Two wires A and B are of the same material. T | vedclass

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(D) Young's modulus $Y$ is given by $Y = \frac{F L}{A \Delta l}$,where $A = \pi r^2$.Thus,the force $F$ is given by $F = \frac{Y A \Delta l}{L} = \fra

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$8:1$

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(D) Young's modulus $Y$ is given by $Y = \frac{F L}{A \Delta l}$,where $A = \pi r^2$.Thus,the force $F$ is given by $F = \frac{Y A \Delta l}{L} = \frac{Y (\pi r^2) \Delta l}{L}$.Since the material is the same,$Y$ is constant. Since the increase in length $\Delta l$ is the same for both wires,we have $F \propto \frac{r^2}{L}$.Therefore,the ratio of the forces is $\frac{F_A}{F_B} = \left( \frac{r_A}{r_B} ight)^2 	imes \left( \frac{L_B}{L_A} ight)$.Given $\frac{L_A}{L_B} = \frac{1}{2}$ and the ratio of diameters is $2:1$,so $\frac{r_A}{r_B} = \frac{2}{1}$.Substituting these values: $\frac{F_A}{F_B} = (2)^2 	imes (2) = 4 	imes 2 = 8$.Thus,the ratio is $8:1$.

Two wires $A$ and $B$ are of the same material. Their lengths are in the ratio $1:2$ and their diameters are in the ratio $2:1$. When stretched by forces $F_A$ and $F_B$ respectively,they experience an equal increase in their lengths. Then the ratio $F_A/F_B$ is:

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