Wires W _ 1 and W _ 2 are made of same material having the breaking stress of 1 . 25 × 10 ^ 9 N m ^ - 2 . W _ 1 and W _ 2 have cross-sectional area of 8 × 10 ^ - 7 m ^ 2 and 4 × 10 ^ - 7 m ^ 2 , respectively. Masses of

Correct answer: 0 ( F _ 1 ) _ max = stress × area = 1 . 25 × 8 × 10 ^ 2 = 1000N ( F _ 2 ) _ max = 500N ( F _ 2 ) _ max = ( 10 + M ) g = 500N M = 40kg

JEE Main 202127 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Evening Shift
Subject: Physics
Chapter: Mechanical Properties of Solids

JEE Main 2021 · 27 August · Evening Shift · Physics · Mechanical Properties of Solids

Wires W _ 1 and W _ 2 are made of same material having the breaking...

Wires ${W}_{1} and {W}_{2}$ are made of same material having the breaking stress of $1 . 25 \times {10}^{9}N{m}^{- 2} . {W}_{1}$ and {W}_{2} have cross-sectional area of $8 \times {10}^{- 7}{m}^{2} and 4 \times {10}^{- 7}{m}^{2}$ , respectively. Masses of $20kg and 10kg$ hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is _____ $kg$ (Use \leftg = 10m{s}^{- 2})

Wires W _ 1 and W _ 2 are made of same material having the breaking stress of 1 . 25 × 10 ^ 9 N m ^ - 2 . W _ 1 and W _

Correct answer

Step-by-step solution

${\left({F}_{1}\right)}_{\text{max}} = stress \times area$

$= 1 . 25 \times 8 \times {10}^{2} = 1000N$

${\left({F}_{2}\right)}_{\text{max}} = 500N$

${\left({F}_{2}\right)}_{\text{max}} = ( 10 + M ) g = 500N$

$M = 40kg$

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