A) \[25\text{ }kg\]
B) \[5\text{ }kg\]
C) \[12.5\text{ }kg\]
D) \[\frac{1}{25}\,\,kg\]
Correct Answer: A
Solution :
\[n=\frac{P}{2\ell }\,\sqrt{\frac{T}{m}}\] No. of loops = No. of antinodes \[{{n}_{1}}=\frac{5}{2\ell }\sqrt{\frac{9g}{m}},\] \[{{n}_{2}}=\frac{3}{2\ell }\sqrt{\frac{Mg}{m}}\] \[\because \] \[{{n}_{1}}={{n}_{2}}\Rightarrow \,\frac{5}{2\ell }\sqrt{\frac{9g}{m}}=\frac{3}{2\ell }\sqrt{\frac{Mg}{m}}\Rightarrow M=25\,kg\]You need to login to perform this action.
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