Two wires as shown in the figure below, made | vedclass

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(B) Breaking stress is defined as $	ext{Stress} = \frac{F_{max}}{A}$, which implies $F_{max} = 	ext{Stress} 	imes A$.For the lower wire:$F_L = (12

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(B) Breaking stress is defined as $	ext{Stress} = \frac{F_{max}}{A}$, which implies $F_{max} = 	ext{Stress} 	imes A$.For the lower wire:$F_L = (12 	imes 10^8 	ext{ N/m}^2) 	imes (0.004 	imes 10^{-4} 	ext{ m}^2) = 480 	ext{ N}$.The weight supported by the lower wire is $(m_{pan} + 10)g = 480 	ext{ N}$.$(m_{pan} + 10) 	imes 10 = 480 \Rightarrow m_{pan} + 10 = 48 \Rightarrow m_{pan} = 38 	ext{ kg}$.For the upper wire:$F_U = (12 	imes 10^8 	ext{ N/m}^2) 	imes (0.008 	imes 10^{-4} 	ext{ m}^2) = 960 	ext{ N}$.The weight supported by the upper wire is $(m_{pan} + 10 + 30)g = 960 	ext{ N}$.$(m_{pan} + 40) 	imes 10 = 960 \Rightarrow m_{pan} + 40 = 96 \Rightarrow m_{pan} = 56 	ext{ kg}$.To ensure both wires remain safe, we must choose the smaller mass, which is $38 	ext{ kg}$.

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