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(A) The critical velocity $v_c$ of a liquid flowing through a tube is given by the formula: $v_c = \frac{N_R \eta}{\rho r}$,where $N_R$ is the Reynold

Vedclass · Accepted Answer

$4:49$

Vedclass · Answer

(A) The critical velocity $v_c$ of a liquid flowing through a tube is given by the formula: $v_c = \frac{N_R \eta}{ho r}$,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.Given that the tubes have equal radii $(r_1 = r_2)$ and assuming the critical Reynolds number $N_R$ is the same for both liquids,the ratio of critical velocities is:$\frac{v_1}{v_2} = \frac{\eta_1}{\eta_2} 	imes \frac{ho_2}{ho_1}$Given ratios:$\frac{\eta_1}{\eta_2} = \frac{52}{49}$$\frac{ho_1}{ho_2} = \frac{13}{1} \implies \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}$Thus,the ratio of their critical velocities is $4:49$.

Two different liquids are flowing in two tubes of equal radius. The ratio of coefficients of viscosity of liquids is $52:49$ and the ratio of their densities is $13:1$,then the ratio of their critical velocities will be

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