**Question 9.19:**

A mild steel wire of length 1.0 m and cross-sectional area
0.50 × 10

^{–2}cm^{2 }is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

__Solution__:

Length of the steel wire = 1.0 m

Area of cross-section,

*A*= 0.50 × 10

^{–2}cm

^{2 }= 0.50 × 10

^{–6}m

^{2}

A mass 100 g is suspended from its midpoint.

*m*= 100 g = 0.1 kg

Hence, the wire dips, as shown in the given figure.

Original length = XZ

Depression =

*l*

The length after mass

*m*, is attached to the wire = XO + OZ

Increase in the length of the wire:

Δ

*l*= (XO + OZ) – XZ

Where,

Expanding and neglecting higer terms, we get:

∆l = l

^{2}/ 0.5

Strain = Increase in length / Original length

Let

*T*be the tension in the wire.

∴

*m*g = 2

*T*cos

*θ*

Using the figure, it can be written as:

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