1
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to :
A
1 m/s
B
$${1 \over 8}$$ m/s
C
$${1 \over 4}$$ m/s
D
$${1 \over 2}$$ m/s
2
JEE Main 2020 (Online) 7th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm2) is 90 ms-1. If the Young's modulus of wire is 16 $$ \times $$ 1011 Nm-2, the extension of wire over its natural length is :
A
0.03 mm
B
0.04 mm
C
0.02 mm
D
0.01 mm
3
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound ? [Given reference intensity of sound as 10–12 W/m2 ]
A
20 cm
B
10 cm
C
40 cm
D
30 cm
4
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound (v) in air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column, l1 = 30 cm and l2 = 70 cm. Then, v is equal to -
A
338 ms–1
B
384 ms–1
C
379 ms–1
D
332 ms–1
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