Question

The momentum [P], area[A] and time [T] are taken as fundamental quantity then the dimensional formula of work will be.

Answer 1

Given : The momentum [P], area[A] and time [T] are taken as fundamental quantity.

$\\:\implies \: \rm [W] = k. [p]^{a} [A]^{b} [T]^{c} ---(i)$

k is a dimensionaless constant.

• Dimensions of work = $\rm [M^{1} L^{2} T^{-2} ]$

• Dimensions of momentum = $\rm [M^{1} L^{1} T^{-1} ]$

• Dimensions of Area = $\rm [M^{0} L^{2} T^{0} ]$

• Dimensions of Time = $\rm [M^{0} L^{0} T^{1} ]$

$\\\longrightarrow\rm [M^{1} L^{2} T^{-2} ]= [M^{1} L^{1} T^{-1} ]^{a} [M^{0} L^{2} T^{0} ]^{b} [M^{0} L^{0} T^{1} ]^{c}$

$\\\longrightarrow\rm [M^{1} L^{2} T^{-2} ]= [M^{a} L^{a} T^{-a} ] [M^{0} L^{2b} T^{0} ] [M^{0} L^{0} T^{c} ]$

$\\\longrightarrow\rm [M^{1} L^{2}L^{-2}] = [M^{(a)} L^{(a+2b)} T^{(-a+c)} ]$

Now,

$\\\mapsto \: a = 1 ---(ii)$

$\\\mapsto 2 = a+2b$

$\\\mapsto 2 = 1+2b$

$\\\mapsto b = \dfrac{1}{2} ---(iii)$

$\\\mapsto -a+c=-2$

$\\\mapsto -1+c=-2$

$\\\mapsto c=-1---(iv)$

substituting the values of eq (ii), (iii) and (iv) in eq (i)

$\\\dashrightarrow\bf [W] = k. p^{1} A^{\frac{1}{2}} t^{-1} }$

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Answer 2

Answer:

The momentum [P], area[A] and time [T] are taken as fundamental quantity:

$\footnotesize\longrightarrow [W] \propto [P]^{\alpha} \: ...(i)$

$\footnotesize\longrightarrow [W] \propto [A]^{\beta} \: ...(ii)$

$\footnotesize\longrightarrow [W] \propto [T]^{\gamma} \: ...(iii)$

From equation (I), (II) and (III):

$\footnotesize\longrightarrow [W] \propto [P]^{\alpha}[A]^{\beta} [T]^{\gamma}$

$\footnotesize\longrightarrow [W] = k [P]^{\alpha}[A]^{\beta} [T]^{\gamma}$

Here, K is the proportionality constant.

Dimension of Work:

$\footnotesize\longrightarrow [W] = [M^1L^2T^{-2}]$

Dimension of Momentum

$\footnotesize\longrightarrow [P] = [M^1L^1T^{-1}]$

Dimension of Area:

$\footnotesize\longrightarrow [A] = [M^0L^2T^{0}]$

Dimension of Time:

$\footnotesize\longrightarrow [T] = [M^0L^0T^{1}]$

Now,

$\footnotesize\longrightarrow [M^1L^2T^{-2}] = [M^1L^1T^{-1}]^{\alpha}[M^0L^2T^{0}]^{\beta} [M^0L^0T^{1}]^{\gamma}$

$\footnotesize\longrightarrow [M^1L^2T^{-2}] = [M]^{\alpha}[L]^{ (\alpha + 2\beta)} [T]^{( - \alpha + \gamma)}$

On comparing the powers of both LHS and RHS:

$\footnotesize\longrightarrow \red{\rm \alpha = 1}$

$\footnotesize\longrightarrow \alpha + 2 \beta = 2$

$\footnotesize\longrightarrow 1 + 2 \beta = 2$

$\footnotesize\longrightarrow 2 \beta = 2 - 1$

$\footnotesize\longrightarrow 2 \beta = 1$

$\footnotesize\longrightarrow \red{ \rm\beta = \dfrac{1}{2} }$

$\footnotesize\longrightarrow - \alpha + \gamma = - 2$

$\footnotesize\longrightarrow - 1 + \gamma = - 2$

$\footnotesize\longrightarrow \gamma = - 2 + 1$

$\footnotesize\longrightarrow \red{ \rm \gamma = -1}$

Therefore,

$\footnotesize\longrightarrow \underline{\boxed{\orange{ \bf [W] = [P]^{1}[A]^{\frac{1}{2}} [T]^{-1}}}}$