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IV. Numerical Problems · Q1

Q.In a submarine equipped with sonar, the time delay between the generation of a pulse and its echo after reflection from an enemy submarine is observed to be 80 s. If the speed of sound in water is 1460 m s−1^{-1}. What is the distance of enemy submarine?

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✓ Free question

Step 1. Given: round-trip time t=80t=80 s; speed of sound in water v=1460v=1460 m s−1^{-1}.

Step 2. Since the sonar pulse travels to the submarine and its echo travels back, the distance dd is covered TWICE in time tt: 2d=v t2d=v\,t.

Step 3. d=v t2=1460 m s−1×80 s2=116800 m2=58400 md=\dfrac{v\,t}{2}=\dfrac{1460\ \text{m s}^{-1}\times80\ \text{s}}{2}=\dfrac{116800\ \text{m}}{2}=58400\ \text{m}.

Step 4. Converting to km: 58400 m=58.40 km58400\ \text{m}=58.40\ \text{km}.

✓Final answer

Distance of the enemy submarine =58.40=58.40 km

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