Question

# A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10¯² cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

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Solution

## Given: The length of steel wire is 1.0m, cross-sectional area is 0.50× 10 −2 cm 2 and the mass of 100 g is suspended from the midpoint of the wire. By Pythagoras theorem, AD=DB= AC 2 + CD 2 = l 2 + x 2 . The increase in length can be calculated as, ΔL=( AD+DB )−AB =2 l 2 + x 2 −2l =2l[ ( 1+ x 2 l 2 ) −1 ] Using binomial theorem in above expression, we get ( 1+x ) n =1+nx. 2l[ ( 1+ x 2 l 2 ) -1 ]=2l x 2 2 l 2 = x 2 l The longitudinal strain is given as, Δl 2l = x 2 2 l 2 Applying the equilibrium condition in the above figure, we get Tcosθ+Tcosθ=mg 2Tcosθ=mg T= mg 2 x x 2 + l 2 T= mgl 2x The stress in the wire is given as, Stress= T A = mgL 2xA The Young’s modulus is given as, Y= stress strain . By substituting the values of stress and strain in Young’s modulus, we get Y= mgl 2xA x 2 2 l 2 Y= mg l 3 A x 3 x=l [ mg YA ] 1 3 By substituting the values in the above equation, we get x= 1 2 [ 0.1×9.8 2× 10 11 ×0.50× 10 −2 × 10 −4 ] 1 3 =1.074× 10 −2  m =1.074 cm Thus, the depression at the midpoint is 1.074 cm.

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