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 :

Laws of Motion PYQ — JEE Main 2019 · 12 Apr · Shift 2 · Q65 with Solution

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Correct answer: B

For a uniform spring, the spring constant is inversely proportional to its length:

$k \propto \frac{1}{L}$

So if a spring of length $l$ has spring constant $k$ , then for a piece of length $L$ ,

$k' = k\frac{l}{L}$

The spring is cut into two parts of lengths $l_1$ and $l_2$ such that

$l_1 = n l_2$

For the piece of length $l_1$ ,

$k_1 = k\frac{l}{l_1}$

For the piece of length $l_2$ ,

$k_2 = k\frac{l}{l_2}$

$\frac{k_1}{k_2} = \frac{k\frac{l}{l_1}}{k\frac{l}{l_2}} = \frac{l_2}{l_1}$

Using $l_1 = n l_2$ ,

$\frac{k_1}{k_2} = \frac{l_2}{n l_2} = \frac{1}{n}$

$\boxed{\frac{k_1}{k_2} = \frac{1}{n}}$

So the correct answer is Option B.