MHT CET · Physics · Laws of Motion
A spring has length \(L\) and force constant K . It is cut into two springs of length \(L_1\) and \(L_2\) such that \(\mathrm{L}_1=\mathrm{NL}_2\) ( N is an integer). The force constant of spring of length \(L_1\) is
- A \((\mathrm{N}+1) \mathrm{K}\)
- B \(\frac{\mathrm{K}}{\mathrm{N}}(1+\mathrm{N})\)
- C 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,
\(\mathrm{L}_1=\mathrm{NL}_2...(i)\)
Spring constant, \(\mathrm{k} \propto \frac{1}{\mathrm{~L}}\)
\(\therefore \quad \mathrm{k}_2=\mathrm{Nk}_1\)
Before cutting, the springs are connected in series, so
\(\begin{aligned}
\therefore \quad \frac{1}{\mathrm{~K}} & =\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2} \\
\frac{1}{\mathrm{~K}} & =\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{Nk}_1} \\
\frac{1}{\mathrm{~K}} & =\frac{\mathrm{N}+1}{\mathrm{Nk}_1} \\
\therefore \quad \mathrm{k}_1 & =\frac{\mathrm{K}}{\mathrm{~N}}(1+\mathrm{N})
\end{aligned}\)
\(\mathrm{L}_1=\mathrm{NL}_2...(i)\)
Spring constant, \(\mathrm{k} \propto \frac{1}{\mathrm{~L}}\)
\(\therefore \quad \mathrm{k}_2=\mathrm{Nk}_1\)
Before cutting, the springs are connected in series, so
\(\begin{aligned}
\therefore \quad \frac{1}{\mathrm{~K}} & =\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2} \\
\frac{1}{\mathrm{~K}} & =\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{Nk}_1} \\
\frac{1}{\mathrm{~K}} & =\frac{\mathrm{N}+1}{\mathrm{Nk}_1} \\
\therefore \quad \mathrm{k}_1 & =\frac{\mathrm{K}}{\mathrm{~N}}(1+\mathrm{N})
\end{aligned}\)
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