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Step-by-Step Solution
Step 1: Identify the Formula for Rate of Heat (Power)
The rate at which heat is developed in a wire due to an applied voltage $V$ is given by:
$P = \dfrac{V^2}{R}$
where $R$ is the electrical resistance of the wire.
Step 2: Express the Resistance in Terms of Wire Dimensions
The resistance $R$ of a wire of length $L$, with resistivity $\rho$, and radius $r$ (cross-sectional area $A = \pi r^2$) is:
$R = \dfrac{\rho L}{A} = \dfrac{\rho L}{\pi r^2}$
Step 3: Determine the Dependence of Power on Dimensions
From Step 1, $P \propto \dfrac{1}{R}$. Combining with the expression for $R$, we have:
$P \propto \dfrac{1}{ \dfrac{\rho L}{\pi r^2} } = \dfrac{\pi r^2}{\rho L}$
Since $\pi$ and $\rho$ are constants for a given material, we see that:
$P \propto \dfrac{r^2}{L}$
Step 4: Apply the Changes (Length Halved, Radius Doubled)
Let the initial length be $L$ and initial radius be $r$. If the length is halved to $L/2$, and the radius is doubled to $2r$, then:
New power $P_{\text{new}} \propto \dfrac{(2r)^2}{\,L/2\,} = \dfrac{4r^2}{\,L/2\,} = \dfrac{4r^2}{1}\cdot \dfrac{2}{L} = \dfrac{8r^2}{L}$
Compared to the original $P \propto \dfrac{r^2}{L}$, this is an 8-fold increase.
Step 5: Conclusion
Therefore, the rate of heat developed in the wire under the new dimensions increases 8 times.
