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Step-by-Step Solution
Step 1: Identify the Physical Quantities and Their Dimensions
• Linear momentum $p$ has the dimension: $[ M L T^{-1} ]$.
• Surface tension $S$ has the dimension: $[ M T^{-2} ]$.
• Moment of inertia $I$ has the dimension: $[ M L^{2} ]$.
• Planck’s constant $h$ has the dimension: $[ M L^{2} T^{-1} ]$.
Step 2: Express Linear Momentum in Terms of the Fundamental Units
Assume the linear momentum $p$ is proportional to a product of powers of $S$, $I$, and $h$:
$ p \propto S^{a} \, I^{b} \, h^{c}.$
Step 3: Write the Dimensional Equation
Taking dimensions on both sides, we get:
$[ M L T^{-1} ] = [ M T^{-2} ]^{a} \, [ M L^{2} ]^{b} \, [ M L^{2} T^{-1} ]^{c}.$
Step 4: Combine the Exponents of M, L, and T
On the right-hand side:
For M: $M^{\,a + b + c}$
For L: $L^{\,2b + 2c}$
For T: $T^{-2a - c}$
Hence, the dimensional expression is
$ M^{\,a + b + c} \, L^{\,2b + 2c} \, T^{-2a - c} = M^{\,1} \, L^{\,1} \, T^{-1}.$
Step 5: Equate the Exponents of M, L, and T
We get three simultaneous equations:
$a + b + c = 1 \quad$ (from exponent of $M$)
$2b + 2c = 1 \quad$ (from exponent of $L$)
$-2a - c = -1 \quad$ (from exponent of $T$)
Step 6: Solve for a, b, and c
From $2b + 2c = 1$, we get $b + c = \tfrac{1}{2}.$
From $-2a - c = -1$, we get $2a + c = 1.$
Now use $a + b + c = 1$ along with the two equations above to solve for $a$, $b$, and $c$:
$a = \tfrac{1}{2},$
$b = \tfrac{1}{2},$
$c = 0.$
Step 7: Substitute the Values Back
Substitute $a = \tfrac{1}{2}$, $b = \tfrac{1}{2}$, and $c = 0$ into
$p \propto S^{a} \, I^{b} \, h^{c}$:
$ p \propto S^{\tfrac{1}{2}} \, I^{\tfrac{1}{2}} \, h^{0} = S^{\tfrac{1}{2}} \, I^{\tfrac{1}{2}}.$
Thus, the required dimensional formula for linear momentum is
$ S^{\tfrac{1}{2}} \, I^{\tfrac{1}{2}} \, h^{0}.$
