NEET Solved Paper 2017 Question 1
Question: A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k?. Then they are connected in parallel and force constant is k?. Then k? : k?? is
Options:
A) 1 : 14
B) 1 : 6
C) 1 : 9
D) 1 : 11
Show Answer
Answer:
Correct Answer: D
Solution:
Spring constant $ \propto \frac{1}{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 $
BETA
