A cylindrical rod made of aluminum has a leng | vedclass

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(D) Given: Length $l = 1 \,m$, Diameter $d = 10 \,cm = 0.1 \,m$, Radius $r = 0.05 \,m$, Force $F = 100 \,kN = 10^5 \,N$, Young's modulus $Y = 70 \,GPa

Vedclass · Accepted Answer

$1.81 	imes 10^{-4} \,m$

Vedclass · Answer

(D) Given: Length $l = 1 \,m$, Diameter $d = 10 \,cm = 0.1 \,m$, Radius $r = 0.05 \,m$, Force $F = 100 \,kN = 10^5 \,N$, Young's modulus $Y = 70 \,GPa = 70 	imes 10^9 \,Pa$.Using the formula for Young's modulus: $Y = \frac{F \cdot l}{A \cdot \Delta l}$, where $A = \pi r^2$.Rearranging for elongation $\Delta l$: $\Delta l = \frac{F \cdot l}{\pi r^2 Y}$.Substituting the values: $\Delta l = \frac{10^5 	imes 1}{3.14159 	imes (0.05)^2 	imes 70 	imes 10^9}$.$\Delta l = \frac{10^5}{3.14159 	imes 0.0025 	imes 70 	imes 10^9} = \frac{10^5}{549.78 	imes 10^6} \approx 1.81 	imes 10^{-4} \,m$.

$A$ cylindrical rod made of aluminum has a length of $1 \,m$ and a diameter of $10 \,cm$. The rod is subjected to a tensile force of $100 \,kN$. Calculate the elongation in the rod. (Young's modulus of aluminum $= 70 \,GPa$)

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