A) 1 : 14
B) 1 : 6
C) 1 : 9
D) 1 : 11
Correct Answer: D
Solution :
Spring constant\[\propto \frac{\text{1}}{\text{length}}\] \[k\propto \frac{1}{l}\] i.e, \[{{k}_{1}}=6k\] \[{{k}_{2}}=3k\] \[{{k}_{3}}=2k\] In series \[\frac{1}{k'}=\frac{1}{6k}+\frac{1}{3k}+\frac{1}{2k}\] \[\frac{1}{k'}=\frac{6}{6k}\] \[k'=k\] \[k''=6k+3k+2k\] \[k''=11k\] \[\frac{k'}{k''}=\frac{1}{11}\]i. e \[k':k''=1:11\]
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