JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

A spring whose unstretched length is \[l\] has a force constant k. The spring is cut into two pieces of un stretched lengths \[{{l}_{1}}\]and \[{{l}_{2}}\] where, \[{{l}_{1}}=n{{l}_{2}}\]and n is an integer. The ratio \[{{k}_{1}}/{{k}_{2}}\]of the corresponding force constants, \[{{k}_{1}}\]and \[{{k}_{2}}\]will be : [JEE Main 12-4-2019 Afternoon]

A) \[\frac{1}{{{n}^{2}}}\]

B) \[{{n}^{2}}\]

C) \[\frac{1}{n}\]

D) \[n\]

Correct Answer: C

Solution :

[c]\[{{k}_{1}}=\frac{C}{{{\ell }_{1}}}\]

\[{{k}_{2}}=\frac{C}{{{\ell }_{2}}}\]

\[\frac{{{k}_{1}}}{{{k}_{2}}}=\frac{C{{\ell }_{2}}}{{{\ell }_{1}}C}{{\ell }_{2}}=\frac{{{\ell }_{2}}}{n{{\ell }_{2}}}=\frac{1}{n}\]