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The elastic limit of an elevator cable is 2 × $10^{9}$ N/ $m^{2}$ . The maximum upward acceleration that an elevator of mass 2 × $10^{3}$ kg can have when supported by a cable whose cross-sectional area is $10^{- 4}$ $m^{2}$ , provided the stress in cable would not exceed half of the elastic limit would be

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$50 {ms}^{- 2}$

$10 {ms}^{- 2}$

Not possible to move up

$40 {ms}^{- 2}$

answer is C.

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Detailed Solution

From the free-body diagram of the elevator,

$T - mg = ma \Rightarrow T = m (g + a)$

Stress in cable is $α = \frac{T}{A} = \frac{m (g + a)}{A}$

From the given condition, $α \leq \frac{α_{\max}}{2}$

$\Rightarrow \frac{m (g + a)}{A} \leq \frac{σ_{\max}}{2}$

$\Rightarrow g + a \leq \frac{2 \times 10^{9}}{2} \times \frac{10^{- 4}}{2 \times 10^{3}} = 50$

$\Rightarrow a = 40 {ms}^{- 2}$

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