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A one metre steel wire of negligible mass and area of cross-section 0.01 cm^(-2) is kept on a smooth horizontal table with one end fixed.A ball of mass 1 kg is attached to the other end. The ball and the wire are rotating with an angular velocity of omega. If the elongation of the wire is 2 mm, then omega is (Young's modulus of steel = 2 xx 10^(11) Nm^(-2) ) |
| Answer» <html><body><p>5 rad `s^(-1)`<br/>10 rad `s^(-1)`<br/>15 rad `s^(-1)`<br/><a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> rad `s^(-1)`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Given, elongation of the wire, `Delta`l= 2mm <br/> = `2 xx 10^(-3)` m<br/> Mass of the ball, m=1 kg <br/> Length of wire, l = 1 m <br/> area of cross - sectional of wire, A= 0.01`cm^(2) = 0.01 xx 10^(-4)` m<br/> Young.s modulus of steel, R = `2 xx 10^(11) Nm^(-2)` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/tension-1241727" style="font-weight:bold;" target="_blank" title="Click to know more about TENSION">TENSION</a> force in wire , T = m `<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>^(2)`l <br/> `therefore ` Stress = `("Tension")/("Area") = (m omega^(2)l)/(A )` <br/> Strain = `( Delta l)/(l ) = ("stress")/("Young.s modulus")` <br/> ` Delta l=(m omega^(2) l^(2))/(YA)` <br/>or`omega = sqrt((YA Delta)/(<a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>^(2))` <br/> putting the given values, we get <br/> `=sqrt((2 x 10^(11) xx 0.01 xx 10^(-4) xx 2 xx 10^(-3))/(1 xx (1)^2))` <br/> `omega = 20 rad/"sec"^(-1)`</body></html> | |