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(C) According to Newton's law of viscosity,the shear stress $	au$ is proportional to the velocity gradient $\frac{du}{dy}$.$	au = \mu \frac{du}{dy}$

Vedclass · Accepted Answer

$4.5$

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(C) According to Newton's law of viscosity,the shear stress $	au$ is proportional to the velocity gradient $\frac{du}{dy}$.$	au = \mu \frac{du}{dy}$Since $	au = \frac{F}{A}$,where $F$ is the applied force and $A$ is the surface area,we have $\frac{F}{A} = \mu \frac{u}{y}$.This implies $F = \left( \frac{A \mu}{y} ight) u$,which means $F \propto u$.Therefore,$\frac{F_1}{u_1} = \frac{F_2}{u_2}$.Given $F_1 = 800$ $N$,$u_1 = 1.5$ $cm/s$,and $F_2 = 2.4$ $kN = 2400$ $N$.Substituting these values: $\frac{800}{1.5} = \frac{2400}{u_2}$.Solving for $u_2$: $u_2 = \frac{2400 	imes 1.5}{800} = 3 	imes 1.5 = 4.5$ $cm/s$.

$A$ Newtonian fluid fills the clearance between a shaft and a sleeve. When a force of $800$ $N$ is applied to the shaft,parallel to the sleeve,the shaft attains a speed of $1.5$ $cm/s$. If a force of $2.4$ $kN$ is applied instead,the shaft would move with a speed of ......... $cm/s$.

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