First slide

Work done by Diff type of forces

Question

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 $k'$ . If they are connected in parallel and force constant is $k''$ , then $k'$ : $k''$ is [NEET 2017]

Moderate

A

1 : 6
B

1 : 9
C

1 : 11
D

1 : 14

Solution

When the spring is cut into pieces, they will have the new force constant. The spring is divided into 1 : 2 : :3 ratio. When the pieces are connected in series, the resultant force constant will be given by
$\frac{1}{k^{'}} = \frac{1}{k_{1}} + \frac{1}{k_{2}} + \frac{1}{k_{3}}$
$\frac{1}{k^{'}} = \frac{1}{x} + \frac{1}{2 x} + \frac{1}{3 x} \Rightarrow k^{'} = \frac{6 x}{11}$
In parallel, the net force constant,
$k^{''} = x + 2 x + 3 x = 6 x$
The required ratio $\frac{k^{'}}{k^{''}} = \frac{6 x / 11}{6 x} = 1 : 11$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App