A rolling wheel of $$12$$ kg is on an inclined plane at position $$P$$ and connected to a mass of $$3$$ kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom $$Q$$ of the inclined plane $$PQ$$ will be $$\frac{1}{2}\sqrt{xgh}$$ m s$$^{-1}$$. The value of $$x$$ (rounded off to the nearest integer) is ______.

Concept:
Loss in gravitational PE = Gain in total KE (translation + rotation)

Step 1: Change in Potential Energy

• Wheel (12 kg) moves down by h: 12gh
• Block (3 kg) moves up by h: −3gh

Net loss in PE=12gh − 3gh = 9gh

Step 2: Total Kinetic Energy at bottom

• Block:

$$\frac{1}{2}(3)v^2$$

• Wheel (translation):

$$\frac{1}{2}(12)v^2$$

• Wheel (rotation):

$$\frac{1}{2}I\omega^2 = \frac{1}{2}(12r^2)\left(\frac{v}{r}\right)^2 = \frac{1}{2}(12)v^2$$

Step 3: Apply energy conservation

9gh = $$\frac{1}{2}(3 + 12 + 12)v^2$$

9gh=$$\ \frac{\ 1}{2}(27)v^2$$

$$\Rightarrow v^2 = \frac{2gh}{3}$$ 

v = $$\sqrt{\frac{2gh}{3}}$$ = $$\frac{1}{2}\sqrt{\frac{8gh}{3}}$$

Step 4: Compare with given form

v = $$\frac{1}{2}\sqrt{xgh} \Rightarrow x = \frac{8}{3} \approx 3$$

Final Answer:

3​