The elastic limit of an elevator cable is 2×109N/m2. The maximum upward acceleration that an elevator of mass 2×103kg can have when supported by a cable whose cross-sectional area is 10−4m2provided the stress in cable would not exceed half of the elastic limit would be

From the free-body diagram of the elevator,T−mg=maT=m(g+a)Stress in cable is σ=TA=m(g+a)AFrom the given condition, σ≤σmax2m(g+a)A≤σmax2g+a≤2×1092×10−42×103=5010+a=50a=40ms-2

The elastic limit of an elevator cable is 2×109N/m2. The maximum upward acceleration that an elevator of mass 2×103kg can have when supported by a cable whose cross-sectional area is 10−4m2provided the stress in cable would not exceed half of the elastic limit would be

From the free-body diagram of the elevator,T−mg=maT=m(g+a)Stress in cable is σ=TA=m(g+a)AFrom the given condition, σ≤σmax2m(g+a)A≤σmax2g+a≤2×1092×10−42×103=5010+a=50a=40ms-2

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The elastic limit of an elevator cable is $2 \times 10^{9} N / m^{2}$ . The maximum upward acceleration that an elevator of mass $2 \times 10^{3} kg$ 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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$40 {ms}^{- 2}$

$50 {ms}^{- 2}$

Not possible to move up

$10 {ms}^{- 2}$

answer is C.

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Detailed Solution

From the free-body diagram of the elevator,

$\begin{array}{l} T - mg = ma \\ T = m (g + a) \end{array}$

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

From the given condition, $σ \leq \frac{σ_{max}}{2}$

$\begin{array}{l} \frac{m (g + a)}{A} \leq \frac{σ_{max}}{2} \\ g + a \leq \frac{2 \times 10^{9}}{2} \times \frac{10^{- 4}}{2 \times 10^{3}} = 50 \\ 10 + a = 50 \\ a = 40 m s^{- 2} \end{array}$