Wires A and B are made from the same material | vedclass

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(B) The energy stored in a stretched wire is given by $U = \frac{1}{2} F \Delta L = \frac{F^2 L}{2AY}$,where $F$ is the tension,$L$ is the length,$A$

Vedclass · Accepted Answer

$3:4$

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(B) The energy stored in a stretched wire is given by $U = \frac{1}{2} F \Delta L = \frac{F^2 L}{2AY}$,where $F$ is the tension,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since $A = \pi r^2$,we have $U \propto \frac{L}{r^2}$ because $F$ and $Y$ are constant for both wires.Given: $L_A = 3 L_B$ and $d_A = 2 d_B$ (which implies $r_A = 2 r_B$).Therefore,the ratio of energy stored is $\frac{U_A}{U_B} = \left( \frac{L_A}{L_B} ight) 	imes \left( \frac{r_B}{r_A} ight)^2$.Substituting the values: $\frac{U_A}{U_B} = (3) 	imes \left( \frac{1}{2} ight)^2 = 3 	imes \frac{1}{4} = \frac{3}{4}$.

Wires $A$ and $B$ are made from the same material. $A$ has twice the diameter and three times the length of $B$. If the elastic limits are not reached,when each is stretched by the same tension,the ratio of energy stored in $A$ to that in $B$ is

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