Question

An oil – tanker explodes on the surface of the sea. At that instant, an aircraft and a submarine were present vertically above and below the oil – tanker respectively at equal distances from it. Find the ratio(air:water) of the time taken by sound waves to reach them. (Speed of sound in air = $340m{s}^{-1}$ and speed of sound in seawater= $1360m{s}^{-1}$)

• A. $1:2$
• B. $2:1$
• C. $1:4$
• D. $4:1$

Answer 1

The correct option is D

$4:1$

Let the aircraft and the submarine be ‘x’ m above and below the tanker respectively.
Speed of sound in air = $340m{s}^{-1}$
Time taken by sound to reach the aircraft is $t_{air}=\frac{x}{340}=\frac{x}{340}s$
Speed of sound in seawater= $1360m{s}^{-1}$
Time taken by sound to reach the submarine $t_{water}=\frac{x}{1360}=\frac{x}{1360}s$
Ratio of the time taken
$\frac{t_{air}}{t_{water}}=\frac{\left(\frac{x}{340}\right)}{\left(\frac{x}{1360}\right)}=\frac{1360}{340}=4$

Hence the ratio of time taken in air and water is $4:1$