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Step-by-Step Solution
Step 1: Express mass in terms of T, C, and h
Let the dimension of mass (M) be expressed as
$M \propto T^x C^y h^z$.
We want to find the exponents $x$, $y$, and $z$.
Step 2: Write the known dimensions of T, C, and h
Time (T): Dimension is $T^1$. In base units: $[T] = T^1$.
Velocity (C): Dimension is length per time, i.e., $L^1T^{-1}$.
Angular momentum (h): Dimension is $ML^2T^{-1}$
(since angular momentum = moment of inertia × angular velocity = $MR^2 × T^{-1}$).
Therefore, when we write $T^x C^y h^z$ in base dimensions (M, L, T), we get:
$T^x \times (L T^{-1})^y \times (M L^2 T^{-1})^z.$
Step 3: Combine dimensions in terms of M, L, and T
Collecting like terms, the expression
$T^x \times (L T^{-1})^y \times (M L^2 T^{-1})^z$
becomes
$M^z \, L^{\,y + 2z} \, T^{\,x - y - z}.$
Step 4: Match dimensions to that of mass
The dimension of mass is $M^1 L^0 T^0$. Equating powers of M, L, and T, we get:
Power of $M$: $z = 1$
Power of $L$: $y + 2z = 0$
Power of $T$: $x - y - z = 0$
Step 5: Solve the system of equations
From $z = 1$, substitute into the other equations:
$y + 2(1) = 0 \implies y + 2 = 0 \implies y = -2.$
$x - (-2) - 1 = 0 \implies x + 2 - 1 = 0 \implies x = -1.$
Step 6: Write the final dimensional form of mass
Hence,
$M = T^{-1} C^{-2} h^{1},$
or simply
$[M] = [T^{-1} C^{-2} h].$
