Question

When a long hollow steel pipe is struck with a hammer at one end, two sounds are heard at the other end. If the time interval between the two sounds heard is 0.53 s, then the length of the pipe is
(Speed of sound in steel is 5100 $ms^{-1}$ and speed of sound in air is 330 $ms^{-1}$)

• A. 153 m
• B. 174 m
• C. 187 m
• D. 270 m

Hint:

Speed of sound order: Solids>Liquids>Gases. Always subtract the smaller time ($L/v_{fast}$) from the larger time ($L/v_{slow}$).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Sound travels faster in solids (steel) than in gases (air). Thus, the sound traveling through the steel pipe arrives first, and the sound traveling through the air inside the pipe arrives later.
Step 2: Formulation:
Let $L$ be the length of the pipe. Time taken in air, $t_a = \frac{L}{v_{air}}$. Time taken in steel, $t_s = \frac{L}{v_{steel}}$. Given time difference $\Delta t = t_a - t_s = 0.53$ s.
Step 3: Calculation:
\[ \frac{L}{330} - \frac{L}{5100} = 0.53 \] \[ L \left( \frac{1}{330} - \frac{1}{5100} \right) = 0.53 \] \[ L \left( \frac{5100 - 330}{330 \times 5100} \right) = 0.53 \] \[ L \left( \frac{4770}{1683000} \right) = 0.53 \] \[ L = \frac{0.53 \times 1683000}{4770} \] \[ L = \frac{0.53 \times 330 \times 5100}{4770} \] \[ L = \frac{174.9 \times 5100}{4770} \approx \frac{175 \times 5100}{4770} \] Calculation check: $0.53 \times 330 = 174.9$. $174.9 / 4770 \approx 0.03666$. $0.03666 \times 5100 \approx 187$.
Step 4: Final Answer:
The length of the pipe is 187 m.