Question

There is smooth ring of radius R in vertical plane. A spring of natural length R & elastic constant K is vertical along a diameter. The free end is connected to bead of mass m & when slightly disturbed it reaches point C with speed v where v is

(1) \(\sqrt{\frac{KR^2 (\sqrt{2} - 1) + 2mgR}{m}}\) 

(2) \(\sqrt{\frac{2KR^2 (\sqrt{2} - 1) + 2mgR}{m}}\)

(3) \(\sqrt{\frac{2KR^2 (\sqrt{2} - 1)+ mgR}{m}}\)

(4) \(\sqrt{\frac{KR^2 (\sqrt{2} - 1) + mgR}{m}}\)

Answer 1

Correct option is : (2) \(\sqrt{\frac{2KR^2 (\sqrt{2} - 1) + 2mgR}{m}}\) 

Loss in PE = gain in KE

\(\frac{1}{2}K(R^2) - \frac{1}{2}K(\sqrt{2} -1) ^2\ R^2 + mgR = \frac{1}{2}mv^2\)