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A stretched wire of lenth 110 cm is divided into three segments whose frequencies are in ratio `1 : 2 : 3.` Their length must beA. 20 cm, 30 cm, 60 cm,B. 60 cm, 30 cm, 20 cmC. 60 cm, 20 cm, 30 cm,D. 30 cm, 60 cm, 20 cm |
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Answer» Correct Answer - b Given `l_(1)+l_(2)+l_(3)=110` cm and `n_(1)l_(1)=n_(2)l_(2)=n_(3)l_(3)` (`because` T and d are same) `n_(1):n_(2):n_(3)::1 : 2: 3` `because n_(1)/n_(2)=1/2=l_(2)/l_(1)rArr l_(2)=l_(1)/2` and `because n_(1)/n_(3)=1/3=l_(3)/l_(1)rArr l_(3)=l_(1)/3` `therefore l_(1) +l_(1)/2+l_(1)/3=110` `rArr l_(1) = 60 ` cm So, ` l_(1) = 60 ` cm, `l_(2) = 30` cm, ` l_(3) = 20` cm |
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