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 the force constant is K′′. Then, K′K′′ is
A
1:9
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B
1:11
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C
1:14
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D
1:6
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Solution
The correct option is B1:11 Let us assume, the length of the spring is l. When we cut the spring in the ratio of lengths 1 : 2 : 3, we get three springs of lengths l6,2l6 and 3l6 with force constants, ∴K1=Kll1=Kll6=6K K2=Kll2=Kl2l6=3K K3=Kll3=Kl3l6=2K
When 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