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Question: Answered & Verified by Expert
Wires $W_1$ and $W_2$ are made of same material having the breaking stress of $1.25 \times 10^9 \mathrm{~N} / \mathrm{m}^2$. $\mathrm{W}_1$ and $\mathrm{W}_2$ have cross-sectional area of $8 \times 10^{-7}$ $\mathrm{m}^2$ and $4 \times 10^{-7} \mathrm{~m}^2$, respectively. Masses of 20 $\mathrm{kg}$ and $10 \mathrm{~kg}$ 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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$)
PhysicsMechanical Properties of SolidsJEE Main
Options:
  • A $40 \mathrm{~kg}$
  • B $60 \mathrm{~kg}$
  • C $80 \mathrm{~kg}$
  • D $20 \mathrm{~kg}$
Solution:
2952 Upvotes Verified Answer
The correct answer is: $40 \mathrm{~kg}$


$\begin{aligned} & \text { B. } S_1=\frac{T_{1 \max }}{8 \times 10^{-7}} \Rightarrow T_{1 \max }=8 \times 1.25 \times 100 \\ & =1000 \mathrm{~N} \\ & \text { B. } S_2=\frac{\mathrm{T}_{2 \max }}{4 \times 10^{-7}} \Rightarrow \mathrm{T}_{2 \max }=4 \times 1.25 \times 100 \\ & =500 \mathrm{~N} \\ & \mathrm{~m}=\frac{500-100}{10}=40 \mathrm{~kg}\end{aligned}$

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