Question

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Answer 1

Explanation:

The length of rod AB = L=2 m

Suppose a weight W is hung at C at a distance x from A.

Ts =Tension in steel wire

Tb =Tension in brass wire

Stress in steel wire=Ts×As

Stress in brass wire=Tb×Ab

We are given:

As = 0.1×10−4m^2

Ab = 0.2×10−4 m^2

So, for equal stresses

Ts×As = Tb×Ab

Ts×0.1×10^−4 = Tb×0.2×10^−4

Ts/Tb=2 or Ts=2 Tb

Since the system is in equilibrium, the moments of force Ts and Tb about C will be equal.

Thus,Ts×x=Tb(2−x)

2 Tb×x=Tb(2−x)

2x=2−x3x=2

x=2/3 = 0.67 m

Thus the weight should be hung at a distance 0.67 m from point A.

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