NEET Solved Paper 2014 Question 20
Question: If $ n _1,n _2 $ and $ n _3 $ are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by [AIPMT 2014]
Options:
A) $ \frac{1}{n}=\frac{1}{n _1}+\frac{1}{n _2}+\frac{1}{n _3} $
B) $ \frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n _1}}+\frac{1}{\sqrt{n _2}}+\frac{1}{\sqrt{n _3}} $
C) $ \sqrt{n}=\sqrt{n _1}+\sqrt{n _2}+\sqrt{n _3} $
D) $ n=n _1+n _2+n _3 $
Show Answer
Answer:
Correct Answer: A
Solution:
- Problem solving strategy in this problem, the fundamental frequencies of each part could be find. The fundamental frequency of the complete wire could be find. The one should check the each option for the given values. For Ist part, $ n _1=\frac{v}{2l _1}\Rightarrow l _1=\frac{v}{2n _1} $ For IInd part, $ n _2=\frac{v}{2l _2}\Rightarrow l _2=\frac{v}{2n _2} $ For IIIrd part, $ n _3=\frac{v}{2l _3}\Rightarrow l _3=\frac{v}{2n _3} $ For the complete wire $ n=\frac{v}{2l}\Rightarrow l=\frac{v}{2n} $ We have $ l=l _1+l _2+l _3 $ $ \frac{v}{2n}={{\frac{v}{2n}}_1}+\frac{v}{2n _2}+\frac{v}{2n _3} $ $ \frac{1}{n}=\frac{1}{n _1}+\frac{1}{n _2}+\frac{1}{n _3} $
