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

1 k g - m^{2}

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3.2 k g - m^{2}

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2.5 k g - m^{2}

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1.5 k g - m^{2}

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Solution

The correct option is D $1.5 k g - m^{2}$
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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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

The correct option is D 1.5 kg−m2On releasing from rest the block falls through 5 m height in 2 sec. 5=0+12a(2)2[As S=ut+12at2]∴a=2.5m/s2 Substituting the value of a in the formula a=g1+ImR2 and by solving we get ⇒2.5=101+I2×(0.5)2⇒I=1.5kg−m2