Question

For steel Y=2xx10^(11) Nm^(-2). The force required to double the length of a steel wire of area 1 cm^(2) is

Answer 1

$\displaystyle\Large\boxed {\sf {F=2\times 10^7\ N}}$

Solution

We have,

$\displaystyle\longrightarrow\sf{F=AY\cdot\dfrac {\Delta L}{L}}$

To double the length of the wire, the wire should extend the same amount as its length, i.e.,

$\displaystyle\sf{\Delta L=L}$

Then,

$\displaystyle\longrightarrow\sf{F=AY}$

Given,

• $\displaystyle\sf {A=1\ cm^2=10^{-4}\ m^2}$

• $\displaystyle\sf {Y=2\times10^{11}\ N\ m^{-2}}$

Then,

$\displaystyle\longrightarrow\sf{F=2\times10^{11}\times10^{-4}\ N}$

$\displaystyle\longrightarrow\sf {\underline {\underline {F=2\times10^7\ N}}}$