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A light rod of length 200 cm 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 steel and is of cross-section 0.1cm2 and the other of brass of cross section 0.2cm2. Along the rod at which distance a weight may be hung to produce equal stresses in both the wires ?

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a

43m from steel wire

b

43 m from brass wire

c

1 m from steel wire

d

14m from brass wire

answer is A.

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Detailed Solution

As stresses are equal, T1A2=T2A1 or T1T2=A1A2or T1T2=0⋅10⋅2=12 or T2=2T1……………(1)For equilibrium of rod,T1+T2 =W……………………(2)Solving eqs. (i) and (ii), we getT1=W3 and T2=2W3Let x be the distance of weight W from steel wire. Taking moment about weight ,T1X=T2(2−x)or W3×x=2W3×(2−x)Solving for x. we get x=43m

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A light rod of length 200 cm 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 steel and is of cross-section 0.1cm2 and the other of brass of cross section 0.2cm2. Along the rod at which distance a weight may be hung to produce equal stresses in both the wires ?