A spring has length L and force constant K. It is cut into two springs of length L₁ and L₂ such that L₁ = NL₂ ( N is an integer). The force constant of spring of length L₁ is (a) (N + 1)K (b) K/N(1 + N) (c) K (d) K/N+1

Answer: (b). Force constant ∝ (1)/(length): kL = constant. L₁ + L₂ = L with L₁ = NL₂ ⇒ L₁ = N/N+1L k₁ = KL/L₁ = K/N(1 + N), so the answer is (b).

A spring has length L and force constant K. It is cut into two… | MHT CET 2024 (15 May, Shift 2)

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Wire Combinations & Special Cases

A spring has length L and force constant K. It is cut into two springs of length $L_1$ and $L_2$ such that $L_1 = NL_2$ ( N is an integer). The force constant of spring of length $L_1$ is

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Answer: (b)

Force constant $\propto \frac{1}{\text{length}}$ : $kL =$ constant.
$L_1 + L_2 = L$ with $L_1 = NL_2 \Rightarrow L_1 = \frac{N}{N+1}L$
$k_1 = K\frac{L}{L_1} = \frac{K}{N}(1 + N)$ , so the answer is (b).

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