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Step-by-Step Detailed Solution
Step 1: Express the given relation for wave speed
The question states that the speed of a wave in water is given by:
$v = \lambda^{a}\, g^{b}\, \rho^{c}$.
Here, $v$ is the speed of the wave, $\lambda$ is the wavelength, $g$ is the acceleration due to gravity, and $\rho$ is the density of water.
Step 2: Write the dimensional formulas for each physical quantity
Dimension of speed ($v$): $[M^0\,L^1\,T^{-1}]$
Dimension of wavelength ($\lambda$): $[L]$
Dimension of acceleration due to gravity ($g$): $[L\,T^{-2}]$
Dimension of density ($\rho$): $[M\,L^{-3}]$
Step 3: Set up the dimensional equation
According to dimensional analysis, the dimensions on both sides of the equation must match, so we write:
$[M^0 L^1 T^{-1}] = [L]^a \times [L\,T^{-2}]^b \times [M\,L^{-3}]^c$.
Step 4: Combine dimensions on the right-hand side
Combine the dimensional factors of each term on the right:
$[L]^a = [L^a]$
$[L\,T^{-2}]^b = [L^b\,T^{-2b}]$
$[M\,L^{-3}]^c = [M^c\,L^{-3c}]$
When multiplied together, the resulting dimension on the right side becomes:
$[M^c\,L^{(a + b - 3c)}\,T^{-2b}]$.
Step 5: Match exponents of $M$, $L$, and $T$
Match each fundamental unit's exponent from the left-hand side ($M^0 L^1 T^{-1}$) to the right-hand side ($M^c\,L^{a+b-3c}\,T^{-2b}$):
Matching powers of $M$: $0 = c$, so $c = 0$.
Matching powers of $L$: $1 = a + b - 3c$. Substituting $c = 0$ gives $1 = a + b$.
Matching powers of $T$: $-1 = -2b$, which gives $b = \tfrac{1}{2}$.
Step 6: Solve for $a$
From $a + b = 1$ and $b = \tfrac{1}{2}$, we get:
$a + \tfrac{1}{2} = 1 \;\; \Rightarrow \;\; a = \tfrac{1}{2}$.
Final Values
The exponents are:
$a = \tfrac{1}{2},\quad b = \tfrac{1}{2},\quad c = 0.$
Thus, the correct combination is $\left(\tfrac{1}{2},\,\tfrac{1}{2},\,0\right)$.
