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A light rod of length 3m is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of Aluminium and is of cross-section 6×103m2 and other is of copper of cross section 2×103m2 . Find the position along the rod at which a weight may be hung to produce

i) Equal stresses in both wires and YA=7×1010N/m2

ii) Equal strain in both wires Ycu=11×1010N/m2

a
0.55 m, 1.55 m
b
0.75 m, 1.03 m
c
1.23 m, 0.86 m
d
1.62 m, 0.43 m

detailed solution

Correct option is B

i) Equal stressa1=6×10−3m2 Here a2=2×10−3m2⇒T1a1=T2a2 ⇒T1T2=a1a2=6×10-32×10-3=3 ∴T1=3T2................(1)In equilibrium, moment of all the forces acting on the rod about a point is zero.Let the centre of gravity acts at a distance of x from string 1 and 3-x from string 2                 T1x=T2(3−x)----(2) from (1) and (2)  3T2x=T2(3−x) 3x=3-x 4x=3 x=34=0.75m                     ii) Equal strain, e2=e2T1a1y1=T2a2y2, T1T2=a1y1a2y2=6×10-3×7×10102×10-3×11×1010⇒T1T2=2111....................(3)From (2) and (3) we get T1x=T2(3-x) ⇒T1T2=(3-x)x ⇒2111=(3-x)x ⇒21x=33-11x ⇒32x=33  ⇒x=3332=1.03 m

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