Question

What mass must be suspended from a steel wire 2m long and 1 mm diameter to stretch it by 1mm?

Answer 1

Answer: 4 kg

Explanation:

Young's Modulus, $Y = \frac {Stress}{strain}$

where, stress and strain are given by:

$stress=\frac{F}{A}=\frac{mg}{\pi r^2}$

Where, F is the force, A is the area, m is the mass, g is the acceleration due to gravity and r is the radius.

$Strain=\frac{\Delta l}{l}$

where, l is the actual length and $\Delta l$ is the change in length.

It is given that $l = 2 m$

$\Delta l= 1 mm =10^{-3}m$

Diameter, $d = 1 mm =10^{-3}m$

Radius, $r = 0.5\times 10^{-3}m$

Acceleration due to gravity, g=9.8 m/s²

Young's Modulus of Steel is $Y=200 \times 10^9 Pa$

Substitute the values:

$200 \times 10^9 Pa=\frac{\frac{m\times 9.8m/s^2}{3.14\times (0.5\times10^{-3}m)^2}}{\frac{0.5\times 10^{-3}m}{2m}}=\frac{\frac{9.8m}{0.785\times 10^{-6}}}{0.25\times10^{-3}}\\ \Rightarrow m=\frac{200 \times 10^9 \times 0.25 \times10^{-3}\times 0.785\times10^{-6}}{9.8}kg=4.0 kg$

Thus, 4 kg mass must be suspended from a steel wire 2 m long and 1 mm diameter to stretch it by 1 mm.

Answer 2

Explanation:

where, stress and strain are given by:

stress=\frac{F}{A}=\frac{mg}{\pi r^2}stress=

A

F

=

πr

2

mg

Where, F is the force, A is the area, m is the mass, g is the acceleration due to gravity and r is the radius.

Strain=\frac{\Delta l}{l}Strain=

l

Δl

where, l is the actual length and \Delta lΔl is the change in length.

It is given that l = 2 ml=2m

\Delta l= 1 mm =10^{-3}mΔl=1mm=10

−3

m

Diameter, d = 1 mm =10^{-3}md=1mm=10

−3

m

Radius, r = 0.5\times 10^{-3}mr=0.5×10

−3

m

Acceleration due to gravity, g=9.8 m/s²

Young's Modulus of Steel is Y=200 \times 10^9 PaY=200×10

9

Pa

Substitute the values:

\begin{gathered}200 \times 10^9 Pa=\frac{\frac{m\times 9.8m/s^2}{3.14\times (0.5\times10^{-3}m)^2}}{\frac{0.5\times 10^{-3}m}{2m}}=\frac{\frac{9.8m}{0.785\times 10^{-6}}}{0.25\times10^{-3}}\\ \Rightarrow m=\frac{200 \times 10^9 \times 0.25 \times10^{-3}\times 0.785\times10^{-6}}{9.8}kg=4.0 kg\end{gathered}

200×10

9

Pa=

2m

0.5×10

−3

m

3.14×(0.5×10

−3

m)

2

m×9.8m/s

2

=

0.25×10

−3

0.785×10

−6

9.8m

⇒m=

9.8

200×10

9

×0.25×10

−3

×0.785×10

−6

kg=4.0kg

Thus, 4 kg mass must be suspended from a steel wire 2 m long and 1 mm diameter to stretch it by 1 mm.

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