The breaking stress of aluminium is 7.5 times | vedclass

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(C) The breaking stress of a wire hanging vertically is given by the formula: $	ext{Breaking Stress} = ho \cdot g \cdot L$,where $ho$ is the dens

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$2.83$

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(C) The breaking stress of a wire hanging vertically is given by the formula: $	ext{Breaking Stress} = ho \cdot g \cdot L$,where $ho$ is the density,$g$ is the acceleration due to gravity,and $L$ is the length of the wire.Given:Breaking stress $= 7.5 	imes 10^7 \,N m^{-2}$Density $ho = 2.7 	imes 10^3 \,kg m^{-3}$$g = 9.8 \,m s^{-2}$Rearranging the formula to solve for $L$:$L = \frac{	ext{Breaking Stress}}{ho \cdot g}$$L = \frac{7.5 	imes 10^7}{2.7 	imes 10^3 	imes 9.8}$$L = \frac{7.5 	imes 10^7}{26.46 	imes 10^3}$$L \approx 0.2834 	imes 10^4 \,m = 2.834 	imes 10^3 \,m$.Rounding to the nearest value,$L \approx 2.83 	imes 10^3 \,m$.

The breaking stress of aluminium is $7.5 \times 10^7 \,N m^{-2}$. The greatest length of aluminium wire that can hang vertically without breaking is ......... $\times 10^3 \,m$ (Density of aluminium is $2.7 \times 10^3 \,kg m^{-3}$ and $g = 9.8 \,m s^{-2}$)

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