2} \bigg)}^{x} \: { \bigg(ML{T}^{ - 1} \bigg)}^{y} \: { \bigg(M \bigg)}^{z}$

$\rm \implies \: L \propto \: {M}^{(x + y + z)} \: {L}^{(x + y)} \: {T}^{( - 2x - y)}$

So, comparing the sides , following equations can be said :

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

Solving these equations, we get :

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

So, dimensional formula for length is :

$\boxed{ \bold{ \therefore \: L \propto \: {F}^{ \frac{ - 1}{2} } \: {P}^{ \frac{3}{2} } \: {M}^{ - 1} }}$

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Let time be written as :

$\rm \therefore \: T\propto \: {F}^{x} \: {P}^{y} \: {M}^{z}$

$\rm \implies \: T \propto \: { \bigg(ML{T}^{ - 2} \bigg)}^{x} \: { \bigg(ML{T}^{ - 1} \bigg)}^{y} \: { \bigg(M \bigg)}^{z}$

$\rm \implies \: T \propto \: {M}^{(x + y + z)} \: {L}^{(x + y)} \: {T}^{( - 2x - y)}$

So, after comparing both sides , we get :

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

After solving, we get :

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

So, dimensional formula becomes:

$\boxed{ \bold{ \therefore \: L \propto \: {F}^{ - 1} \: {P}^{ 1 } \: {M}^{ 0} }}$

Hope It Helps.