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(D) The formula for the extension in length $(\Delta L)$ of a wire is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the or

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(D) The formula for the extension in length $(\Delta L)$ of a wire is given by $\Delta L = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the original length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since $A = \pi r^2$,where $r$ is the radius of the wire,we have $\Delta L = \frac{FL}{\pi r^2 Y}$.Given that $F$,$L$,and $Y$ are constant,the extension is inversely proportional to the square of the radius: $\Delta L \propto \frac{1}{r^2}$.Since the diameter $D = 2r$,the relationship also holds for the diameter: $\Delta L \propto \frac{1}{D^2}$.Let $\Delta L_1 = 2.4\, cm$ and $D_1 = 0.4\, mm$. If the diameter is doubled,$D_2 = 2D_1$.Then,$\frac{\Delta L_2}{\Delta L_1} = \left( \frac{D_1}{D_2} \right)^2 = \left( \frac{D_1}{2D_1} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.Therefore,$\Delta L_2 = \frac{\Delta L_1}{4} = \frac{2.4\, cm}{4} = 0.6\, cm$.

When a weight of $10\, kg$ is suspended from a copper wire of length $3\, m$ and diameter $0.4\, mm$,its length increases by $2.4\, cm$. If the diameter of the wire is doubled,then the extension in its length will be ........ $cm$.

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