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In a resonance tube with tuning fork of frequency `512 Hz`, first resonance occurs at water level equal to `30.3cm` and second resonance ocuurs at `63.7cm`. The maximum possible error in the speed of sound isA. `5.12 cm//s`B. `102.4 cm//s`C. `204.8 cm//s`D. `153.6 cm//s` |
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Answer» Correct Answer - C For first the resonance `l_(1) + r = (lamda)/(4)` But `v = f lamda` `:. v = f4 (l_(1) + e)` `rArr l_(1) + e = (v)/(4 f)`…(i) For the second resonance `l_(2) + e = (3 lamda)/(4)` `:. v = f(4)/(3) (l_(2) +e)` `rArr l_(2) + e =(3v)/(4f)` ....(ii) From Eqs. (i) and (ii), we get `v = 2f(l_(2) -l_(1))` `:. Delta v = 2f(Delta l_(2) + Delta l_(1))` =`2 xx 512 xx (0.1 + 0.1) cm//s` =`204.8 cm//s`. |
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