Q122 / 180 · AIIMS 2014 · Physics
A steel wire with cross-section $3 \mathrm{~cm}^2$ has elastic limit $2.4 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}$. The maximum upward acceleration that can be given to a $1200 \mathrm{~kg}$ elevator supported by thiscable wire if the stress is not to exceed one-third of the elastic limit is (Take $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ )
- A$12 \mathrm{~m} \mathrm{~s}^{-2}$
- Correct answerB$10 \mathrm{~m} \mathrm{~s}^{-2}$
- C$8 \mathrm{~m} \mathrm{~s}^{-2}$
- D$7 \mathrm{~m} \mathrm{~s}^{-2}$
Solution
Maximum tension an elevator can tolerate is
$\begin{aligned}
T & =\frac{1}{3} \text { stress } \times \text { area of cross-section } \\
& =\frac{1}{3} \times\left(2.4 \times 10^8\right) \times\left(3 \times 10^{-4}\right)=2.4 \times 10^4 \mathrm{~N}
\end{aligned}$
If $a$ is the maximum upward acceleration of elevator then $T=m(g+a)$
or $2.4 \times 10^4=1200(10+a)$
On solving, $a=10 \mathrm{~m} \mathrm{~s}^{-2}$.
Hence, the correct answer is option: 'B'.