A) 25 kg
B) 5 kg
C) 12.5 kg
D) 1/25 kg
Correct Answer: A
Solution :
The frequency of vibration of a string \[n=\frac{p}{2l}\sqrt{\frac{T}{m}}\] Also number of loops = Number of antinodes. Hence, with 5 antinodes and hanging mass of 9 kg. We have p = 5 and T = 9g Þ \[{{n}_{1}}=\frac{5}{2l}\sqrt{\frac{9g}{m}}\] With 3 antinodes and hanging mass M We have p = 3 and T = Mg Þ \[{{n}_{2}}=\frac{3}{2l}\sqrt{\frac{Mg}{m}}\] \[\because \] n1 = n2 Þ \[\frac{5}{2l}\sqrt{\frac{9g}{m}}=\frac{3}{2l}\sqrt{\frac{Mg}{m}}\]Þ \[M=25\] kg.
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