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Rotational Motion question

2025 · 2 Apr · Shift 2 · Q73
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Rotational Motion question

2025 · 2 Apr · Shift 2 · Q73

JEE MainPhysicsRotational MotionNumerical+4 / −1
JEE Main 2025 (Online) 2nd April Evening Shift Physics - Rotational Motion Question 8 EnglishA wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 N as shown in figure. The established torque produces an angular acceleration of 2rad/s22 \mathrm{rad} / \mathrm{s}^22rad/s2. Moment of intertia of the wheel is ‾kg  m2\underline{\hspace{2cm}}\mathrm{kg} \mathrm{}\,\, \mathrm{m}^2​kgm2. (Acceleration due to gravity =10 m/s2=10 \mathrm{~m} / \mathrm{s}^2=10 m/s2 )
Numerical answer
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Correct answer: 1

  1. Given data
  • Radius of wheel: r=0.2 mr = 0.2\,\text{m}r=0.2m
  • Pulling force: F=10 NF = 10\,\text{N}F=10N
  • Angular acceleration produced: α=2 rad s−2\alpha = 2\,\text{rad s}^{-2}α=2rad s−2

We need to find the moment of inertia III.

  1. Torque due to the pulling force

Since the string is wrapped around the rim, the force acts tangentially at radius rrr.

So the torque is

τ=Fr\tau = F rτ=Fr

Substituting values:

τ=10×0.2=2 N m\tau = 10 \times 0.2 = 2\,\text{N m}τ=10×0.2=2N m

  1. Use rotational equation of motion

For rotational motion,

τ=Iα\tau = I\alphaτ=Iα

Hence,

I=ταI = \frac{\tau}{\alpha}I=ατ​

Substitute:

I=22=1 kg m2I = \frac{2}{2} = 1\,\text{kg m}^2I=22​=1kg m2

  1. Final answer

1\boxed{1}1​

So, the moment of inertia of the wheel is 1 kg m21\,\text{kg m}^21kg m2.

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