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Question

(A) The formula for critical velocity $(v_c)$ is given by $v_c = \frac{N_R \eta}{\rho r}$,where $N_R$ is the Reynolds number,$\eta$ is the coefficient

Vedclass · Accepted Answer

$4:49$

Vedclass · Answer

(A) The formula for critical velocity $(v_c)$ is given by $v_c = \frac{N_R \eta}{ho r}$,where $N_R$ is the Reynolds number,$\eta$ is the coefficient of viscosity,$ho$ is the density,and $r$ is the radius of the tube.Since the radius $(r)$ and the Reynolds number $(N_R)$ are the same for both liquids,the critical velocity is proportional to $\frac{\eta}{ho}$.Therefore,the ratio of critical velocities is $\frac{v_1}{v_2} = \frac{\eta_1}{\eta_2} 	imes \frac{ho_2}{ho_1}$.Given $\frac{\eta_1}{\eta_2} = \frac{52}{49}$ and $\frac{ho_1}{ho_2} = \frac{13}{1}$,we have $\frac{ho_2}{ho_1} = \frac{1}{13}$.Substituting these values: $\frac{v_1}{v_2} = \frac{52}{49} 	imes \frac{1}{13} = \frac{4}{49}$.

Two different liquids flow through tubes of the same radius. If the ratio of their coefficients of viscosity is $52:49$ and the ratio of their densities is $13:1$,what is the ratio of their critical velocities?

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