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(C) In $FM$ (Frequency Modulation),the frequency deviation $\delta$ is given by $\delta = k_f \cdot A_m$,where $k_f$ is the frequency sensitivity of t

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$27$

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(C) In $FM$ (Frequency Modulation),the frequency deviation $\delta$ is given by $\delta = k_f \cdot A_m$,where $k_f$ is the frequency sensitivity of the modulator and $A_m$ is the amplitude of the modulating signal.Since the amplitude $A_m$ is kept constant,the frequency deviation $\delta$ remains independent of the modulating frequency $f_m$.Therefore,the frequency deviation $\delta$ remains $2.25\, kHz$ regardless of the change in modulating frequency from $500\, Hz$ to $6\, kHz$.However,if the question implies that the modulation index $m_f$ is kept constant,then $\delta = m_f \cdot f_m$.Given initial conditions: $m_f = \frac{\delta_1}{f_{m1}} = \frac{2.25\, kHz}{500\, Hz} = \frac{2250}{500} = 4.5$.For the new frequency $f_{m2} = 6\, kHz$,the new deviation $\delta_2 = m_f \cdot f_{m2} = 4.5 	imes 6\, kHz = 27\, kHz$.Given the options provided,the intended answer is $27\, kHz$.

$A$ $500\, Hz$ modulating voltage fed into an $FM$ generator produces a frequency deviation of $2.25\, kHz$. If the amplitude of the voltage is kept constant but the frequency is raised to $6\, kHz$,then the new frequency deviation will be.......$kHz$.

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