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Home » A mild steel wire of length and cross-sectional area is stretched, well within its elastic limit, horizontally between two pillars. A mass of is suspended from the mid-point of the wire. Calculate the depression at the midpoint.
A mild steel wire of length 1.0 \mathrm{~m} and cross-sectional area 0.50 \times 10^{-2} \mathrm{~cm}^{2} is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 \mathrm{~g} is suspended from the mid-point of the wire. Calculate the depression at the midpoint.
CBSE | Class 11 | Mechanical Properties of Solids | NCERT | Physics
A mild steel wire of length 1.0 \mathrm{~m} and cross-sectional area 0.50 \times 10^{-2} \mathrm{~cm}^{2} is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 \mathrm{~g} is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

Water pressure at the bottom is given as,

p=1000 a t m=1000 \times 1.013 \times 10^{5} Pa

p=1.01 \times 10^{8} \mathrm{~Pa}

Initial volume of the steel ball is given as V=0.30 \mathrm{~m}^{3}

Bulk modulus of steel is known as B=1.6 \times 10^{11} \mathrm{Nm}^{-2}

Let \Delta V be the change in the volume of the ball on reaching the bottom of the trench.

Bulk modulus, \mathrm{B}=\mathrm{p} /(\Delta \mathrm{V} / \mathrm{V})

\Delta \mathrm{V}=\mathrm{pV} / \mathrm{B}

=[1.01\times10^{8}\times0.30]/(1.6\times 10^{11})
=1.89\times10^{-4} m^{-3}

Hence, on reaching the bottom of the trench, volume of the ball changes by 1.89\times10^{-4} m^{-3}.

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