2.A bat echolocates and receives an echo 10 milliseconds later. Assuming a
speed of sound in air of 340 m/s, how far away is the object?
3.4 m
1.7 m
6.8 m
5.1 m

Bats use echolocation to detect objects by emitting ultrasonic pulses and measuring echo return time, where distance equals half the total path since sound travels to the object and back. For a 10 ms echo delay and sound speed of 340 m/s, the object is 1.7 m away.[execute_python]

Echolocation Distance Formula

The core formula derives from speed = distance/time, adjusted for round trip: d=v×t2d=2v×t, where $v=340$ m/s is speed of sound and $t$ is echo time. Convert 10 ms to seconds: $t=0.01$ s. Total distance traveled by sound: $340\times 0.01=3.4$ m. One-way object distance: $3.4/2=1.7$ m.[execute_python]

Option Analysis

• 3.4 m: Matches total round-trip distance (340 m/s × 0.01 s) but ignores division by 2 for one-way path, overestimating by factor of two.

• 1.7 m: Correct answer, as it applies full formula d=340×0.012=1.7d=2340×0.01=1.7 m.[execute_python]

• 6.8 m: Doubles correct total distance (perhaps using 20 ms), incorrect for given 10 ms.

• 5.1 m: No direct calculation yields this; possibly confuses speed (e.g., 510 m/s total path), not matching parameters.

This problem tests acoustics in bat biology, common in CSIR NET Life Sciences for integrating physics with animal physiology. Speed of 340 m/s assumes standard air conditions.