Question

Write the dimensions of $a$ and $b$ in the relation: $P=\frac{b-x^2}{at}$
Where $P$ is power, $x$ is distance and $t$ is time.

• A. $\left[M^0{L}^2{T}^0\right],\left[M^{-1}{L}^0{T}^2\right]$
• B. $\left[M^0{L}^2{T}^0\right],\left[M^{-1}{L}^1{T}^2\right]$
• C. $\left[M^{-1}{L}^1{T}^2\right],\left[M^0{L}^2{T}^0\right]$
• D. $\left[M^{-1}{L}^0{T}^2\right],\left[M^0{L}^2{T}^0\right]$

Answer 1

Answer: (D)

Dimensions of Power: $P-M{L}^2{T}^{-3}$

Since the given expression is dimensionally correct, each term of the expression must have same dimensions as that of power.

Therefore,$\frac{\left[x^2\right]}{\left[a\right]\left[t\right]}=\frac{L^2}{\left[a\right]T}=M{L}^2{T}^{-3}$

$\Rightarrow \left[a\right]-M^{-1}{L}^0{T}^2$

$\frac{\left[b\right]}{\left[a\right]\left[t\right]}=\frac{\left[b\right]}{\left(M^{-1}{L}^0{T}^2\right)\left(T\right)}=M{L}^2{T}^{-3}$

$\Rightarrow \left[b\right]-L^2$

Hence, Option A is correct.