In the given figure a ball of mass m is connected to an elastic spring of force constant K = mg/L through an inextensible string as shown. Now the ball is released from rest.

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

The maximum velocity of mass m during fall is $\sqrt{3 gL}$

The maximum velocity of mass m during fall is $\sqrt{2 gL}$

The maximum potential energy stored in the spring during the fall is $mgL [2 + \sqrt{3}]$

The maximum potential energy stored in the spring during the fall is $mgL [2 - \sqrt{3}]$

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Using energy conservation between initial moment and equilibrium moment when

$\begin{array}{l} T = mg \\ \frac{1}{2} \frac{mg}{L} \times L^{2} + \frac{1}{2} {mv}^{2} = mg \times 2 L \\ v = \sqrt{3 gL} \end{array}$

If ball falls a maximum distance of L + x.

$\begin{array}{l} mg (L + x) = \frac{1}{2} \frac{mg}{L} x^{2} x = L (1 + \sqrt{3}) ∴ {PE}_{max} = mgL (2 + \sqrt{3}) \end{array}$

Watch 3-min video & get full concept clarity