+3 votes
in JEE by kratos

A steel wire of diameter 0.8 mm and length 1 m is clamped firmly at two points A and B which are 1 m apart and in the same horizontal plane. A body is hung from the middle point of the wire such that the middle point sags 1 cm lower from the original position. Calculate the mass of the body. Given Young’* modulus of the material of wire = 2 × 1012 dynes/cm2.

1 Answer

+4 votes
by kratos
 
Best answer

Let the body be hung from the middle point C so that it sags through 1 cm to the point D as shown in the figure

∴ AD2 = AC2 + CD2 = (50)2 + (1)2

or AD = 50.01 cm

Increase in length = 0.01

Strain = 0.01/50 = 2 x 10-4

Stress = 2 × 1012 × 2 × 10–4

= 4 × 108 dynes/cm2

Tension T = Stress × Area of cross-section

= 4 × 108 × π × (0.08)2

Since the mass m is in equilibrium,

mg = 2T cosθ or m = 2T cosθ/g

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