QJEEM25Ph690f: JEE Main 2025 Physics - Class 11 JEEM25Ph690f Solution: A bead of mass ' ' slides without friction on the wall of a…

When mass is at position A,

Elongation of spring, ${x_1} = 2R - R = R$

(As equilibrium length of spring is R)

${h_1} = 2R$

${v_1} = 0$ (As the bead is released from zero initial speed)

When mass is at position B,

${x_2} = R - R = 0$

${h_2} = R/2,{v_2} = v$

So using energy conservation,

${E_1}^\circ = {E_f}$

$\Rightarrow {1 \over 2}kx_1^2 + {1 \over 2}mv_1^2 + mg{h_1} = {1 \over 2}kx_2^2 + {1 \over 2}mv_2^2 + mg{h_2}$

$\Rightarrow {1 \over 2}k{R^2} + mg(2R) = {1 \over 2}k(o) + {1 \over 2}m{v^2} + mg{R \over 2}$

$\Rightarrow {1 \over 2}k{R^2} + 2mgR - {{mgR} \over 2} = {1 \over 2}m{v^2}$

$\Rightarrow {1 \over 2}m{v^2} = {1 \over 2}k{R^2} + {{3mgR} \over 2}$

$\Rightarrow {v^2} = {{k{R^2}} \over m} + 3gR$

$\Rightarrow v = \sqrt {{{k{R^2}} \over m} + 3gR}$