A carrier wave of peak voltage 14 V is used for transmitting a message signal. The peak voltage of the modulating signal given to achieve a modulation index of 80% will be:

We first recall the standard relation for an amplitude-modulated (AM) wave. For AM, the modulation index (also called the degree of modulation) is defined as the ratio of the peak voltage of the modulating signal to the peak voltage of the unmodulated carrier. Symbolically, the formula is stated as $$ m \;=\; \frac{V_m}{V_c}, $$ where $$m$$ is the modulation index (dimensionless), $$V_m$$ is the peak voltage of the modulating (message) signal, and $$V_c$$ is the peak voltage of the carrier signal.

We are given that the modulation index is 80 %. Converting this percentage to a pure number, we write $$ 80\% \;=\; \frac{80}{100} \;=\; 0.8. $$ So we have $$m = 0.8.$$ The problem also provides the peak voltage of the carrier wave: $$ V_c = 14\ \text{V}. $$

Now we substitute these known values into the formula: $$ m \;=\; \frac{V_m}{V_c}. $$ Inserting $$m = 0.8$$ and $$V_c = 14\ \text{V},$$ we get $$ 0.8 \;=\; \frac{V_m}{14}. $$

To isolate $$V_m,$$ we multiply both sides of the equation by the denominator $$14$$: $$ V_m \;=\; 0.8 \times 14. $$

Performing the multiplication step by step, we observe that $$ 0.8 \times 14 \;=\; \frac{8}{10} \times 14 \;=\; \frac{8 \times 14}{10}. $$ Calculating the numerator, $$ 8 \times 14 = 112, $$ so we have $$ \frac{112}{10} = 11.2. $$

Hence, $$ V_m = 11.2\ \text{V}. $$

Thus, the peak voltage of the modulating signal required to obtain a modulation index of 80 % is 11.2 V.

Hence, the correct answer is Option C.