A) 30 cm and 60 cm
B) 40 cm and 80 cm
C) 60 cm and 90 cm
D) 30 cm and 90 cm
Correct Answer: C
Solution :
Given, \[L=110\text{ }cm\] \[{{v}_{1}}:{{v}_{2}}:{{v}_{3}}=1:2:3\] Frequency \[v=\frac{1}{2l}\sqrt{\frac{T}{m}}\] \[\Rightarrow \] \[v\propto \frac{1}{l}\] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\] \[\frac{1}{2}=\frac{{{l}_{2}}}{{{l}_{1}}}\] \[\therefore \] \[{{l}_{2}}=\frac{{{l}_{1}}}{2}\] ?.(i) \[\frac{{{v}_{1}}}{{{v}_{3}}}=\frac{{{l}_{3}}}{{{l}_{1}}}\] \[\frac{1}{3}=\frac{{{l}_{3}}}{{{l}_{1}}}\] \[{{l}_{3}}=\frac{{{l}_{1}}}{3}\] ?.(ii) \[\therefore \] \[L={{l}_{1}}+{{l}_{2}}+{{l}_{3}}\] \[110={{l}_{1}}+\frac{{{l}_{1}}}{2}+\frac{{{l}_{1}}}{3}\] \[=\frac{6{{l}_{1}}+3{{l}_{1}}+2{{l}_{1}}}{6}=\frac{11{{l}_{1}}}{6}\] \[\therefore \] \[{{l}_{1}}=60cm\] \[{{l}_{2}}=\frac{60}{2}=30cm\] [From Eq. (i)], Therefore, bridges should be placed at \[60cm\]and \[60+30=90\text{ }cm\]from A.
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