Question

NEET 2017 Q:
Class 11th !!

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 : -​

Answer 1

EXPLANATION.

A spring of force constant k is cut into lengths of ratio = 1 : 2 : 3.

They are connected in series and the new force constants = k'.

They are connected in parallel and force constant = k''.

As we know that,

Spring constant ∝ 1/(length).

⇒ k ∝ 1/L.

Let, we consider the length of spring be = l.

A spring of force constant k is cut into lengths of ratio = 1 : 2 : 3.

⇒ l/6, l/3, 1/2

We can write spring constants as,

⇒ k₁ = 6k, k₂ = 3k, k₃ = 2k.

For series combination.

⇒ 1/k' = 1/k₁ + 1/k₂ + 1/k₃.

⇒ 1/k' = 1/6k + 1/3k + 1/2k.

⇒ 1/k' = (2 + 4 + 6)/12k.

⇒ 1/k' = 12/12k.

⇒ 1/k' = 1/k.

⇒ k' = k.

For parallel combination.

⇒ k'' = k₁ + k₂ + k₃.

⇒ k'' = 6k + 3k + 2k.

⇒ k'' = 11k.

⇒ To find : k' : k''.

⇒ k' : k'' = k : 11k.

⇒ k' : k'' = 1 : 11.