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(D) The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.Thus,$R = \rho \frac{L}{\pi d^2

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$16$

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(D) The resistance of a wire is given by $R = ho \frac{L}{A}$,where $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.Thus,$R = ho \frac{L}{\pi d^2 / 4} = \frac{4 ho L}{\pi d^2}$.When a wire is drawn to reduce its diameter,its volume $V = A 	imes L$ remains constant.Since $V = (\pi d^2 / 4) 	imes L$ is constant,$L \propto \frac{1}{d^2}$.Substituting this into the resistance formula: $R \propto \frac{L}{d^2} \propto \frac{1/d^2}{d^2} = \frac{1}{d^4}$.If the diameter is reduced to half $(d' = d/2)$,the new resistance $R'$ is:$R' = R 	imes (d/d')^4 = R 	imes (d / (d/2))^4 = R 	imes (2)^4 = 16R$.Therefore,the resistance becomes $16$ times the original value.

When a piece of aluminium wire of finite length is drawn through a series of dies to reduce its diameter to half its original value,its resistance will become ........ times.

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