The velocity of water waves (v) may depend on their wavelength `lamda`, the density of water `rho` and the acceleration due to gravity g. The method of dimensions gives the relation between these quantities as `v^(2)=kg^(x)lamda^(x)rho^(y)`. The value of x is (Here k is a constant).

Explanation

To solve the problem, we need to find the value of \( x \) in the equation \( v^2 = k g^x \lambda^x \rho^y \) using dimensional analysis. Let's break it down step by step.

### Step 1: Identify the dimensions of each quantity
- The velocity \( v \) has dimensions of \( [L T^{-1}] \).
- The wavelength \( \lambda \) is a length, so its dimensions are \( [L] \).
- The density \( \rho \) is mass per unit volume, so its dimensions are \( [M L^{-3}] \).
- The acceleration due to gravity \( g \) has dimensions of \( [L T^{-2}] \).

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