Mechanical Properties of Solids

JEE Main level, numerical answer, medium

15 of 118

Question 15 of 118

A uniform wire of length ll of weight ww is suspended from the roof with a weight of WW at the other end. The stress in the wire at l3\dfrac{l}{3} distance from the top is (WA+2γwA)\left(\dfrac{W}{A} + \dfrac{2}{\gamma}\dfrac{w}{A}\right), where, AA is the cross sectional area of the wire. The value of γ\gamma is _______.

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Answer: 3

Let TT be the tension in the wire at a distance l3\dfrac{l}{3} from the top.

The length of the wire below this point is l−l3=2l3l - \dfrac{l}{3} = \dfrac{2l}{3}.

The weight of this lower part of the wire is wl×2l3=2w3\dfrac{w}{l} \times \dfrac{2l}{3} = \dfrac{2w}{3}.

The total downward force (tension TT) at this point is the sum of the weight WW attached at the bottom and the weight of the lower part of the wire:

T=W+2w3T = W + \dfrac{2w}{3}

The stress at this point is given by the tension divided by the cross-sectional area AA:

Stress=TA=W+2w3A=WA+23wA\text{Stress} = \dfrac{T}{A} = \dfrac{W + \dfrac{2w}{3}}{A} = \dfrac{W}{A} + \dfrac{2}{3}\dfrac{w}{A}

Comparing this with the given expression for stress, WA+2γwA\dfrac{W}{A} + \dfrac{2}{\gamma}\dfrac{w}{A}, we get:

γ=3\gamma = 3

Answer: 33

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