Question

A 6-kg weight is fastened to the end of a steel wire of un-stretched length 60 cm. It is whirled in a vertical circle and has an angular velocity of 2 revolutions per second at the bottom of the circle. The area of cross-section of the wire is $0.05c{m}^2$. Calculate the elongation of the wire when the weight is at the lowest point of the path. Young's modulus of steel = $2\times {10}^{11}$ Pa.

• A. 4.97 m
• B. 3.77 m
• C. 2.34m
• D. 2.34m

Answer 1

The correct option is B

3.77 m

At the lowest position, $T-mg=m{\omega }^{2}l$
$or,T=mg+m{\omega }^2{l}_0+\Delta lStress=Y\times strain=Y\frac{\Delta l}{l_0}T=A\times stress=A\times Y\left(\frac{\Delta l}{l_0}\right)or,A\times Y\left(\frac{\Delta l}{l_0}\right)=mg+m{\omega }^2(l_0+\Delta l)or,\Delta l=\frac{mg+m{\omega }^2{l}_0}{[\left(\frac{AY}{l_0}\right)-m{\omega }^2]}=3.77\times {10}^{-4}m$