In resonance tube exp. we find l1=25.0 cm and l2=75.0 cm. If there is no error in frequency, what will be maximum permissible error in speed of sound. (Take f0=325Hz) V=2f0(l2−l1)
A
V = ( 329 ± 1.3) m/s
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B
V = ( 326 ± 1.3) m/s
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C
V = ( 325 ± 1.4) m/s
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D
V = ( 325 ± 1.3) m/s
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Solution
The correct option is DV = ( 325 ± 1.3) m/s (dv)=2f0(dl2−dl1) (ΔV)max=maxof[2f0(±Δl2∓Δl2)=2f0(Δl2+Δl1) l1=25.0cm⇒Δl1=0.1cm (place value of last number) l1=75.0cm⇒Δl2=0.1 cm (place value of last number) So max permissible error in speed of sound (ΔV)max = 2(325 Hz) (0.1 cm + 0.1 cm) = 1.3 m/s Value of V = 2f0(l2−l1) = 2(325 Hz) (75.0 cm - 25.0 cm) 325 m/s So V = ( 325 ± 1.3) m/s