The speed of a wave produced in water is give | vedclass

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(A) The given equation is $v = \lambda^a g^b \rho^c$.Using the dimensional analysis method,we write the dimensions of each quantity:$[v] = [L T^{-1}]$

Vedclass · Accepted Answer

$\frac{1}{2}, \frac{1}{2}, 0$

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(A) The given equation is $v = \lambda^a g^b \rho^c$.Using the dimensional analysis method,we write the dimensions of each quantity:$[v] = [L T^{-1}]$$[\lambda] = [L]$$[g] = [L T^{-2}]$$[\rho] = [M L^{-3}]$Substituting these into the equation:$[M^0 L^1 T^{-1}] = [L]^a [L T^{-2}]^b [M L^{-3}]^c$$[M^0 L^1 T^{-1}] = [M^c L^{a+b-3c} T^{-2b}]$Comparing the powers of $M$,$L$,and $T$ on both sides:For $M$: $c = 0$For $T$: $-2b = -1 \Rightarrow b = \frac{1}{2}$For $L$: $a + b - 3c = 1$Substituting $b = \frac{1}{2}$ and $c = 0$ into the equation for $L$:$a + \frac{1}{2} - 3(0) = 1$$a = 1 - \frac{1}{2} = \frac{1}{2}$Thus,the values are $a = \frac{1}{2}$,$b = \frac{1}{2}$,and $c = 0$.

The speed of a wave produced in water is given by $v = \lambda^a g^b \rho^c$. Where $\lambda$,$g$,and $\rho$ are the wavelength of the wave,acceleration due to gravity,and density of water,respectively. The values of $a$,$b$,and $c$ are,respectively:

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