Let the original resistance of the heater coil be $R$.

Resistance of a wire is proportional to its length: $R \propto L$ So if the coil is cut into two equal parts, the resistance of each part becomes: $R' = \frac{R}{2}$

The heater is connected to the same supply voltage $V$. Heat produced in time $t$ is: $H = \frac{V^2}{R}t$

Originally, $H_1 = \frac{V^2}{R}t$

Using only one half of the coil, the new resistance is $\frac{R}{2}$. Hence, $H_2 = \frac{V^2}{R/2}t = \frac{2V^2}{R}t = 2H_1$

Therefore, the heat generated becomes double the original value.

Option check:

• A: four times $\rightarrow$ incorrect
• B: doubled $\rightarrow$ correct
• C: halved $\rightarrow$ incorrect
• D: one fourth $\rightarrow$ incorrect