MHT CET · Physics · Laws of Motion
A spring has length \(l\) and force constant \(K\). If is cut into two springs of length \(l_1\) and \(l_2\) such that, \(l_1=n l_2(n\) is integer). The force constant of the spring of length \(l_2\) is
- A \(\frac{(n+1) K}{n}\)
- B \(K\)
- C \(\frac{K}{(n+1)}\)
- D \(K(1+n)\)
Answer & Solution
Correct Answer
(D) \(K(1+n)\)
Step-by-step Solution
Detailed explanation
If \(l\) is cut into \(l_1\) and \(l_2\), such that
\(l_1+l_2=l\), where, \(n l_2=l_1\)
\(\therefore l_2=\frac{l}{n+1}\)
As we know from the elasticity relation \(K=\frac{E A}{l}\), where \(E\) is Young's modulus, \(A\) area of cross-section and \(l\) is the length of the spring.
\(\therefore K_2=\frac{E A}{l_2}=\frac{(n+1) E A}{l}=(n+1) K\)
\(l_1+l_2=l\), where, \(n l_2=l_1\)
\(\therefore l_2=\frac{l}{n+1}\)
As we know from the elasticity relation \(K=\frac{E A}{l}\), where \(E\) is Young's modulus, \(A\) area of cross-section and \(l\) is the length of the spring.
\(\therefore K_2=\frac{E A}{l_2}=\frac{(n+1) E A}{l}=(n+1) K\)
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