JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Elasticity & SHM

The area of cross section of the rope used to lift a load by a crane is

2.5 \times 1 0^{- 4}

^{2}

. The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be :
(take

g = 10

^{- 2}

)

(A) $6.25 \times 1 0^{- 4}$ m $^{2}$ ✓ Correct
(B) $10 \times 1 0^{- 4}$ m $^{2}$
(C) $1 \times 1 0^{- 4}$ m $^{2}$
(D) $1.67 \times 1 0^{- 4}$ m $^{2}$

Answer: (A)

Solution

The breaking stress (maximum stress the material can withstand) is a material property and remains constant. Breaking stress =

\frac{Force}{Area}

. For the initial case:

σ = \frac{10 \times 1 0 ^{3} \times 10}{2.5 \times 1 0 ^{- 4}}

. For the new case with 25 metric tons:

σ = \frac{25 \times 1 0 ^{3} \times 10}{A}

. Equating:

\frac{10}{2.5 \times 1 0 ^{- 4}} = \frac{25}{A}

, giving

A = 6.25 \times 1 0^{- 4}

^{2}

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Elasticity & SHM

Solution

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