Q. One end of steel wire is fixed to ceiling of an elevator moving up with an acceleration 2 $ms^{- 2}$ and a load of 10 kg hangs from other end. Area of cross-section of the wire is $2\ cm^{2}.$ The longitudinal strain in the wire is (Take g = $10\ ms^{- 2}\ and\ Y = 2 \times 10^{11}\ Nm^{- 2})$
A) $4\ \times 10^{11}$
B) $3\ \times 10^{- 6}$
C) $8\ \times 10^{- 6}$
D) $2\ \times 10^{- 6}$
Explanation
(b) $T = m\ \left( g + a_{0} \right) = \ 10\ (10 + 2) = \ 120\ N\ $
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$\therefore\ \ Stress = \frac{T}{A}$
= $\frac{120}{2\ \times 10^{- 4}} = 60 \times 10^{4}\ Nm^{- 2}$
$\therefore\ \ Y = \frac{stress}{4strain}\ $
$\therefore\ \ strain\ = \frac{stress}{Y}$
= $\frac{60\ \times 10^{4}}{2\ \times 10^{11}} = 30\ \times 10^{- 7} = 3\ \times 10^{- 6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.