CDS (I) 2023 General Knowledge · Question 6 of 120

Correct — D, 850 m. The sound of the cracker does not travel to the cliff and stop; it reflects and comes back, so in the 5 seconds before the echo is heard it has covered twice the distance to the cliff. Total path is speed times time, 340 multiplied by 5, which is 1700 metres. Half of that is the distance to the cliff, so the cliff stands 850 metres away. Written as a formula, d = v times t divided by 2, and forgetting the division by two is the single mistake this item is built to catch.

• (a)1700 m — The whole path the sound travelled, there and back. It is the right multiplication and the missing halving — exactly the slip the question is fishing for, which is why it sits first in the list.
• (b)170 m — Half the numerical value of the speed. It comes out of dividing 340 by 2 and never using the 5 seconds at all, so the time given in the stem does no work.
• (c)85 m — A quarter of the speed's numerical value. Again the 5 seconds is dropped, and the halving is applied twice over, which leaves an answer twenty times too small.

An echo is sound reflected from a large obstacle and heard as a separate repetition of the original. Because the sensation of a sound persists in the ear for about a tenth of a second, the reflection has to arrive at least that late to be told apart from the original — which at roughly 344 metres per second puts the nearest useful reflecting surface about 17.2 metres away. In any echo calculation the sound covers the obstacle distance twice, so the working distance is speed times time, halved.

Two of the three wrong options are arithmetic near-misses rather than conceptual errors, so the discipline that saves marks is to draw the path before touching the numbers. Sound out, sound back, one time interval covering both. It is also worth remembering that the speed of sound is not a constant of nature the way the speed of light is: in air it is about 331 metres per second at 0 degrees Celsius and rises with temperature, reaching roughly 344 at ordinary room warmth. The paper supplies 340 so that no such judgement is needed.

• In an echo problem the sound covers twice the obstacle distance, so d = v times t divided by 2.
• The sensation of a sound persists in the ear for about 0.1 second, which is why a reflection must be delayed at least that long to be heard separately.
• At about 344 metres per second, that 0.1 second sets the minimum obstacle distance for a distinguishable echo at roughly 17.2 metres.
• The speed of sound in air is about 331 metres per second at 0 degrees Celsius and increases as the air warms.
• The same reflection principle is used by SONAR and by ultrasound scanning, where the returning pulse is timed to fix a distance.

• Using the full 5 seconds for a one-way trip and answering 1700 m.
• Assuming the speed of sound is fixed at 340 m/s regardless of the medium or the temperature.
• Forgetting that an echo needs a delay of about 0.1 second, so no echo is heard from a wall a few metres away.

As a numerical with the halving buried in it, or as a conceptual item on the minimum distance needed to hear an echo, or on the principle behind SONAR.

• practice — not a real PYQ

  A ship sends a sonar pulse towards the sea bed and receives the echo after 4 seconds. If the speed of sound in sea water is 1500 m/s, the depth of the sea bed below the ship is

  • (a)1500 m
  • (b)3000 m
  • (c)6000 m
  • (d)750 m

  Answer(b) 3000 m — the pulse covers twice the depth, so depth = 1500 x 4 / 2 = 3000 metres.

• practice — not a real PYQ

  An echo is heard as a separate sound only if the reflecting surface is far enough away, because

  • (a)sound travels faster near a wall
  • (b)the sensation of a sound persists in the ear for about 0.1 second
  • (c)sound cannot be reflected by a nearby surface
  • (d)the reflected sound has a lower frequency

  Answer(b) the sensation of a sound persists in the ear for about 0.1 second — a reflection arriving sooner merges with the original instead of being heard separately.