Discussion Forum

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 x 10⁶ ($\triangle $  L) , where  F is  the load (in N) and $\triangle L$  is the elongation  (in m) . The youngs' modulus  of the material in GPa is 

Answer:

Explanation:

Given that, diameter  of metal bar (d)= 0.15 m

length (L)=0.2 m

 $F=97.2 \times  10^{6}  \times \triangle L$

 We know that, $Y=\frac{FL}{A.\triangle L}$

 $\Rightarrow$     $Y= \frac{97.2\times10^{6}\times0.2\times4}{3.14\times (0.015)^{2}}$

  = $110063.69 \times 10^{6}$≈ 110 GPa