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Q1. The frequency and velocity of sound wave are 600 Hz and 360 m/s respectively. Phase difference between two particles of medium are 60°, the minimum distance between these two particles will be
•  50 cm
•  20 cm
•  15 cm
•  10 cm
Solution
Velocity of wave (v)=360 m/s Frequency, n= 600Hz Phase difference, ∆Φ=60ο If the minimum distance between two points is ∆x, then ∆x=λ/2π×∆Φ Δx=v/2πn×∆Φ Or ∆x=360/(2π×600)×60 ∆x=360/(2π×600)×π/3 ∆x=1/10 m ∆x=10 cm

Q2.An observer is standing 500 m away from a vertically hill. Starting between the observer and the hill a police van having a siren of frequency 1000 Hz moves towards the hill with a uniform speed. If the frequency of the sound heard directly from the siren is 970 Hz, the frequency of the sound heard after reflection from the hill (in Hz) is about, (velocity of sound =330 ms^(-1))
•  1042/span>
•  1032
•  1012
•  1022
Solution
Sound geard directly v_1=v_o (v/(v-v_s )) ∴970=1000(330/(330+v_s )) Or v_s=10.2ms^(-1) The frequency of reflected sound is given by v_2=v_o (v/(v-v_s ))=1000(330/(330-10.2)) =(1000×330)/319.8≈1032Hz

Q3.  An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. What is the percentage increases in the apparent frequency?
•   0
•  0.5%
•  5%
•  20%
Solution
Given, v_o=v/5 ⟹ v_o=320/5=64ms^(-1) When observers moves towards the stationary source, then n^'=((v+v_o)/v)n n^'=((320+64)/320)n n^'=(384/320)n n^'/n=384/320 Hence, percentage increases ((n^'-n)/n)=((384-320)/320×100)% =(64/320×100)%=20%

Q4. With the propagation of a longitudinal wave through a material medium, the quantities transmitted in the propagation direction are
•  Energy, momentum and mass
•  Energy
•  Energy and mass
•  Energy and linear momentum
Solution
D

Q5.When a tuning fork vibrates, the waves produced in the fork are
•  Stationary
•  Progressive
•  Transverse
•  Longitudinal
Solution
A

Q6. A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is
•  4
•  8
• 6
•  12
Solution

Q7.A 1000 Hz sound wave in air strikes the surface of a lake and penetrates into water. If speed of sound in water is 1500ms^(-1),the frequency and wavelength of waves in water are
•  1500 Hz, 1m
•  1500 Hz, 1.5m
•  1000 Hz, 1.5m
•  1000 Hz, 1m
Solution
Frequency remains the same ie 1000 Hz wavelength chances λ_ω=v_ω/V=1500/1000=1.5 m

Q8.If in a resonance tube a oil of density higher than that water is used then at the resonance frequency would
•  Increase
•  Remain same
•  Decrease
•  Slightly increase
Solution
In a resonance tube, water works as a reflector and the resonance frequency is independent of the substance (liquid) which is filled in the tube.

Q9.Three sources of equal intensities with frequencies 400, 401 and 402 vib/s are sounded together. The number of beats/s is
•  0
•  4
•  1
•  2
Solution
Beats are the periodic and repeating functions heard in the intensity of sound, when two sound waves of very similar frequency interface with ine another. Beats = defference in frequencies. Maximum number of beats =402-400=2

Q10. Figure here shown an incident pulse P reflected from a rigid support. Which one of A,B,C,D represents the reflected pulse correctly
Solution
When pulse is reflected from a rigid support, the pulse is inverted both lengthwise and sidewise #### Written by: AUTHORNAME

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: WAVE MOTION QUIZ-2
WAVE MOTION QUIZ-2
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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE
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