91Ó°ÊÓ

When a force (F) is applied to stretch a uniform wire made of a certain material, the length of the wire gets changed. The change in length of the wire is proportional to the original length (\({l_0}\)), and inversely proportional to the cross-sectional area (A), i.e.,

\(\Delta l = \frac{1}{E}\frac{F}{A}{l_0}\).

Here, E is the constant of proportionality and is termed as the elastic modulus or Young’s modulus. The value of Young’s modulus depends on the material of the wire.

In this problem, the value of Young’s modulus of nylon string is \(E{\bf{ = 3 \times 1}}{{\bf{0}}^{\bf{9}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\).

Tension acting on a nylon string is T = 275 N.

The diameter of nylon string is,\(d = 1.00\;{\rm{mm}} = 1.00 \times 1{{\rm{0}}^{ - 3}}\;{\rm{m}}\).

The radius of nylon string is calculated as follows:

\(\begin{array}{c}r = \frac{d}{2}\\ = \frac{{1.00 \times 1{{\rm{0}}^{ - 3}}\;{\rm{m}}}}{2}\\ = 0.50 \times 1{{\rm{0}}^{ - 3}}\;{\rm{m}}\end{array}\)

The original length of the nylon string is\({l_{\rm{o}}} = 30.0\;{\rm{cm}} = 30.{\rm{0}} \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}\).

The area of the cross-section of the nylon string is calculated as follows:

\(\begin{array}{c}A = \pi {r^2}\\ = 3.14 \times {\left( {0.50 \times 1{{\rm{0}}^{ - 3}}\;{\rm{m}}} \right)^2}\\ = 0.785 \times 1{{\rm{0}}^{ - 6}}\;{{\rm{m}}^2}\end{array}\)

The change in length of the nylon string due to tension applied on it is as follows:

\(\begin{array}{c}\Delta l = \frac{1}{E}\frac{F}{A}{l_0}\\ = \frac{1}{{\left( {3 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}} \right)}}\frac{{\left( {275\;{\rm{N}}} \right)}}{{\left( {0.785\;{{\rm{m}}^2}} \right)}}\left( {30.{\rm{0}} \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}} \right)\\ = 3.50 \times {10^{ - 2}}\;{\rm{m}}\end{array}\)

Thus, the length of the nylon string is increased by \(3.50 \times {10^{ - 2}}\;{\rm{m}}\) from its untensioned length of 30.0 cm.