JEE Main 2025 Apr 2 Shift 1, Physics Q3: Work Energy Theorem And Its Applications

The cord is pulled with constant force, so the work done by the force as $1\text{m}$ of cord unwinds equals the kinetic energy gained by the wheel (work-energy theorem; the wheel starts from rest).

$$W=F\cdot d=20\times 1=20\text{J}$$

For a rim-loaded wheel (mass concentrated at the rim, spokes massless):

$$I=M{R}^2=10\times (0.1{)}^2=0.1{\text{kg m}}^2$$

Equating $W$ to rotational kinetic energy:

$$\frac{1}{2}I{\omega }^2=20\Rightarrow {\omega }^2=\frac{2\times 20}{0.1}=400$$

$$\omega =20\text{rad/s}$$

$\therefore$ Option (1) is correct.