A spring of force constant k is cut into lengths of ratio $$1$$: $$2$$: $$3$$ . They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k". Then k' : k" is

Question

Infinity Learn · Accepted Answer

Let us assume, the length of spring be l. When we cut the spring into ratio of length $$1$$: $$2$$: $$3$$, we get three springs of lengths $$l6$$,$$2l6$$ and $$3l6$$ with force constant,∴  $$k1=kll1=kll/6=6kk2=kll2=kl2l/6=3kk3=kll3=kl3l/6=2kWhen$$ connected in series, $$1k$$'$$=16k+13k+12k=1+2+36k=1k$$∴k'$$=k$$ When connected in parallel, k''$$=6k+3k+2k=11k$$ k'k''$$=k11k=111$$

A spring of force constant k is cut into lengths of ratio 1: 2: 3 . They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k". Then k' : k" is

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