Suppose mass and radius of each disc are m and R respectively. Also potential energy at points B and D is zero i.e., they are on reference line.

Given final kinetic energy for each disc is same, say it is K.

Applying energy conservation principle,

For surface AB,

$${1 \over 2}{I_2}\omega _1^2 + mg \times 30 = K$$ ..... (i)

For surface CD,

$${1 \over 2}{I_2}\omega _2^2 + mg \times 27 = K$$ ..... (ii)

From eqns. (i) and (ii), we get

$${1 \over 2}{I_2}\omega _1^2 + mg \times 30 = {1 \over 2}{I_2}\omega _2^2 + mg \times 27$$ ...... (iii)

Here, $${\omega _1} = {{{v_1}} \over R}$$, $${\omega _2} = {{{v_2}} \over R}$$, v₁ = 3 m s^$$-$$1, v₂ = ?

I₁ = I₂ = Moment of inertia of disc about the point of contact $$ = {1 \over 2}m{R^2} + m{R^2} = {3 \over 2}m{R^2}$$

From eqn. (iii), $${1 \over 2}\left( {{3 \over 2}m{R^2}} \right) \times {\left( {{3 \over R}} \right)^2} + m \times 10 \times 30$$

$$ = {1 \over 2}\left( {{3 \over 2}m{R^2}} \right) \times {\left( {{{{v_2}} \over R}} \right)^2} + m \times 10 \times 27$$

$${{27} \over 4} + 300 = {3 \over 4}v_2^2 + 270$$

$${3 \over 4}v_2^2 = {{27} \over 4} + 30 \Rightarrow 3v_2^2 = 147$$

$$v_2^2 = 49$$ $$\therefore$$ v₂ = 7 m s^$$-$$1