Q 11-13-144JEE MainJEE Main 2019 (12 Apr, Shift 2)Easy
A spring whose unstretched length is $l$ has a force constant $k$. The spring is cut into two pieces of unstretched lengths $l_1$ and $l_2$ where, $l_1 = nl_2$ and $n$ is an integer. The ratio $k_1/k_2$ of the corresponding force constants, $k_1$ and $k_2$ will be:
Answer: (D) $\dfrac1n$
For pieces of the same spring, $kl$ is constant, so $k \propto \dfrac1l$:
$$\frac{k_1}{k_2} = \frac{l_2}{l_1} = \frac1n$$
Solution by Sreeraj P, M.Sc Physics