Wires W_{1} and W_{2} are made of the same ma | vedclass

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(B) Let $m$ be the mass placed in the pan. The tension in wire $W_{2}$ is $T_{2} = (m + 10)g$. The tension in wire $W_{1}$ is $T_{1} = (m + 10 + 20)g

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Let $m$ be the mass placed in the pan. The tension in wire $W_{2}$ is $T_{2} = (m + 10)g$. The tension in wire $W_{1}$ is $T_{1} = (m + 10 + 20)g = (m + 30)g$.Breaking stress $\sigma_{b} = 1.25 	imes 10^{9} \, N/m^{2}$.For wire $W_{2}$: $T_{2,max} = \sigma_{b} 	imes A_{2} = (1.25 	imes 10^{9}) 	imes (4 	imes 10^{-7}) = 500 \, N$.$(m + 10) 	imes 10 = 500 \Rightarrow m + 10 = 50 \Rightarrow m = 40 \, kg$.For wire $W_{1}$: $T_{1,max} = \sigma_{b} 	imes A_{1} = (1.25 	imes 10^{9}) 	imes (8 	imes 10^{-7}) = 1000 \, N$.$(m + 30) 	imes 10 = 1000 \Rightarrow m + 30 = 100 \Rightarrow m = 70 \, kg$.Since the wire $W_{2}$ will break first when $m = 40 \, kg$,the maximum mass that can be placed is $40 \, kg$.

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