(b) $m_{f} = \frac{\delta}{f_{m}} = \frac{2250}{500} = 4.5$ $\therefore$ New deviation $= 2\left( m_{f}\ f_{m} \right) = 2 \times 4.5 \times 6 = 54\ kHz$
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Ph.D. in Physics with 19+ years of teaching experience.