A) \[\frac{1500}{23}cm,\frac{2000}{23}cm\]
B) \[\frac{1500}{23}cm,\frac{500}{23}cm\]
C) \[\frac{1500}{23}cm,\frac{300}{23}cm\]
D) \[\frac{300}{23}cm,\frac{1500}{23}cm\]
Correct Answer: A
Solution :
[a] From formula, \[f=\frac{1}{x}\sqrt{\frac{T}{m}}\Rightarrow \frac{1}{f}\propto l\] \[\therefore \,\,{{l}_{1}}:{{l}_{2}}:{{l}_{3}}=\frac{1}{{{f}_{1}}}:\frac{1}{{{f}_{2}}}:\frac{1}{{{f}_{3}}}={{f}_{2}}{{f}_{3}}:{{f}_{1}}{{f}_{3}}:{{f}_{1}}{{f}_{2}}\] \[\left[ Given:{{f}_{1}}:{{f}_{2}}:{{f}_{3}}=1:3:5 \right]\] =15: 5: 3 Therefore the positions of two bridges below the wire are \[\frac{15\times 100}{15+5+3}cm\,\,\,and\,\,\,\frac{15\times 100+5\times 100}{15+5+3}cm\] i.e., \[\frac{1500}{23}cm,\frac{2000}{23}cm\]
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