One end of a steel rod is clamped to the roof and the other end is attached to a mass of \(1000\ \text{kg}\) as shown in the figure. The length of the rod is \(50\ \text{cm}\) and its cross-sectional area is \(1000\ \text{mm}^2\). The change in the length of the rod due to the weight of the mass is
\[
(\text{Young's modulus of steel }=2\times10^{11}\ \text{N m}^{-2})
\]
\[
(\text{acceleration due to gravity }=10\ \text{m s}^{-2})
\]
Show Hint
For elongation of a wire or rod under a load,
\[
\Delta L=\frac{FL}{AY},
\]
where \(F\) is the applied force, \(L\) is the original length, \(A\) is the cross-sectional area, and \(Y\) is Young's modulus.
Step 1: Write the given data.
Mass suspended from the rod:
\[
m=1000\ \text{kg}
\]
Length of the rod:
\[
L=50\ \text{cm}=0.5\ \text{m}
\]
Cross-sectional area:
\[
A=1000\ \text{mm}^2
\]
Since
\[
1\ \text{mm}^2=10^{-6}\ \text{m}^2,
\]
\[
A=1000\times10^{-6}
\]
\[
A=10^{-3}\ \text{m}^2.
\]
Young's modulus:
\[
Y=2\times10^{11}\ \text{N m}^{-2}.
\]
Acceleration due to gravity:
\[
g=10\ \text{m s}^{-2}.
\]
Step 2: Calculate the stretching force.
The force acting on the rod is the weight of the suspended mass:
\[
F=mg.
\]
\[
F=1000\times10.
\]
\[
F=10^4\ \text{N}.
\]
Step 3: Use Young's modulus formula.
Young's modulus is
\[
Y=\frac{\text{Stress}}{\text{Strain}}
=
\frac{\frac{F}{A}}{\frac{\Delta L}{L}}.
\]
Therefore,
\[
\Delta L=\frac{FL}{AY}.
\]
Substituting the values,
\[
\Delta L
=
\frac{(10^4)(0.5)}
{(10^{-3})(2\times10^{11})}.
\]
Step 4: Simplify the expression.
\[
\Delta L
=
\frac{5000}{2\times10^8}.
\]
\[
\Delta L
=
2.5\times10^{-5}\ \text{m}.
\]
Converting into millimetres,
\[
\Delta L
=
2.5\times10^{-5}\times10^3.
\]
\[
\Delta L
=
2.5\times10^{-2}\ \text{mm}.
\]
\[
\Delta L
=
0.025\ \text{mm}.
\]
Step 5: Final conclusion.
Hence, the increase in length of the rod is
\[
\boxed{0.025\ \text{mm}}
\]
Therefore, the correct option is
\[
\boxed{(1)}
\]
