WBJEE · Physics · Laws of Motion
Two massless springs of force constants \(\mathrm{K}_1\) and \(\mathrm{K}_2\) are joined end to end. The resultant force constant \(\mathrm{K}\) of the system is
- A \(\mathrm{K}=\frac{\mathrm{K}_1+\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2}\)
- B \(\mathrm{K}=\frac{\mathrm{K}_1-\mathrm{K}_2}{\mathrm{~K}_1 \mathrm{~K}_2}\)
- C \(\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}\)
- D \(\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1-\mathrm{K}_2}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}\)
Step-by-step Solution
Detailed explanation
In series \(\mathrm{K}_{\text {eff }}=\frac{K_1 K_2}{K_1+K_2}\)
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