MHT CET · Physics · Laws of Motion
A simple spring has length ' \(l\) ' and force constant ' \(\mathrm{K}\) ', It is cut in to two springs of length ' \(l_1\) ' and ' \(l_2\) ' such that \(l_1=\mathrm{n} l_2\) ( \(\mathrm{n}\) is an integer). The force constant of spring of length ' \(l_1\) ' is
- A \(\mathrm{K}(1+\mathrm{n})\)
- B \(\frac{\mathrm{K}(\mathrm{n}+1)}{\mathrm{n}}\)
- C \(\mathrm{K}\)
- D \(\frac{\mathrm{K}}{(\mathrm{n}+1)}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{K}(\mathrm{n}+1)}{\mathrm{n}}\)
Step-by-step Solution
Detailed explanation
Given
\(l_1=\mathrm{n} l_2\)
\(\therefore \quad \mathrm{k}_2=\mathrm{nk}_1\)
Before cutting, the springs are connected in series, so
\(\therefore \quad \frac{1}{\mathrm{k}}=\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2}\)
\(\frac{1}{\mathrm{~K}}=\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{nk}_1}\)
\(\frac{1}{\mathrm{~K}}=\frac{\mathrm{n}+1}{\mathrm{nk}_1}\)
\(\therefore \quad k_1-\frac{K(n+1)}{n}\)
\(l_1=\mathrm{n} l_2\)
\(\therefore \quad \mathrm{k}_2=\mathrm{nk}_1\)
Before cutting, the springs are connected in series, so
\(\therefore \quad \frac{1}{\mathrm{k}}=\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2}\)
\(\frac{1}{\mathrm{~K}}=\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{nk}_1}\)
\(\frac{1}{\mathrm{~K}}=\frac{\mathrm{n}+1}{\mathrm{nk}_1}\)
\(\therefore \quad k_1-\frac{K(n+1)}{n}\)
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