Question

A wire of unifrm cross section and negligible mass is suspended from a rigid support .The length of wire is 1.2m. A load of 4kg is attached to the wire at 40cm from the point of suspension.Another load of 4kg is attached to the wire at 80cm from the point of suspension.A third load 4kg is suspended at 120cm from the point of suspension. The ratio of the increase in the length of the upper one-thid portion of the wire to the lower one-third portion of the wire is

Answer 1

Answer:

Ratio of extension in two parts of the wire is given as

$\Delta L_1 : \Delta L_2 = 3 : 1$

Explanation:

As we know that the elasticity of the wire is defined as the ratio of stress and strain

so here we have

$E = \frac{stress}{strain}$

now we have

$E = \frac{T L}{A \Delta L}$

now we have

$\Delta L = \frac{T L}{A E}$

now for upper one third part of wire the tension in the wire is

$T = (4 + 4 + 4)(10) = 120 N$

so we have

$\Delta L_1 = \frac{(120 N) L}{A E}$

similarly for lowest one third part of the wire tension is given as

$T = 4 \times 10 = 40 N$

so we have

$\Delta L_2 = \frac{(40 N) L}{A E}$

now ratio of extension in two parts of the wire is given as

$\Delta L_1 : \Delta L_2 = \frac{120}{40} = 3 : 1$

#Learn

Topic : Elasticity

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