A block of mass 2 kg hangs from the rim of a wheel of radius 0.5 m. On releasing

Question

A block of mass 2 kg hangs from the rim of a wheel of radius 0.5 m. On releasing 2 kg from rest the block falls through 5 m height in 2 s. The moment of inertia of the wheel will be

Options:

A . 1 kg−m2

B . 3.2 kg−m2

C . 2.5 kg−m2

D . 1.5 kg−m2

Answer: Option D
:
D
On releasing from rest the block falls through 5 m height in 2 sec.

5 = 0 + \frac{1}{2} a (2)^{2} [A s S = u t + \frac{1}{2} a t^{2}] ∴ a = 2.5 m / s^{2}

Substituting the value of a in the formula

a = \frac{g}{1 + \frac{I}{m R^{2}}}

and by solving we get

\Rightarrow 2.5 = \frac{10}{1 + \frac{I}{2 \times (0.5)^{2}}} \Rightarrow I = 1.5 k g - m^{2}

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