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A spring of force constant k is cut into...

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 :

A

`1:9`

B

`1:11`

C

`1:14`

D

`1:6`

Text Solution

Verified by Experts

The correct Answer is:
B

Spring constant `K prop 1/l`
Where l = natural length of spring
`K=c/l` c=constant
It is cut into lengths of ratio 1:2:3
then ratio of spring constant,
`c/1:c/2:c/3implies 1/1:1/2:1/3`
`K_1:K_2:K_3implies6:3:2`
Now,
parallel combination
`K..=6K+3K+2KimpliesK..=11K`
series combination
`(1)/(K.)=(1)/(6K)+(1)/(2K)+(1)/(3K)`
`(1)/(K.)=(1)/(6K)+((3+2))/(6K)implies(1)/(K.)=(1)/(K)`
K.=K
`(K.)/(K..)=(K)/(11K)implies (K.)/(K..)=(1)/(11)`
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