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A radio can tune into any station in the 6 MHz to 10 MHz band. Find thecorresponding wavelength band?
Question

A radio can tune into any station in the 6 MHz to 10 MHz band. Find thecorresponding wavelength band?

A.

100 m to 60 m

B.

50 m to 30 m

C.

40 m to 25 m

D.

36 m to 100 m

Correct option is B

We use the relationship between wavelength and frequency of an electromagnetic wave:λ×f=cWhere: λ=Wavelength of the wave f=Frequency of the wave c=3×108 m/s=Speed of light in vacuum\text{We use the \textbf{relationship between wavelength and frequency} of an electromagnetic wave:} \\\quad \lambda \times f = c \\\\\text{Where:} \\\quad \bullet\ \lambda = \text{Wavelength of the wave} \\\quad \bullet\ f = \text{Frequency of the wave} \\\quad \bullet\ c = 3 \times 10^8\, \text{m/s} = \text{Speed of light in vacuum}

Given: Lower frequency f1=6 MHz=6×106 Hz Upper frequency f2=10 MHz=10×106 HzCalculate minimum wavelength (at highest frequency)λ2=cf2=3×10810×106=30 m maximum wavelength (at lowest frequency)λ1=cf1=3×1086×106=50 mFinal Answer:Wavelength band=50 m to 30 m\textbf{Given:} \\\quad \bullet\ \text{Lower frequency } f_1 = 6\, \text{MHz} = 6 \times 10^6\, \text{Hz} \\\quad \bullet\ \text{Upper frequency } f_2 = 10\, \text{MHz} = 10 \times 10^6\, \text{Hz} \\\\\textbf{Calculate minimum wavelength (at highest frequency)} \\\quad \lambda_2 = \frac{c}{f_2} = \frac{3 \times 10^8}{10 \times 10^6} = 30\, \text{m} \\\\\textbf{ maximum wavelength (at lowest frequency)} \\\quad \lambda_1 = \frac{c}{f_1} = \frac{3 \times 10^8}{6 \times 10^6} = 50\, \text{m} \\\\\textbf{Final Answer:} \\\boxed{\text{Wavelength band} = 50\, \text{m to } 30\, \text{m}}​​​

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