The speed of a wave produced in water is given by v=\(λ^ag^bρ^c\). Where λ,g and ρ are wavelength of wave, acceleration due to gravity and density of water respectively. The values of a, b and c respectively, are:
- 1,1,0
- \(\frac{1}{2},0,\frac{1}{2}\)
- \(\frac{1}{2},\frac{1}{2},0\)
- 1,-1,0
The Correct Option is C
Solution and Explanation
Understanding the Problem
We are given that the speed of a wave (\(v\)) depends on wavelength (\(\lambda\)), acceleration due to gravity (\(g\)), and density (\(\rho\)). We need to find the exponents \(a\), \(b\), and \(c\) in the equation:
\( v = \lambda^a g^b \rho^c \)
Dimensional Analysis
1. Dimensional Formulas:
- \( [v] = [L T^{-1}] \) (velocity)
- \( [\lambda] = [L] \) (wavelength)
- \( [g] = [L T^{-2}] \) (acceleration)
- \( [\rho] = [M L^{-3}] \) (density)
2. Substitute into the Equation:
\( [L T^{-1}] = [L]^a [L T^{-2}]^b [M L^{-3}]^c \)
3. Simplify:
\( [L T^{-1}] = [L^{a+b-3c} M^c T^{-2b}] \)
4. Equate Powers:
- \( a + b - 3c = 1 \) (from \(L\))
- \( -2b = -1 \) (from \(T\))
- \( c = 0 \) (from \(M\))
Solving the Equations
1. From the \(T\) equation:
\( -2b = -1 \)
\( b = \frac{1}{2} \)
2. From the \(M\) equation:
\( c = 0 \)
3. Substitute \(b\) and \(c\) into the \(L\) equation:
\( a + \frac{1}{2} - 3(0) = 1 \)
\( a + \frac{1}{2} = 1 \)
\( a = 1 - \frac{1}{2} = \frac{1}{2} \)
Final Answer
The values of \(a\), \(b\), and \(c\) are:
- \( a = \frac{1}{2} \)
- \( b = \frac{1}{2} \)
- \( c = 0 \)
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