Question 1 of 9 What is the distance between node and antinode in a stationary wave in terms of wavelength? A λ/2 B λ/4 C λ D 2λ 💡 Explanation Adjacent nodes (or antinodes) are λ/2 apart, so a node and its neighboring antinode are half of that: λ/4. Question 2 of 9 Which of the following is a mechanical wave? A Radio waves B Light waves C Ultra violet D Sound waves 💡 Explanation Mechanical waves need a material medium to travel; sound waves propagate through the vibration of particles in air, water, or solids, unlike radio waves, light, and ultraviolet, which are electromagnetic waves that can travel through vacuum. Question 3 of 9 Which of the following is an example of transverse wave? A Sound wave in air B Waves inside the water C Wave produced in the stem of vibrating tuning fork D Wave produced in the stretched string 💡 Explanation In a transverse wave, particles vibrate perpendicular to the direction of wave travel, as seen in a stretched string; sound waves (in air, water, or a tuning fork's stem) are longitudinal, with particles vibrating parallel to the wave's direction. Question 4 of 9 When sound travels from air to water, which remains constant? A wavelength B frequency C velocity D phase angle 💡 Explanation Frequency is determined by the source and stays unchanged as the wave crosses into a new medium; wavelength and velocity change because water's density and elastic properties differ from air's. Question 5 of 9 What is the distance between two consecutive particles in a wave which are in the same phase? A λ/4 B λ/2 C 3λ/4 D λ 💡 Explanation Particles in phase repeat their motion identically at intervals of one full wavelength, so the nearest two particles vibrating exactly in phase are separated by λ. Question 6 of 9 Distance between two consecutive nodes in a standing wave is A λ B λ/2 C λ/4 D 2λ 💡 Explanation Nodes occur at every half-wavelength interval along a standing wave, so consecutive nodes (and likewise consecutive antinodes) are always λ/2 apart. Question 7 of 9 The frequency of tuning fork is 384 Hz and velocity of sound in air is 352 m/s. How far has the sound traversed while the fork completes 36 vibrations? A 33 m B 36 m C 352 m D 72 m 💡 Explanation Time for 36 vibrations [latex]t = \dfrac{36}{384\text{ Hz}} = 0.09375\text{ s}[/latex], so distance [latex]= v \times t = 352 \times 0.09375 = 33\text{ m}[/latex]. Question 8 of 9 When sound wave is refracted from air to water, which of the following will remain unchanged? A frequency B wave length C wave velocity D wave number 💡 Explanation Frequency is set by the source and doesn't change when a wave crosses a boundary; wavelength, velocity, and wave number all change because they depend on the medium's properties. Question 9 of 9 The distance between any two consecutive antinodes in a stationary wave is A [latex]\pi[/latex] B [latex]\pi/2[/latex] C [latex]\pi/4[/latex] D [latex]2\pi[/latex] 💡 Explanation Antinodes in a stationary wave repeat at half-wavelength intervals, matching the standard result that consecutive antinodes (like consecutive nodes) are separated by half a wavelength. ← Prev Check Next → Finish & See Score 0 of 9 checked ← Back 1 / 9 1 2 3 4 5 6 7 8 9
