JEE PHYSICS MCQ: WAVES

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Application +4 | -1

A tuning fork of frequency

340 \mathrm{~Hz}

is vibrated just above a cylindrical tube of length

120 \mathrm{~cm}

. Water is slowly poured in the tube. If the speed of sound in air is

340 \mathrm{~m} / \mathrm{s}

. Find the minimum height of water required for resonance.

(v=

340 \mathrm{~m} / \mathrm{s}

)

View answer and explanation

Correct option: A

45 cm

Option A (45 cm) is correct because it represents the minimum height of water required for resonance, calculated using the formula for a closed pipe.

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3\,\,kHz

0.5

f\mathrm{ }Hz

V\mathrm{ }m/sec

0.2\,\,V

y = \frac{3}{{{{\left( {2x + 3t} \right)}^2} + 4}},

(– \mathrm{ve}) \,x

y = f(kx +\omega t),

X

0.01\,\,s

5 \mathrm{~m}

245 \mathrm{~g}

1 \mathrm{~kg}

g=980 \mathrm{~cm} \mathrm{~s}^{-2}

T

V