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Length of a spring of force constant k in its unstretched condition is l. The spring is cut into two parts which have their unstretched lengths in the ratio $l_{1} : l_{2} = q : 1$ . Force constants of the two parts k₁ and k₂ are then:

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$k \frac{(q - 1)}{q}, k \frac{(q + 1)}{q}$

$k \frac{(q + 1)}{q}, kq$

$k (q + 1), kq$

$k \frac{(q + 1)}{q}, k (q + 1)$

answer is C.

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Detailed Solution

$l_{1} + l_{2} = l As l_{1} = {ql}_{2}$
$∴ l_{2} = \frac{l}{q + 1} and l_{1} = \frac{ql}{q + 1} k_{1} = \frac{mg}{l_{1}} = \frac{mg (q + 1)}{ql} = \frac{k (q + 1)}{q} k_{2} = \frac{mg}{l_{2}} = \frac{mg}{l_{2}} (q + 1) = k (q + 1)$

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