Question

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:

• A. 1,1,0
• B. \(\frac{1}{2},0,\frac{1}{2}\)
• C. \(\frac{1}{2},\frac{1}{2},0\)
• D. 1,-1,0

Answer 1

The Correct Option is C

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:

• \( a = \frac{1}{2} \)
• \( b = \frac{1}{2} \)
• \( c = 0 \)