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Consider a fuse wire of length $l$ and radius $r$. The time of heating ($t$) for passing the maximum current will depend on

This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$t \propto r^4 l^0$

Fuse Wire Heating Time Dependence Explained

The time ($t$) it takes for a fuse wire to heat up and melt depends on the heat generated by the current passing through it and the heat required to melt the wire's material.

Deriving Heating Time Dependence

1. Heat Generated ($H$): According to Joule's law, the heat generated in a conductor is given by $H \propto I^2 R t$, where $I$ is the current, $R$ is the resistance, and $t$ is the time.

2. Resistance ($R$): The resistance of a fuse wire is calculated as $R = \frac{\rho l}{A}$, where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area. Since the area $A = \pi r^2$, we have $R \propto \frac{l}{r^2}$.

3. Heat Required for Melting ($H_{melt}$): The heat required to melt the fuse wire is proportional to its mass ($m$). The mass is given by $m = \text{density} \times \text{Volume}$. The volume ($V$) of the wire is $V = A \cdot l = \pi r^2 l$. Therefore, $H_{melt} \propto m \propto V \propto r^2 l$.

4. Time Calculation: The fuse melts when the heat generated equals the heat required for melting ($H = H_{melt}$). So, we have the proportionality:

$I^2 R t \propto r^2 l$

Substituting the proportionality for $R$ ($R \propto \frac{l}{r^2}$):

$I^2 \left( \frac{l}{r^2} \right) t \propto r^2 l$

Now, we solve for $t$:

$t \propto \frac{r^2 l}{I^2 (l/r^2)}$

$t \propto \frac{r^2 l \cdot r^2}{I^2 l}$

$t \propto \frac{r^4 l}{I^2 l}$

The length term $l$ cancels out:

$t \propto \frac{r^4}{I^2}$

Final Dependence

The question asks for the dependence when passing the *maximum current*. This implies $I$ is a fixed value. Therefore, the time of heating ($t$) is proportional to the fourth power of the radius ($r^4$) and has no dependence on the length ($l^0$).

$t \propto r^4 l^0$

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