Search any question & find its solution
Question:
Answered & Verified by Expert
In a tensile test on a metal bar of diameter $0.015 \mathrm{~m}$ and length $0.2 \mathrm{~m}$, 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 $\mathrm{m}$ ). The Young's modulus of the material in GPa is
Options:
Solution:
2763 Upvotes
Verified Answer
The correct answer is:
110
Given that, diameter of metal bar $(d)=0.015 \mathrm{~m}$
$$
\text { length } \begin{aligned}
(L) & =0.2 \mathrm{~m} \\
F & =97.2 \times 10^6 \times \Delta L
\end{aligned}
$$
We know that, $Y=\frac{F L}{A \cdot \Delta L}$
$$
\begin{aligned}
\Rightarrow \quad Y & =\frac{97.2 \times 10^6 \times 0.2 \times 4}{3.14 \times(0.01)^2} \\
& =110063.69 \times 10^6 \approx 110 \mathrm{GPa}
\end{aligned}
$$
$$
\text { length } \begin{aligned}
(L) & =0.2 \mathrm{~m} \\
F & =97.2 \times 10^6 \times \Delta L
\end{aligned}
$$
We know that, $Y=\frac{F L}{A \cdot \Delta L}$
$$
\begin{aligned}
\Rightarrow \quad Y & =\frac{97.2 \times 10^6 \times 0.2 \times 4}{3.14 \times(0.01)^2} \\
& =110063.69 \times 10^6 \approx 110 \mathrm{GPa}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.