Question

In a tensile test on a metal bar of diameter $0.015\, m$ and length $0.2 \,m$, the relation between the load and elongation within the proportional limit is found to be $F=97.2 \times 10^{6}(\Delta\, L)$, where $F$ is the load (in $N$ ) and $\Delta\, L$ is the elongation (in $m$ ). The Young's modulus of the material in $GPa$ is

• A. 75.5
• B. 85.6
• C. 98.7
• D. 110

Answer 1

Answer: (D)

Given that, diameter of metal bar $(d)=0 \cdot 015 \,m$
length $(L) =0 \cdot 2 \,m $
$F =97 \cdot 2 \times 10^{6} \times \Delta\, L$
We know that, $Y=\frac{F L}{A \cdot \Delta L}$
$Y=\frac{97.2 \times 10^{6} \times 0.2 \times 4}{3.14 \times(0.015)^{2}}$
$=110063.69 \times 10^{6} \approx 110\, GPa$