A 10-meter-long steel rod (Young's modulus of $2 \times 10 ^ { 11 }$ Pa) having a 0.8-cm radius is supporting a 500-kg mass. The change in the length of the rod is
A) 2.3 mm.
B) 4.5 mm.
C) 1.2 mm.
D) 0.5 mm.

Question

Quizplus · Accepted Answer

The change in length ($\Delta L$) can be calculated using the formula $\Delta L = \frac{FL}{AE}$, where $F$ is the force (weight of the mass), $L$ is the original length, $A$ is the cross-sectional area, and $E$ is Young's modulus. Here, $F = mg = 500 \times 9.8$ N, $L = 10$ m, $A = \pi \times (0.008)^2$ m$^2$, and $E = 2 \times 10^{11}$ Pa. Substituting these values gives $\Delta L \approx 1.2$ mm.

A 10-Meter-Long Steel Rod (Young's Modulus Of 2×10112 \times 10 ^ { 11 }2×1011

A 10-Meter-Long Steel Rod (Young's Modulus Of $2 \times 10 ^ { 11 }$