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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Sound Waves

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Sound Waves

To detect tumours in soft tissues, ultrasonic waves with a...

To detect tumours in soft tissues, ultrasonic waves with a frequency of 4.5 MHz4.5 \text{ MHz} are utilized. The speed of sound is 1.5 km s−11.5 \text{ km s}^{-1} in tissue and 340 m s−1340 \text{ m s}^{-1} in air. Determine the wavelength of these ultrasonic waves in air and in tissue, respectively.

Options

  1. A3.3×10−4 m3.3 \times 10^{-4} \text{ m}, 7.6×10−5 m7.6 \times 10^{-5} \text{ m}
  2. B7.6×10−5 m7.6 \times 10^{-5} \text{ m}, 3.3×10−4 m3.3 \times 10^{-4} \text{ m}
  3. C7.6×10−4 m7.6 \times 10^{-4} \text{ m}, 3.3×10−5 m3.3 \times 10^{-5} \text{ m}
  4. D1.5×10−3 m1.5 \times 10^{-3} \text{ m}, 3.4×10−4 m3.4 \times 10^{-4} \text{ m}

Correct answer

B. 7.6×10−5 m7.6 \times 10^{-5} \text{ m}, 3.3×10−4 m3.3 \times 10^{-4} \text{ m}

Step-by-step solution

Given frequency, f=4.5 MHz=4.5×106 Hzf = 4.5 \text{ MHz} = 4.5 \times 10^{6} \text{ Hz}

Speed of sound in air, va=340 m s−1v_{a} = 340 \text{ m s}^{-1}

Speed of sound in tissue, vt=1.5 km s−1=1500 m s−1v_{t} = 1.5 \text{ km s}^{-1} = 1500 \text{ m s}^{-1}

The wavelength of the ultrasonic waves in air is given by:

λa=vaf=3404.5×106 m\lambda_{a} = \dfrac{v_{a}}{f} = \dfrac{340}{4.5 \times 10^{6}} \text{ m}

λa≈75.5×10−6 m≈7.6×10−5 m\lambda_{a} \approx 75.5 \times 10^{-6} \text{ m} \approx 7.6 \times 10^{-5} \text{ m}

The wavelength of the ultrasonic waves in tissue is given by:

λt=vtf=15004.5×106 m\lambda_{t} = \dfrac{v_{t}}{f} = \dfrac{1500}{4.5 \times 10^{6}} \text{ m}

λt≈333.3×10−6 m≈3.3×10−4 m\lambda_{t} \approx 333.3 \times 10^{-6} \text{ m} \approx 3.3 \times 10^{-4} \text{ m}

Thus, the wavelengths in air and tissue are 7.6×10−5 m7.6 \times 10^{-5} \text{ m} and 3.3×10−4 m3.3 \times 10^{-4} \text{ m} respectively.
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