Some heat is developed in a uniform metallic wire across the ends of which a constant voltage V is applied. The heat developed will be tripled if

both the radius and the length of the wire are tripled

The correct answer is Option (4) → both the radius and the length of the wire are tripled

Heat developed in a resistor (wire) under constant voltage: $H \propto I^2 R t$, and $I = V/R$, so $H \propto V^2 / R$

Resistance of wire: $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$

So, $H \propto \frac{V^2}{R} \propto \frac{V^2 \pi r^2}{L} \propto \frac{r^2}{L}$

We want $H_\text{new} = 3 H_\text{old}$, i.e., $\frac{r_\text{new}^2}{L_\text{new}} = 3 \frac{r^2}{L}$

Check options:

1. Radius tripled ($r_\text{new} = 3r$), $L$ same → $H \propto 9 r^2 / L = 9 H_\text{old}$

2. Length tripled ($L_\text{new} = 3L$), $r$ same → $H \propto r^2 / 3L = H_\text{old}/3$

3. Both radius and length reduced to 1/3: $r_\text{new} = r/3$, $L_\text{new} = L/3$ → $H \propto (r^2/9)/(L/3) = r^2 /3L = H_\text{old}/3$

4. Both radius and length tripled: $r_\text{new} = 3r$, $L_\text{new} = 3L$ → $H \propto (9 r^2)/(3 L) = 3 r^2 / L = 3 H_\text{old}$ → correct

Correct option: both the radius and the length of the wire are tripled.