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JEE Main 2019 (12 Apr, Shift 2) · Inertia, First & Second Laws

Inertia, First & Second Laws question from JEE Main 2019 (12 Apr, Shift 2)

Q31JEE Main 2019 · 12 APR · SHIFT 2Medium

Inertia, First & Second Laws

A spring whose unstretched length is l has a force constant k. The spring is cut into two pieces of unstretched lengths l1 and l2 where, l1 = nl2 and n is an integer. The ratio k1/k2 of the corresponding force constant, k1 and k2 will be:

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Answer: (b)

For pieces of the same spring, $kl = \text{constant}$ , so $k \propto \frac{1}{l}$ .
$\frac{k_1}{k_2} = \frac{l_2}{l_1} = \frac{l_2}{nl_2} = \frac{1}{n}$
So the answer is (b) $\frac{1}{n}$ .

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