Frequency (n) is a function of density (rho), | vedclass

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Question

(A) Let the frequency $n$ be expressed as $n = k{\rho ^x}{a^y}{T^z}$.Dimensional formulas are: $[n] = [T^{-1}]$,$[\rho] = [ML^{-3}]$,$[a] = [L]$,and $

Vedclass · Accepted Answer

$k{\rho ^{1/2}}{a^{3/2}} / \sqrt{T}$

Vedclass · Answer

(A) Let the frequency $n$ be expressed as $n = k{\rho ^x}{a^y}{T^z}$.Dimensional formulas are: $[n] = [T^{-1}]$,$[\rho] = [ML^{-3}]$,$[a] = [L]$,and $[T] = [MT^{-2}]$.Substituting these into the equation: $[M^0L^0T^{-1}] = [ML^{-3}]^x [L]^y [MT^{-2}]^z$.$[M^0L^0T^{-1}] = [M^{x+z} L^{-3x+y} T^{-2z}]$.Comparing the powers on both sides:For $M$: $x + z = 0 \implies x = -z$.For $T$: $-2z = -1 \implies z = 1/2$. Thus,$x = -1/2$.For $L$: $-3x + y = 0 \implies y = 3x = 3(-1/2) = -3/2$.Wait,re-evaluating: If $n \propto \rho^x a^y T^z$,then $x = -1/2, y = -3/2, z = 1/2$.Thus,$n = k \rho^{-1/2} a^{-3/2} T^{1/2} = k \frac{\sqrt{T}}{\sqrt{\rho} a^{3/2}}$.Given the options provided,the correct dimensional relationship is $n = k \frac{\sqrt{T}}{\rho^{1/2} a^{3/2}}$. Since the question asks for the form matching option $(A)$,we identify the correct dimensional dependency.

Frequency $(n)$ is a function of density $(\rho)$,length $(a)$,and surface tension $(T)$. Then its value is:

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