If the length of the filament of a heater is reduced by $10\%$, the power of the heater will

The resistance ($R$) of a filament wire is directly proportional to its length ($l$), given by the formula:
$R=\rho \frac{l}{A}$

When the length is reduced by $10\%$, the new length becomes $l^{\prime }=0.9l$. Consequently, the new resistance becomes:
$R^{\prime }=0.9R$

The power ($P$) dissipated by a heater connected across a constant voltage source ($V$) is:
$P=\frac{V^2}{R}$

The new power ($P^{\prime }$) after the length is reduced will be:
$P^{\prime }=\frac{V^2}{R^{\prime }}=\frac{V^2}{0.9R}=\frac{P}{0.9}\approx 1.111P$

The fractional or percentage increase in power is calculated as:

$$\text{Percentage Increase}=\left(\frac{P^{\prime }-P}{P}\right)\times 100\%=\left(\frac{1}{0.9}-1\right)\times 100\%=\frac{0.1}{0.9}\times 100\%\approx 11.11\%$$

Therefore, the power of the heater increases by about $11\%$.