NCERT Solutions for Class 9th Science Chapter 10 Chapter opener — Think It Over

Book page 184 Updated on2026-09-08

Q1.
Two astronauts are repairing the arm of a space station together during a spacewalk. Can they talk to each other and hear the sounds of metal clanking as they do on the Earth?
Answer

No — not directly. Outer space is a near vacuum, and a sound wave has nothing to travel in there.

Why it happens: sound is a series of compressions and rarefactions — regions where the particles of a medium are pushed closer together and pulled further apart. A particle can only pass the disturbance on by colliding with a neighbouring particle. Where there are almost no particles, there are no collisions, so there is nothing to carry the disturbance. This is exactly what the vacuum bell jar experiment (Fig. 10.7) shows: as the air is pumped out, the bell is still seen ringing but the sound fades to nothing.

On the Earth the same clank is heard easily because the air between the two astronauts is a perfectly good medium.

Did you know? Astronauts talk over radio sets built into their suits. Radio waves are electromagnetic waves, not mechanical waves, so they need no medium and cross the vacuum happily. If two helmets are pressed together, the clank can be heard — the solid helmet material then acts as the medium.
Q2.
How do most bats use sound to locate their prey in the dark at night?
Answer

Most bats send out short bursts of ultrasonic waves (frequency above 20 kHz), and then listen for the echoes that bounce back from objects and prey. This is called echolocation.

  • The bat emits a pulse of ultrasound.
  • The pulse is reflected by an insect, a wall or a branch.
  • The bat's ears pick up the returning echo.
  • From the time delay the bat judges how far the object is; from the direction and strength of the echo it judges where the object is and how big it is.
distance to prey = v × t / 2  (the sound travels there and back)
For an echo returning in 0.02 s in air, distance = 340 m s⁻¹ × 0.02 s ÷ 2 = 3.4 m
Why ultrasound and not ordinary sound: a shorter wavelength reflects cleanly from small objects. At 40 kHz in air, λ = 344 m s⁻¹ ÷ 40000 s⁻¹ ≈ 0.0086 m ≈ 0.9 cm — small enough to bounce off a mosquito-sized insect. A 100 Hz sound has λ ≈ 3.4 m and would simply flow around it.
Did you know? Dolphins, whales and some birds also echolocate, and the same principle is used by humans in sonar and in ultrasonography.
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