A spring has length l and force constant k. It is cut into 2 springs of length l₁ and l₂ such that l₁ = nl₂ ( n =an integer). The force constant of the spring of length l₂ is (k ∝ 1/l) (a) k(n+1) (b) (n+1)k/n (c) k (d) (k)/((n+1))

Answer: (a). k ∝ 1/l, so k l is constant: k₂ l₂ = k l. l = l₁ + l₂ = n l₂ + l₂ = (n+1)l₂ k₂ = (k l)/(l₂) = (n+1)k So the answer is (a) k(n+1).

A spring has length l and force constant k. It is cut into 2 springs… | MHT CET 2026 (13 May, Shift 2)

Q8MHT CET 2026 · 13 MAY · SHIFT 2Medium

Spring Systems

A spring has length $l$ and force constant $k$ . It is cut into 2 springs of length $l_1$ and $l_2$ such that $l_1 = nl_2$ ( $n$ =an integer). The force constant of the spring of length $l_2$ is $\left(k \propto \frac{1}{l}\right)$

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Answer: (a)

$k \propto \frac{1}{l}$ , so $k l$ is constant: $k_2 l_2 = k l$ .
$l = l_1 + l_2 = n l_2 + l_2 = (n+1)l_2$
$k_2 = \frac{k l}{l_2} = (n+1)k$
So the answer is (a) $k(n+1)$ .

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Spring Systems question from MHT CET 2026 (13 May, Shift 2)

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