1.

A metal block of area 0.10 m^2 is connected to a 0.010 kg mass via a string that passes over an ideal pulley(considered massless and frictionless) as in Fig 10.15 A liquid with a film thickness of 0.30 mm is placed between the block and the table . When released the block moves to the right with a constant speed of 0.085 ms^-1. FInd the coefficienyt of viscosity of the liquid.

Answer» <html><body><p></p>Solution :The metal <a href="https://interviewquestions.tuteehub.com/tag/block-18865" style="font-weight:bold;" target="_blank" title="Click to know more about BLOCK">BLOCK</a> moves to the <a href="https://interviewquestions.tuteehub.com/tag/right-1188951" style="font-weight:bold;" target="_blank" title="Click to know more about RIGHT">RIGHT</a> because of the <a href="https://interviewquestions.tuteehub.com/tag/tension-1241727" style="font-weight:bold;" target="_blank" title="Click to know more about TENSION">TENSION</a> in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus the shear force F is `F=T=mg=0.010 <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> times 9.8 ms^-2=9.8 times 10^-2 N` <br/> Shear stress on the fluid =`F//A=(9.8 times 10^-2)/0.10 N//m^2` <br/> Strain rate `=v/l=0.08/(0.30 times 10^-3)` <br/> `n=("stress")/("strainrate") s^-1` <br/> `=((9.8 times 10^-2 N)(0.30 times 10^-3 m))/((0.085 ms^-1)(0.10 m^2))` <br/> `=3.46 times 10^-3 Pas`</body></html>


Discussion

No Comment Found

Related InterviewSolutions